Differential equation: Difference between revisions

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AI-assisted (RonzzWikiCowriter): make the homogeneous/particular structure explicit — state the y = y_p + y_h result in the classification section and label homogeneous and non-homogeneous cases inside Methods 2 and 3 (first order) and Method 2 (second order), adding a forced-equation demonstration on the same operator; add PDE Method 2, travelling waves via the method of characteristics (transport equation), with a river-pollutant numeric example and the wave-equation remark. (via update-pag...
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'''A differential equation''' is an equation whose unknown is a function and which also involves that function's derivatives (rates of change). Where an ordinary equation such as $x^2=9$ is solved by numbers, a differential equation such as $y'+2y=0$ is solved by functions $y(x)$. Laws of nature state how quantities change, so differential equations describe pendulums, cooling drinks, growing populations, and discharging capacitors. This article covers the standard analytical solution methods, by class of equation, each stated in general and then demonstrated on a concrete numerical example, and the numerical, series, and qualitative routes used when no exact formula exists. It treats '''ordinary differential equations''' (one independent variable) and, briefly, '''partial differential equations''' (several).
'''A differential equation''' is an equation whose unknown is a function and which also involves that function's derivatives (rates of change). Where an ordinary equation such as $x^2=9$ is solved by numbers, a differential equation such as $y'+2y=0$ is solved by functions $y(x)$.


== A first example: slopes and a family of solutions ==
A surprising number of laws of nature can be described by differential equations: pendulums, cooling drinks, growing populations, and discharging capacitors.
 
== General and specific solution ==


The simplest differential equation prescribes the slope of a function $y(x)$:
The simplest differential equation prescribes the slope of a function $y(x)$:
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$$\frac{dy}{dx}=2x$$
$$\frac{dy}{dx}=2x$$


Integration inverts differentiation, so integrating both sides gives
Integrating both sides gives


$$\int\frac{dy}{dx}\,dx=\int 2x\,dx\qquad\Longrightarrow\qquad y(x)=x^{2}+C$$
$$\int\frac{dy}{dx}\,dx=\int 2x\,dx\qquad\Longrightarrow\qquad y(x)=x^{2}+C$$


Every $C$ works, since $\frac{d}{dx}\left(x^2+C\right)=2x$; the solutions form the parabola family $y=x^2+C$, the '''general solution'''.
Where $C$ is an unknown constant. Therefore, the solution is not one single function, but a family of functions, known as the '''general solution'''.


If $y(0)=3$, then
If we know the value of $y$ at a particular $x$, for instance, $y(0)=3$, then a '''particular solution''' function can be identified


$$3=0^2+C\qquad\Longrightarrow\qquad C=3\qquad\Longrightarrow\qquad y=x^2+3$$
$$3=0^2+C\qquad\Longrightarrow\qquad C=3\qquad\Longrightarrow\qquad y=x^2+3$$


A prescribed value such as this is an '''initial condition'''.
A prescribed value such as $y(a)=b$ allowing us to pin down a particular solution is called an '''initial condition'''.


== Classifying differential equations ==
== Classifying differential equations ==


Three features decide how to solve an equation: its '''order''', its '''linearity''', and how many independent variables it involves.
A differential equation can be classified by three criteria: its '''order''', its '''linearity''', and how many independent variables it involves.


=== Order ===
=== Order ===


The order is the order of the highest derivative present. $dy/dx=2x$ is first order; Newton's second law,
The order is the order of the highest derivative present.  
 
$\frac{dy}{dx}=2x$ is first order, whereas Newton's second law,


{{#content:Q1583}}
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is second order ($x(t)$ position of mass $m$, $F$ net force). Integration introduces one arbitrary constant per integration, so the general solution of an nth-order equation carries $n$ constants, fixed by $n$ initial conditions. For equations of the special form $y^{(n)}=f(x)$ the constants appear exactly as the integration constants of $n$ successive integrations; the free-fall example in the second-order section below works this out for $n=2$.
is second order ($x(t)$ position of mass $m$, $F$ net force).


=== Linearity and homogeneity ===
=== Linearity and homogeneity ===
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$$\frac{dy}{dx}+p(x)\,y=q(x)$$
$$\frac{dy}{dx}+p(x)\,y=q(x)$$


and is '''homogeneous''' when $q(x)=0$. The equations $dy/dx=y^2$ and $d^2\theta/dt^2+\sin\theta=0$ are nonlinear (square of $y$; sine of $\theta$).
It is said to be '''homogeneous''' when $q(x)=0$.  
 
<blockquote>
The equations $dy/dx=y^2$ and $d^2\theta/dt^2+\sin\theta=0$ are nonlinear (square of $y$; sine of $\theta$).
</blockquote>
 
An important property of linear homogeneous equation is the '''superposition principle'''. If $y_1,y_2$ solve a homogeneous linear equation, so does $c_1y_1+c_2y_2$.


If $y_1,y_2$ solve a homogeneous linear equation, so does $c_1y_1+c_2y_2$ (the '''superposition principle'''): substituting the combination adds the two expressions that already vanish. For a nonlinear equation the combination does not generally solve it: if $y_1'=y_1^2$ and $y_2'=y_2^2$, then
For a nonlinear equation, the same principle usually does not apply: if $y'=y^2$ has two solutions $y_1'=y_1^2$ and $y_2'=y_2^2$


$$(y_1+y_2)'=y_1^2+y_2^2\neq (y_1+y_2)^2$$
$$(y_1+y_2)'=y_1^2+y_2^2\neq (y_1+y_2)^2$$


so $y_1+y_2$ does not solve $y'=y^2$.
so $y_1+y_2$ does not solve $y'=y^2$.
Superposition also joins the homogeneous and non-homogeneous problems of one linear equation. Write the left-hand side as $L(y)$, so the equation reads $L(y)=q(x)$, with $L(y)=0$ its homogeneous form. If $y_p$ is any single solution of $L(y)=q$ (a '''particular solution''') and $y_h$ runs through all solutions of $L(y)=0$, then every solution of the original equation is
$$y=y_p+y_h$$
because $L(y_p+y_h)=L(y_p)+L(y_h)=q+0=q$, and conversely any two solutions of the non-homogeneous equation differ by a solution of the homogeneous one. The constants of integration therefore live entirely in $y_h$: the general solution of a linear equation is one particular solution plus the whole homogeneous family. This is why each linear method below is presented in two parts, the homogeneous case first.


=== Ordinary and partial ===
=== Ordinary and partial ===


An '''ordinary differential equation''' (ODE) has one independent variable. A '''partial differential equation''' (PDE) has several, with partial derivatives. For example, the temperature $u(x,t)$ of an insulated metal bar, which depends on position $x$ and time $t$, obeys the heat equation
An '''ordinary differential equation''' (ODE) has one independent variable. A '''partial differential equation''' (PDE) has several, with partial derivatives. All the examples we have seen above are ordinary differential equations. For an example of a partial differential equation, we can take the heat equation describing the temperature $u$ of an insulated metal bar, according to position $x$ on the metal bar and time $t$


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where $\alpha$ is the thermal diffusivity.
where $\alpha$ is the thermal diffusivity.


== Slope fields ==
== Direction field ==


A first-order equation can be written
A first-order equation can be written
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assigning to each point $(x,y)$ the slope $f(x,y)$ a solution must have there. Drawing short segments of that slope gives a '''direction field'''; solution curves run tangent to it.
assigning to each point $(x,y)$ the slope $f(x,y)$ a solution must have there. Drawing short segments of that slope gives a '''direction field'''. Solution curves must run tangent to the direction field for every point they pass through.


[[File:Slope field of exponential growth.png|thumb|Direction field of $dy/dx=y$. Credit: jjbeard (public domain).]]
[[File:Slope field of exponential growth.png|thumb|Direction field of $dy/dx=y$. Credit: jjbeard (public domain).]]
Numerical methods such as [[Euler's method]] follow the field: read the slope, step a short distance along it, repeat.<ref>{{#cite:Q1576}}</ref>


== Solving differential equations ==
== Solving differential equations ==


There is no formula that solves every differential equation. The practical route is to recognise the class of the equation and apply that class's method. For first-order ODEs there are two standard classes, separable and linear, with the constant-coefficient case of the latter solved fastest by an exponential trial (Method 3), the same idea that grows into the characteristic equation of order two; for second-order ODEs there are direct integration and constant-coefficient linear equations; for linear PDEs there are separation of variables and the method of characteristics (travelling waves). Each linear method below follows the structure derived above: solve the homogeneous equation (its solutions carry the constants), then add one particular solution of the non-homogeneous one. Each method is stated for its general case and then applied to a concrete numerical example. When an equation fits none of these classes, it is treated by the numerical, series, or qualitative methods summarised at the end of this section.
Just like there is no general formula solving all algebraic equations, there is no general method permitting the solution of all differential equations. However, some standard methods exist for solving a restricted class of simple differential equations.


=== First-order ODEs ===
=== First-order ODEs ===
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==== Method 1: separable equations ====
==== Method 1: separable equations ====


'''General case.''' A first-order equation is '''separable''' when the right-hand side factors into a function of $x$ alone times a function of $y$ alone:
'''General case.''' A first-order equation is separable when it can be rewritten as the equality of the derivative to the product of two functions, one containing only $x$, one containing only $y$.


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Divide both sides by $h(y)$ and integrate: all the $y$'s land on one side and all the $x$'s on the other, and if the two integrals can be evaluated the resulting relation between $y$ and $x$ is the general solution.
after which both integrals can be evaluated directly.


'''Example: exponential growth and decay.''' When a quantity changes at a rate proportional to its own size — a bank balance earning interest, a population with unlimited food, a radioactive sample — the equation is separable with $g(t)=k$, $h(y)=y$:
'''Example: exponential growth and decay.''' A quantity whose rate of change is proportional to its own size, such as an unchecked population or a radioactive sample, obeys


{{#content:Q1584}}
{{#content:Q1584}}


Divide by $y$ and integrate:
Separating variables and integrating,
 
$$\int\frac{dy}{y}=\int k\,dt\qquad\Longrightarrow\qquad \ln|y|=kt+C$$
 
Exponentiating, $|y|=e^C e^{kt}$. The sign of $y$ never changes, so absorbing it into the constant and writing $y(0)=y_0$,


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$$\int\frac{dy}{y}=\int k\,dt\;\Longrightarrow\;\ln|y|=kt+C_1\;\Longrightarrow\;y=Ce^{kt}$$


Numbers: €1000 at 5% interest compounded continuously ($k=0.05\ \text{yr}^{-1}$) gives $y(t)=1000\,e^{0.05t}$, and
The initial condition $y(0)=y_0$ fixes $C=y_0$, giving


$$y(10)=1000\,e^{0.5}\approx 1648.7$$
{{#content:Q1585}} (1)


Doubling time: $1000\,e^{0.05t}=2000\Rightarrow t=\ln 2/0.05\approx 13.9$ years. For $k<0$ the same solution describes decay; the '''half-life''' $y=y_0/2$ is $t_{1/2}=(\ln 2)/(-k)$.
Observing (1) we notice this is very well an exponential growth (k>0)/decay(k<0).


'''Example: Newton's law of cooling.''' A body hotter than its surroundings cools at a rate proportional to the temperature gap, which makes the equation separable:
'''Example: Newton's law of cooling.''' A hot object in a cooler room loses heat through its surface, and the larger the temperature gap, the faster it cools: The temperature of the object is modelled by


{{#content:Q1586}}
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Separate and integrate:
Separating variables and integrating,


$$\int\frac{dT}{T-T_a}=\int-k\,dt\qquad\Longrightarrow\qquad \ln|T-T_a|=-kt+C$$
$$\int\frac{dT}{T-T_a}=-k\int dt\;\Longrightarrow\;\ln|T-T_a|=-kt+C\;\Longrightarrow\;T-T_a=Ce^{-kt}$$


Exponentiating and folding the (constant-sign) factor $T-T_a$ into the constant, with $T(0)=T_0$:
so with $T(0)=T_0$, hence $C=T_0-T_a$,


$$T(t)=T_a+(T_0-T_a)e^{-kt}$$
$$T(t)=T_a+(T_0-T_a)e^{-kt}$$
The gap $T-T_a$ decays exponentially; $T$ itself does not. Example: a drink at $80\,^{\circ}\mathrm{C}$ in a $20\,^{\circ}\mathrm{C}$ room, $k=0.1\ \text{min}^{-1}$:
$$T(t)=20+60\,e^{-0.1t}$$
Reaches $40\,^{\circ}\mathrm{C}$ when $20+60e^{-0.1t}=40$, i.e. $t=10\ln 3\approx 11$ min.<ref>{{#cite:Q1576}}</ref>


==== Method 2: linear first-order equations (integrating factor) ====
==== Method 2: linear first-order equations (integrating factor) ====


'''General case.''' Many first-order equations are linear but not separable. A linear first-order equation has the form
'''General case.''' For the linear equation


$$\frac{dy}{dx}+p(x)\,y=q(x)$$
$$y'+p(x)\,y=q(x)$$


with $q(x)$ the forcing. The homogeneous case $q=0$ is the separable equation of Method 1 with $h(y)=y$, and its solutions are
introduce the integrating factor $\mu=e^{\int p\,dx}$, chosen so that $\mu'=p\mu$; multiplying by $\mu$ collapses the left-hand side into a single derivative:


$$y_h=C\,e^{-\int p(x)\,dx}$$
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For the non-homogeneous case multiply both sides by the '''integrating factor'''
Integrating both sides,


$$\mu(x)=e^{\int p(x)\,dx}$$
$$\mu y=\int\mu\,q\,dx+C\;\Longrightarrow\;y=\frac{1}{\mu}\int\mu\,q\,dx+\frac{C}{\mu}$$


so that the left-hand side collapses into a single derivative:
If you notice, the first term $y=\frac{1}{\mu}\int\mu\,q\,dx$$\frac{C}{\mu}$ provides one particular solution to the non-homogeneous equation, and the second term, $C/\mu=Ce^{-\int p\,dx}$, is a solution to the homogeneous counter par of the original equation: $y'+p(x)\,y=0$. It generalises the solution to the entire solution family.


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This is in fact a general principle: $y=y_p+y_h$, i.e., the solution family of a non-homogeneous equation is the sum of one particular solution plus the solution to its homogeneous counterpart.


Both sides then integrate directly, giving the general solution
To see the method in action, consider


$$y(x)=\frac{1}{\mu(x)}\left(\int \mu(x)\,q(x)\,dx + C\right)$$
$$ \frac{dy}{dx} + \frac{1}{x}\,y = x^2, \qquad x>0. $$


which is the homogeneous solution $y_h=C/\mu$ plus the particular solution $y_p=(1/\mu)\int\mu q\,dx$, the structure $y=y_p+y_h$ of the classification section.
Here $P(x)=1/x$, so


'''Worked demonstration:''' solve $y'+y=e^{-x}$. Here $p=1$, $\mu=e^x$, and
$$ \mu(x)=e^{\int \frac{1}{x}\,dx}=e^{\ln x}=x. $$


$$e^x y'+e^x y=1\qquad\Longrightarrow\qquad (e^x y)'=1\qquad\Longrightarrow\qquad e^x y=x+C\qquad\Longrightarrow\qquad y=(x+C)e^{-x}$$
Multiply the whole equation by $\mu(x)=x$:


{{#content:Q1607}}
$$ x\frac{dy}{dx} + y = x^3. $$


The condition $y(0)=2$ gives $C=2$.<ref>{{#cite:Q1576}}</ref><ref>{{#cite:Q1577}}</ref>
The left side is exactly the derivative of the product $x\,y$:


'''Example: falling with air resistance.''' A falling body of mass $m$ is pulled down by gravity $mg$ and slowed by air drag proportional to its speed, $-bv$. Newton's second law gives the linear first-order equation
$$ \frac{d}{dx}\bigl(xy\bigr) = x^3. $$


$$m\frac{dv}{dt}=mg-bv\qquad\Longrightarrow\qquad \frac{dv}{dt}+\frac{b}{m}\,v=g$$
Integrate with respect to $x$:


Here $p=b/m$ and $q=g$, both constant, so $\mu=e^{(b/m)t}$ and
$$ xy = \int x^3\,dx = \frac{x^4}{4} + C. $$


$$\frac{d}{dt}\left(e^{(b/m)t}v\right)=g\,e^{(b/m)t}\qquad\Longrightarrow\qquad e^{(b/m)t}v=\frac{mg}{b}e^{(b/m)t}+C$$
Finally, divide by $x$ to obtain the general solution:


hence, with $v(0)=0$,
$$ y = \frac{x^3}{4} + \frac{C}{x}. $$


$$v(t)=\frac{mg}{b}\left(1-e^{-(b/m)t}\right)$$
A quick substitution verifies that this function family satisfies the original equation. The arbitrary constant $C$ can be determined later if an initial condition $y(x_0)=y_0$ is given.
 
As $t$ grows the exponential fades and the speed approaches the constant terminal velocity $mg/b$; in the language of the classification section the terminal velocity is the particular solution and the fading exponential the homogeneous part. Numbers: a skydiver of $m=70\ \mathrm{kg}$ with $b=14\ \mathrm{kg\,s^{-1}}$ has $mg/b = 70\times9.8/14 = 49\ \mathrm{m\,s^{-1}}$ and $b/m = 0.2\ \mathrm{s^{-1}}$, so
 
$$v(t)=49\left(1-e^{-0.2t}\right)\ \mathrm{m\,s^{-1}}$$
 
After 5 seconds $v=49(1-e^{-1})\approx 31\ \mathrm{m\,s^{-1}}$; after 10 seconds $v=49(1-e^{-2})\approx 42\ \mathrm{m\,s^{-1}}$; the terminal $49\ \mathrm{m\,s^{-1}}$ is approached but never quite reached.<ref>{{#cite:Q1576}}</ref>


==== Method 3: constant-coefficient linear equations (trial solutions) ====
==== Method 3: constant-coefficient linear equations (trial solutions) ====
'''General case.''' When the coefficient $p(x)$ is the constant $a$, the linear first-order equation has constant coefficients, and the homogeneous and non-homogeneous cases are solved separately and then added, as in the classification section.
'''Homogeneous case.''' The equation $y'+a\,y=0$ is solved by an exponential trial. Try $y=C e^{bx}$: substituting gives $(b+a)\,C e^{bx}=0$, and since the exponential never vanishes the only possible exponent is $b=-a$, so the homogeneous solution is
$$y_h=C e^{-ax}$$
Exponential growth $y'=k\,y$ is the special case $a=-k$, giving $y_h=C e^{kt}$ with no integration at all.
'''Non-homogeneous case.''' For the forced equation $y'+a\,y=q(x)$, the general solution is the particular-plus-homogeneous sum
$$y=y_p+y_h=C e^{-ax}+y_p$$
because substituting $y_p+y_h$ leaves $q(x)+0=q(x)$: the homogeneous part already vanishes on its own, so it can be added to any particular solution without spoiling it. When $q(x)$ is a constant, an exponential, a sine or cosine, or a polynomial, a particular solution of the same form can be guessed and its coefficient fixed by substitution — this is the '''method of undetermined coefficients'''. A trial that duplicates the homogeneous solution is multiplied by $x$ instead. The same exponential trial reappears, as the characteristic equation, for the constant-coefficient equations of order two below.
'''Example: an account with steady withdrawals.''' An account earning 10% interest compounded continuously, from which €100 is withdrawn each year, is forced exponential growth:
$$y'=0.1\,y-100$$
The homogeneous part, found by the trial $y=C e^{bt}$, is $y_h=C e^{0.1t}$. The withdrawal term is constant, so guess the constant particular solution $y_p=A$; substituting gives $0.1A-100=0$, hence $A=1000$. Starting with €5000, the condition $y(0)=1000+C=5000$ gives $C=4000$ (the constant lives, as always, in the homogeneous part), and
$$y(t)=1000+4000\,e^{0.1t}$$
Check: $y'-0.1y=400e^{0.1t}-(100+400e^{0.1t})=-100$, as required. The €1000 is the balance whose annual interest (10% of €1000) exactly offsets the withdrawals: a balance above €1000 grows, and one below it shrinks. After ten years
$$y(10)=1000+4000\,e\approx 11\,873$$
whereas without the withdrawals the €5000 would have grown to $5000e\approx 13\,591$.<ref>{{#cite:Q1576}}</ref>
Textbooks add further first-order classes — equations homogeneous in the sense $y'=f(y/x)$, Bernoulli equations, and exact equations — each solvable by an extra change of variables or, for exact equations, by recognising a total differential.<ref>{{#cite:Q1576}}</ref>


=== Second-order ODEs ===
=== Second-order ODEs ===
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==== Method 1: direct integration ====
==== Method 1: direct integration ====


'''General case.''' When the equation has the form $y''=f(x)$, with the right-hand side depending only on the independent variable, each derivative is undone by one integration:
'''General case.''' For $y''=f(x)$,


$$y''=f(x)\qquad\Longrightarrow\qquad y'=\int f(x)\,dx+C_1\qquad\Longrightarrow\qquad y=\int\!\!\left(\int f(x)\,dx\right)dx+C_1 x+C_2$$
$$y''=f(x)\;\Longrightarrow\;y'=\int f(x)\,dx+C_1\;\Longrightarrow\;y=\int\!\!\left(\int f(x)\,dx\right)dx+C_1x+C_2$$


The two constants are fixed by two conditions, typically the initial value and the initial derivative (see the discussion of order above). The same pattern applies to $y^{(n)}=f(x)$ with $n$ integrations and $n$ constants.
and likewise $y^{(n)}=f(x)$ by $n$ integrations.


'''Example: free fall.''' With only gravity acting, the height $x(t)$ of a falling object obeys $x''=-g$. Integrating twice,
'''Example: free fall.''' A ball released above the ground falls under gravity alone, which accelerates it downward at the constant rate $g\approx 9.8\ \mathrm{m\,s^{-2}}$; Newton's second law gives the second-order equation $x''=-g$, of the form above with $f(x)=-g$. Integrating twice,


$$\frac{dx}{dt}=-gt+v_0\qquad\Longrightarrow\qquad x(t)=-\frac{g}{2}t^2+v_0 t+x_0$$
$$x''=-9.8\;\Longrightarrow\;\frac{dx}{dt}=-9.8t+v_0\;\Longrightarrow\;x(t)=-4.9t^2+v_0t+x_0$$


where $v_0$ and $x_0$ are the speed and height at $t=0$. A ball dropped from rest ($v_0=0$) at height $19.6\ \mathrm{m}$ reaches the ground, $x=0$, when
Numeric scenario: the ball is dropped from rest, $v_0=0$, at height $x_0=19.6\ \mathrm{m}$. It reaches the ground, $x=0$, when


$$0=19.6-4.9\,t^{2}\qquad\Longrightarrow\qquad t=\sqrt{19.6/4.9}=2\ \text{s}$$
$$0=19.6-4.9t^2\;\Longrightarrow\;t=\sqrt{19.6/4.9}=2\ \text{s}$$


The two initial conditions have fixed the whole trajectory.
so the two initial conditions have pinned down the whole trajectory.


==== Method 2: linear equations with constant coefficients ====
==== Method 2: linear equations with constant coefficients ====


'''General case.''' Linear equations with constant coefficients,
'''General case.'''


$$y''+a\,y'+b\,y=f(x)$$
$$y''+a\,y'+b\,y=f(x)$$


model a mass on a spring, a small-angle pendulum, and an RLC circuit. As with every linear equation, the homogeneous and non-homogeneous cases are solved separately and joined by superposition.
'''Homogeneous case ($f=0$).''' The exponential trial $y=e^{rx}$,
 
'''Homogeneous case ($f=0$).''' Try the exponential $y=e^{rx}$ (the trial of Method 3), since $y'=re^{rx}$ and $y''=r^2e^{rx}$, giving


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The factor $e^{rx}$ is never zero, so the exponential solves the equation exactly when $r$ solves the algebraic '''characteristic equation'''
gives the characteristic equation $r^2+ar+b=0$, whose roots determine $y_h$:
 
$$r^2+ar+b=0$$
 
whose roots determine the homogeneous solution:
 
* distinct real roots $r_1\neq r_2$: $y_h=C_1e^{r_1x}+C_2e^{r_2x}$;
* one repeated root $r$: $y_h=(C_1+C_2x)e^{rx}$;
* complex pair $r=\alpha\pm i\beta$: $y_h=e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)$.
 
The two arbitrary constants, fixed by the initial conditions, live here and only here.
 
'''Non-homogeneous case ($f\neq 0$).''' By the classification result, the general solution is


$$y=y_p+y_h$$
* $r_1\neq r_2$ real: $y_h=C_1e^{r_1x}+C_2e^{r_2x}$;
* $r_1=r_2=r$: $y_h=(C_1+C_2x)e^{rx}$;
* $r=\alpha\pm i\beta$: $y_h=e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)$.


with $y_h$ from above and $y_p$ any single solution of the full equation: substituting $y_p+y_h$ gives $f(x)+0$, since the homogeneous part vanishes on its own. When $f(x)$ is a constant, an exponential, a sine or cosine, or a polynomial, $y_p$ is found by undetermined coefficients, exactly as in Method 3, and a trial duplicating $y_h$ is multiplied by $x$.
'''Non-homogeneous case ($f\neq 0$).''' $y=y_h+y_p$, with $y_p$ found by undetermined coefficients as in Method 3.


'''Worked demonstration (homogeneous):''' solve $y''-3y'+2y=0$. The characteristic equation $r^2-3r+2=(r-1)(r-2)=0$ has the two distinct real roots $1$ and $2$, so
'''Worked demonstration (homogeneous).''' $y''-3y'+2y=0$: $r^2-3r+2=(r-1)(r-2)=0$,


{{#content:Q1608}}
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Check: for $y=e^x$, $y''-3y'+2y=(1-3+2)e^x=0$.
Check: $e^x$ gives $(1-3+2)e^x=0$.
 
'''Worked demonstration (non-homogeneous):''' the same operator forced by $2e^{3x}$, i.e. $y''-3y'+2y=2e^{3x}$, keeps the homogeneous solution found above, $y_h=C_1e^x+C_2e^{2x}$, and adds a trial $y_p=Ae^{3x}$. Substituting,


$$y_p''-3y_p'+2y_p=(9-9+2)Ae^{3x}=2Ae^{3x}=2e^{3x}$$
'''Worked demonstration (non-homogeneous).''' $y''-3y'+2y=2e^{3x}$: keep $y_h$ above, try $y_p=Ae^{3x}$:


so $A=1$ and
$$y_p''-3y_p'+2y_p=(9-9+2)Ae^{3x}=2Ae^{3x}\;\Longrightarrow\;A=1$$


$$y=C_1e^x+C_2e^{2x}+e^{3x}$$
$$y=C_1e^x+C_2e^{2x}+e^{3x}$$


Check: substituting $y=e^{3x}$ gives $(9-9+2)e^{3x}=2e^{3x}$; the homogeneous terms $e^x,e^{2x}$ vanish by the earlier check, so the sum solves the forced equation for any $C_1,C_2$.
'''Example: the harmonic oscillator (a mass on a spring).''' A mass attached to a spring is pulled back towards its rest position by a force $-kx$ proportional to the displacement (Hooke's law); once released it oscillates. Newton's second law models the motion,
 
'''Example: the harmonic oscillator (a mass on a spring).''' A mass displaced $x$ from rest is pulled back by $-kx$ (Hooke's law), so Newton's second law gives
 
$$m\frac{d^2x}{dt^2}=-kx\qquad\Longrightarrow\qquad x''+\frac{k}{m}x=0$$


Writing $\omega_0^2=k/m$ gives the '''harmonic oscillator equation'''
$$m\frac{d^2x}{dt^2}=-kx\;\Longrightarrow\;x''+\omega_0^2x=0,\qquad \omega_0=\sqrt{\frac{k}{m}}$$


{{#content:Q1588}}
{{#content:Q1588}}


whose characteristic equation $r^2+\omega_0^2=0$ has the purely imaginary roots $r=\pm i\omega_0$, the complex-pair case with $\alpha=0$. Since $\frac{d^2}{dt^2}\cos\omega_0 t=-\omega_0^2\cos\omega_0 t$, and likewise for sine, superposition gives
The trial $x=e^{rt}$ gives the characteristic equation $r^2+\omega_0^2=0$ with roots $r=\pm i\omega_0$, the complex-pair case with $\alpha=0$, hence


$$x(t)=A\cos\omega_0 t+B\sin\omega_0 t$$
$$x(t)=A\cos\omega_0t+B\sin\omega_0t$$


with $A,B$ fixed by the initial position and velocity.
with $A,B$ fixed by the initial position and velocity.
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[[File:Simple harmonic motion animation.gif|thumb|A mass on a spring: the harmonic oscillator solution is a sinusoid of fixed amplitude. Credit: Evil saltine (public domain).]]
[[File:Simple harmonic motion animation.gif|thumb|A mass on a spring: the harmonic oscillator solution is a sinusoid of fixed amplitude. Credit: Evil saltine (public domain).]]


Numbers: $m=2\ \mathrm{kg}$, $k=8\ \mathrm{N/m}$, so $\omega_0=\sqrt{8/2}=2\ \text{rad/s}$. Pulled $10\ \mathrm{cm}$ out and released from rest, $B=0$ and $x(t)=0.10\cos 2t$ metres. The period is
Numeric scenario: a mass $m=2\ \mathrm{kg}$ hangs on a spring with $k=8\ \mathrm{N\,m^{-1}}$, so $\omega_0=\sqrt{8/2}=2\ \mathrm{rad\,s^{-1}}$ and the displacement obeys $x''+4x=0$. Pulled $0.10\ \mathrm{m}$ from rest and released, the conditions $x(0)=0.10$, $x'(0)=0$ give $A=0.10$, $B=0$:
 
$$x(t)=0.10\cos 2t\ \mathrm{m},\qquad P=\frac{2\pi}{\omega_0}=\pi\approx 3.14\ \text{s}$$
 
After one second $x(1)=0.10\cos 2\approx -0.042\ \mathrm{m}$, and the motion is '''simple harmonic motion'''.<ref>{{#cite:Q1577}}</ref>
 
=== Partial differential equations ===


$$P=\frac{2\pi}{\omega_0}=\pi\approx 3.14\ \text{s}$$
Two standard techniques give closed-form solutions of linear PDEs: separation of variables, for separable problems on bounded domains, and the method of characteristics, for first-order equations. The wave equation is also solved by the second technique, because its operator factors into two first-order parts. Both methods below are stated in general and then applied to a concrete equation.


and after one second
==== Method 1: separation of variables ====


$$x(1)=0.10\cos 2\approx 0.10(-0.416)\approx -0.042\ \text{m}$$
'''General form.''' For a linear homogeneous PDE in two variables on a bounded domain with homogeneous boundary conditions, seek a solution of the separated form


Such fixed-amplitude sinusoidal motion is '''simple harmonic motion'''.<ref>{{#cite:Q1577}}</ref>
$$u(x,t)=X(x)\,T(t)$$


=== Partial differential equations ===
Substituting into the PDE and dividing by $XT$ separates the variables into one ordinary differential equation in $x$ and one in $t$. Since the two sides are functions of different variables, they can be identically equal only if each equals the same constant, the separation constant $-\lambda$. The $x$-equation together with the boundary conditions is an eigenvalue problem: only a discrete sequence of constants $\lambda_n$, with eigenfunctions $X_n(x)$, is admissible. The $t$-equation then has a solution $T_n(t)$ for each $n$, and every product $X_nT_n$ solves the PDE.


Two elementary exact methods cover the standard introductory cases: separation of variables, for diffusion problems on finite domains, and travelling waves (the method of characteristics), for transport. Each is stated in general and then applied to a concrete numerical example.
'''General algebraic solution.''' The PDE is linear and homogeneous, so the separated modes superimpose:


==== Method 1: separation of variables (the heat equation) ====
$$u(x,t)=\sum_n c_n\,X_n(x)\,T_n(t)$$


'''General case.''' For a linear, homogeneous PDE on a simple domain, look for a solution that is a product of functions of the separate independent variables, $u(x,t)=X(x)T(t)$. Substituting splits the PDE into two linked ordinary equations; boundary conditions pick out which solutions survive; and superposition of those basic solutions then gives the general solution.
with the coefficients $c_n$ chosen so that the series equals the initial profile $u(x,0)$; orthogonality of the eigenfunctions $X_n$ determines them.


'''Application.''' Solve the heat equation on a bar of length $L$ with insulated sides and both ends held at $0$:
'''Example: the heat equation.''' The temperature of a bar of length $L$ with insulated sides and both ends held at $0$ obeys


$$\frac{\partial u}{\partial t}=\alpha\frac{\partial^2u}{\partial x^2}$$
$$\frac{\partial u}{\partial t}=\alpha\frac{\partial^2u}{\partial x^2},\qquad u(0,t)=u(L,t)=0$$


Seek $u(x,t)=X(x)T(t)$. Substituting gives $XT'=\alpha X''T$; dividing by $\alpha XT$,
Substituting $u=X(x)T(t)$ gives $XT'=\alpha X''T$, and dividing by $\alpha XT$,


{{#content:Q1622}}
{{#content:Q1622}}


The left side depends only on $t$ and the right only on $x$, so both must equal one and the same constant, written $-\lambda$. Each side is now an ODE:
The $t$-equation $T'=-\alpha\lambda T$ has solution $T=e^{-\alpha\lambda t}$, and the $x$-equation
 
$$T'=-\alpha\lambda T\qquad\Longrightarrow\qquad T=e^{-\alpha\lambda t}$$


$$X''=-\lambda X\qquad\Longrightarrow\qquad X=A\cos(\sqrt{\lambda}\,x)+B\sin(\sqrt{\lambda}\,x)$$
$$X''=-\lambda X\;\Longrightarrow\;X=A\cos(\sqrt\lambda\,x)+B\sin(\sqrt\lambda\,x)$$


The end conditions $u(0,t)=u(L,t)=0$ force $X(0)=X(L)=0$: hence $A=0$ and $\sin(\sqrt\lambda\,L)=0$, so $\sqrt\lambda\,L=n\pi$, $n=1,2,\dots$ Each allowed value $\lambda=(n\pi/L)^2$ gives one basic solution, a mode
together with the boundary conditions forces $X(0)=X(L)=0$: hence $A=0$ and $\sin(\sqrt\lambda\,L)=0$, so $\sqrt\lambda\,L=n\pi$, $n=1,2,\dots$. Each $\lambda=(n\pi/L)^2$ gives one mode


$$u_n(x,t)=\sin\frac{n\pi x}{L}\,e^{-\alpha(n\pi/L)^2t}$$
$$u_n(x,t)=\sin\frac{n\pi x}{L}\,e^{-\alpha(n\pi/L)^2t}$$


The equation is linear and homogeneous, so superposition applies and the general solution is
and the general algebraic solution above becomes


{{#content:Q1611}}
{{#content:Q1611}}


with the $b_n$ fixed by the initial profile $u(x,0)$ (a Fourier sine series). The decay rate $\alpha(n\pi/L)^2$ grows as $n^2$, so higher modes die out first.
with $b_n$ determined by the Fourier sine series of the initial profile $u(x,0)$; the decay rate $\alpha(n\pi/L)^2$ grows as $n^2$.


'''Numbers:''' a $1\ \mathrm{m}$ iron bar, $\alpha\approx 2.3\times10^{-5}\ \mathrm{m^2s^{-1}}$, heated so $u(x,0)=100\sin(\pi x/L)$ (ends at $0\,^{\circ}\mathrm{C}$, centre $100\,^{\circ}\mathrm{C}$). Only the $n=1$ mode is present:
Numeric scenario: a $1\ \mathrm{m}$ iron bar, heated so that its centre is at $100\,^{\circ}\mathrm{C}$ while both ends are held at $0\,^{\circ}\mathrm{C}$, cools by conduction with iron's diffusivity $\alpha\approx 2.3\times10^{-5}\ \mathrm{m^2s^{-1}}$. The initial profile $u(x,0)=100\sin(\pi x/L)$ is exactly the first mode, so only $n=1$ contributes and


$$u(x,t)=100\sin\frac{\pi x}{L}\,e^{-\alpha\pi^2t/L^2}$$
$$u(x,t)=100\sin\frac{\pi x}{L}\,e^{-\alpha\pi^2t/L^2}$$
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so after one hour $u\approx 100e^{-0.82}\approx 44\,^{\circ}\mathrm{C}$, and $50\,^{\circ}\mathrm{C}$ is reached at $t=\ln 2/(2.3\times10^{-4})\approx 3050\ \text{s}\approx 51$ min.<ref>{{#cite:Q1579}}</ref>
so after one hour $u\approx 100e^{-0.82}\approx 44\,^{\circ}\mathrm{C}$, and $50\,^{\circ}\mathrm{C}$ is reached at $t=\ln 2/(2.3\times10^{-4})\approx 3050\ \text{s}\approx 51$ min.<ref>{{#cite:Q1579}}</ref>


==== Method 2: travelling waves (the method of characteristics) ====
==== Method 2: the method of characteristics ====
 
'''General form.''' The method of characteristics solves first-order PDEs by tracing curves along which the PDE reduces to ordinary differential equations. In two independent variables the general quasilinear first-order equation is
 
$$A(x,t,u)\,u_x+B(x,t,u)\,u_t=C(x,t,u)$$
 
A solution $u=u(x,t)$ is a surface in $(x,t,u)$-space. Its tangent plane at each point is spanned by $(1,0,u_x)$ and $(0,1,u_t)$, so a vector $(A,B,C)$ is tangent to the surface exactly when $C=A u_x+B u_t$, the condition expressed by the PDE itself. The solution surface is therefore swept out by the integral curves of the vector field $(A,B,C)$, the characteristic curves, which solve the characteristic system of ordinary differential equations
 
$$\frac{dx}{ds}=A(x,t,u),\qquad \frac{dt}{ds}=B(x,t,u),\qquad \frac{du}{ds}=C(x,t,u)$$
 
Given data on a curve that is not itself characteristic, such as $u(x,0)=u_0(x)$, one characteristic issues from each point of the curve, and integrating the system carries the data across the region the characteristics cover. For the linear homogeneous case


'''General case.''' Some first-order PDEs are solved not by separating variables but by noticing that their solutions travel. The transport equation with constant speed $c$,
$$a(x,t)\,u_x+b(x,t)\,u_t=0$$


$$u_t+c\,u_x=0$$
the $x$- and $t$-equations do not involve $u$, and the third gives $du/ds=0$: the solution is constant along each characteristic. The characteristics form a one-parameter family; let $\psi(x,t)=\text{const}$ be a first integral, a function constant on each member of the family.


states that $u$ is carried along unchanged. Trying a travelling wave $u(x,t)=f(x-ct)$, with $f$ arbitrary, the chain rule gives $u_t=-c\,f'$ and $u_x=f'$, so
'''General algebraic solution.''' Since $u$ is constant on every characteristic and the characteristics are the level sets of $\psi$, the general solution is an arbitrary function of the first integral,


$$u_t+c\,u_x=-c\,f'+c\,f'=0$$
$$u(x,t)=F\bigl(\psi(x,t)\bigr)$$


whatever $f$ is. Hence the general solution is any profile sliding rigidly to the right at speed $c$ (to the left if $c<0$):
with $F$ fixed by the initial data. When the right-hand side of the PDE is nonzero, $u$ changes along a characteristic at the rate $C$ (or of the given source term), so the general solution acquires an integral of that term along the curve.


$$u(x,t)=f(x-ct)$$
'''Example: transport of a pollutant.''' For constant coefficients $c$ the equation $u_t+c\,u_x=0$ has characteristics $dx/dt=c$, the straight lines $x-ct=\text{const}$; hence $\psi=x-ct$, and the general algebraic solution is the travelling wave


with $f$ fixed by the initial profile, $u(x,0)=f(x)$. Equivalently, $u$ is constant on each of the straight '''characteristic''' lines $x-ct=\text{constant}$. The equation is linear and homogeneous, so sums of travelling waves are again solutions. A source term makes it non-homogeneous, $u_t+cu_x=s(x,t)$, and adds, along each characteristic, the accumulated contribution of $s$, the same particular-plus-homogeneous structure as for ODEs:
$$u(x,t)=F(x-ct),\qquad u(x,0)=F(x)$$


$$u(x,t)=f(x-ct)+\int_0^t s\bigl(x-c(t-\tau),\tau\bigr)\,d\tau$$
A river flows steadily at speed $c=2\ \mathrm{m\,s^{-1}}$, and a factory releases a concentrated slug of pollutant at one point; as long as mixing and diffusion are negligible, the current simply carries the whole slug downstream without changing it. The concentration obeys $u_t+2u_x=0$ with the Gaussian initial profile


'''Example: a slug of pollutant in a river.''' A river flows at $2\ \mathrm{m\,s^{-1}}$, and at $t=0$ a release at $x=0$ gives the Gaussian concentration profile $u(x,0)=50\,e^{-(x/10)^2}\ \mathrm{mg\,L^{-1}}$: $50\ \mathrm{mg\,L^{-1}}$ at the release point, falling by a factor $e^{-1}\approx 0.37$ ten metres away. The solution slides the whole profile downstream without changing it:
$$u(x,0)=50\,e^{-(x/10)^2}\ \mathrm{mg\,L^{-1}}$$
 
(peak $50\ \mathrm{mg\,L^{-1}}$ at the release point, falling by $e^{-1}$ ten metres away). The solution above gives


$$u(x,t)=50\,e^{-((x-2t)/10)^2}\ \mathrm{mg\,L^{-1}}$$
$$u(x,t)=50\,e^{-((x-2t)/10)^2}\ \mathrm{mg\,L^{-1}}$$


After one minute the peak, which started at $x=0$, has reached $x=2\times 60=120$ m downstream and still reads $50\ \mathrm{mg\,L^{-1}}$; a monitor $120$ m downstream sees the peak arrive after $60$ s. The transport equation describes pure advection, with no spreading: that is why the heat example above needed the extra second-order term $\alpha u_{xx}$ to spread its pulse out.
After one minute the peak has moved from $x=0$ to $x=ct=120\ \mathrm{m}$, still reading $50\ \mathrm{mg\,L^{-1}}$; pure transport does not spread the slug, which would require the second-order term $\alpha u_{xx}$ of the heat equation. With a source $q(x,t)$, the value accumulates along each characteristic:


'''Remark: the wave equation.''' The second-order wave equation $u_{tt}=c^2u_{xx}$ (a plucked string) is solved by the same travelling-wave idea applied in both directions at once; its general solution is
$$u(x,t)=F(x-ct)+\int_0^t q\bigl(x-c(t-\tau),\tau\bigr)\,d\tau$$
 
'''Example: the wave equation.''' The wave equation
 
$$u_{tt}=c^2u_{xx}$$
 
is second order, yet its operator factors into two first-order transport operators, so the method of characteristics still applies. Introduce the characteristic coordinates
 
$$\xi=x-ct,\qquad \eta=x+ct$$
 
in which the operator becomes $u_{tt}-c^2u_{xx}=-4c^2u_{\xi\eta}$, so the equation reads $u_{\xi\eta}=0$. Hence $u_\xi$ depends on $\xi$ alone, and one further integration gives the general algebraic solution (d'Alembert, 1747):


$$u(x,t)=f(x-ct)+g(x+ct)$$
$$u(x,t)=f(x-ct)+g(x+ct)$$


two arbitrary shapes moving left and right ([[Person:Jean le Rond d'Alembert|d'Alembert]], 1747, described in the history section).<ref>{{#cite:Q1579}}</ref>
a superposition of two travelling waves, one in each direction. The functions $f,g$ are fixed by the initial displacement and velocity: for a string released from rest with initial displacement $\phi(x)$, the conditions $u(x,0)=\phi(x)$ and $u_t(x,0)=0$ give $f=g=\phi/2$, so
 
$$u(x,t)=\frac{\phi(x-ct)+\phi(x+ct)}{2}$$
 
and the initial hump separates into two half-size copies travelling apart at speed $c$.<ref>{{#cite:Q1579}}</ref>


=== When no formula exists ===
=== When no formula exists ===
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== A short history ==
== A short history ==


Differential equations are as old as the calculus itself, because the calculus is the mathematics of change: the laws of physics say how quantities change, and predicting the future means undoing those changes. When Newton published his laws of motion and of gravitation in the ''Principia'' in 1687, the equations he needed were differential equations, and he solved them by geometry and by infinite series. Newton had his own notation for rates of change, but it was Leibniz's $dy/dx$, first written in the 1670s, that survived: it displays the whole equation on the page, and it is the notation used throughout this article.<ref>{{#cite:Q1577}}</ref>
The origins of differential equations coincide with those of the calculus, since the calculus supplies the language in which rates of change are expressed and inverted. Newton's laws of motion and of universal gravitation, published in the ''Philosophiae Naturalis Principia Mathematica'' (1687), are differential equations; Newton treated them by the geometrical and infinite-series methods of his fluxional calculus. Although Newton developed a notation for fluxions, the differential notation $dy/dx$ introduced by Leibniz in the 1670s proved the more enduring: it exhibits the structure of the equation directly and is the notation adopted in this article.<ref>{{#cite:Q1577}}</ref>


[[File:Isaac Newton portrait.jpg|thumb|left|Isaac Newton (portrait after Godfrey Kneller, 1689). The laws of motion and of gravitation published in the ''Principia'' (1687) are differential equations. Credit: James Thronill after Godfrey Kneller (public domain).]]
[[File:Isaac Newton portrait.jpg|thumb|left|Isaac Newton (portrait after Godfrey Kneller, 1689). Newton's laws of motion and of gravitation (''Principia'', 1687) are differential equations. Credit: James Thronill after Godfrey Kneller (public domain).]]


The scattered tricks of the early calculus became a subject when [[Person:Leonhard Euler|Leonhard Euler]] took them up in the middle decades of the 18th century. He recognised that a linear equation with constant coefficients is solved by substituting $y=e^{rx}$, turning calculus into algebra; he developed series solutions; and, for equations that resisted formulas, he invented the step-by-step numerical scheme, described in this article, that still bears his name. Most of the exact methods above descend from his work.<ref>{{#cite:Q1577}}</ref>
The consolidation of these techniques into a systematic theory is due in large measure to [[Person:Leonhard Euler|Leonhard Euler]], whose work in the middle decades of the eighteenth century established the principal exact methods. Euler showed that linear equations with constant coefficients are solved by the substitution $y=e^{rx}$, which reduces the problem to an algebraic equation, and he advanced the theory of series solutions. For equations that admitted no closed-form solution, he introduced the step-by-step numerical procedure, described above as Euler's method, that bears his name. The exact methods presented in this article derive, in large part, from his work.<ref>{{#cite:Q1577}}</ref>


[[File:Leonhard Euler portrait.jpg|thumb|Leonhard Euler (portrait by Jakob Emanuel Handmann, 1753). Credit: Jakob Emanuel Handmann (public domain).]]
[[File:Leonhard Euler portrait.jpg|thumb|Leonhard Euler (portrait by Jakob Emanuel Handmann, 1753). Credit: Jakob Emanuel Handmann (public domain).]]


Meanwhile physics began asking for equations with more than one independent variable. A plucked string takes a shape that depends on position along the string and on time, and in 1747 [[Person:Jean le Rond d'Alembert|Jean le Rond d'Alembert]] wrote down the wave equation for it and solved it, showing that its solutions are two waves travelling in opposite directions. Heat conduction posed a subtler problem, because the initial temperature of a bar can have any shape at all. [[Person:Joseph Fourier|Joseph Fourier]] derived the heat equation from the physics of conduction and, to solve it, had to express an arbitrary initial profile as a sum of sine ripples. His ''Théorie analytique de la chaleur'' of 1822 turned separation of variables into a cornerstone of applied mathematics, and the Fourier series invented for the purpose now appears wherever signals are analysed, from acoustics to image compression.<ref>{{#cite:Q1579}}</ref>
The theory of partial differential equations arose from the demands of eighteenth-century physics. In 1747, [[Person:Jean le Rond d'Alembert|Jean le Rond d'Alembert]] derived the wave equation for the vibrating string and established that its general solution consists of two waves propagating in opposite directions. The problem of heat conduction proved more demanding, because the initial temperature distribution of a conducting body is arbitrary. In his ''Théorie analytique de la chaleur'' (1822), [[Person:Joseph Fourier|Joseph Fourier]] derived the heat equation from the physical principles of conduction and solved it by expanding the initial data into a trigonometric series. This work established separation of variables as a standard technique of mathematical physics, and the Fourier series introduced for the purpose has since become fundamental to the analysis of periodic phenomena, from acoustics to signal processing.<ref>{{#cite:Q1579}}</ref>


The exact formulas, however, have their limits, and the history of the subject since the late 19th century is largely the story of what to do when no formula exists. Studying the three-body problem of celestial mechanics, [[Person:Henri Poincaré|Henri Poincaré]] realised that the shape of the motion can be understood without solving the equations, founding the qualitative theory of dynamical systems. The electronic computer then made the numerical route routine: approximate the solution step by step, as Euler's method does, refining the steps until the error is acceptable. The two strands met in 1963, when the meteorologist [[Person:Edward Lorenz|Edward Lorenz]] found that a simple system of three differential equations, meant to model atmospheric convection, behaved chaotically: the equations were deterministic, yet their solutions were aperiodic and so sensitive to initial conditions that long-term weather prediction is impossible in practice. Each of these later routes, qualitative study, numerical stepping, series, and transforms, is the subject of its own article.<ref>{{#cite:Q1578}}</ref>
The limits of closed-form methods became apparent towards the end of the nineteenth century, and the later history of the subject is concerned principally with equations for which elementary solutions do not exist. In his investigation of the three-body problem of celestial mechanics, [[Person:Henri Poincaré|Henri Poincaré]] demonstrated that qualitative properties of the motion, such as its equilibria, stability, and long-term behaviour, can be characterised without solving the equations, thereby founding the qualitative theory of dynamical systems. The subsequent development of electronic computing made numerical approximation, of which Euler's method is the simplest instance, a routine and general technique. The two strands converged in 1963, when [[Person:Edward Lorenz|Edward Lorenz]], studying a simplified system of three ordinary differential equations that models atmospheric convection, established the phenomenon of deterministic chaos: although the equations are deterministic, their solutions are aperiodic and depend so sensitively on initial conditions that long-term weather prediction is not feasible in practice. These later approaches, qualitative analysis, numerical approximation, and series and transform methods, are treated in dedicated articles.<ref>{{#cite:Q1578}}</ref>


== References ==
== References ==

Latest revision as of 11:56, 7 September 2026

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A differential equation is an equation whose unknown is a function and which also involves that function's derivatives (rates of change). Where an ordinary equation such as $x^2=9$ is solved by numbers, a differential equation such as $y'+2y=0$ is solved by functions $y(x)$.

A surprising number of laws of nature can be described by differential equations: pendulums, cooling drinks, growing populations, and discharging capacitors.

General and specific solution

The simplest differential equation prescribes the slope of a function $y(x)$:

$$\frac{dy}{dx}=2x$$

Integrating both sides gives

$$\int\frac{dy}{dx}\,dx=\int 2x\,dx\qquad\Longrightarrow\qquad y(x)=x^{2}+C$$

Where $C$ is an unknown constant. Therefore, the solution is not one single function, but a family of functions, known as the general solution.

If we know the value of $y$ at a particular $x$, for instance, $y(0)=3$, then a particular solution function can be identified

$$3=0^2+C\qquad\Longrightarrow\qquad C=3\qquad\Longrightarrow\qquad y=x^2+3$$

A prescribed value such as $y(a)=b$ allowing us to pin down a particular solution is called an initial condition.

Classifying differential equations

A differential equation can be classified by three criteria: its order, its linearity, and how many independent variables it involves.

Order

The order is the order of the highest derivative present.

$\frac{dy}{dx}=2x$ is first order, whereas Newton's second law,

m\frac{d^{2}x}{dt^{2}}=F

is second order ($x(t)$ position of mass $m$, $F$ net force).

Linearity and homogeneity

An equation is linear when the unknown and its derivatives appear only to the first power and never multiplied together. A linear first-order equation can always be written

$$\frac{dy}{dx}+p(x)\,y=q(x)$$

It is said to be homogeneous when $q(x)=0$.

The equations $dy/dx=y^2$ and $d^2\theta/dt^2+\sin\theta=0$ are nonlinear (square of $y$; sine of $\theta$).

An important property of linear homogeneous equation is the superposition principle. If $y_1,y_2$ solve a homogeneous linear equation, so does $c_1y_1+c_2y_2$.

For a nonlinear equation, the same principle usually does not apply: if $y'=y^2$ has two solutions $y_1'=y_1^2$ and $y_2'=y_2^2$

$$(y_1+y_2)'=y_1^2+y_2^2\neq (y_1+y_2)^2$$

so $y_1+y_2$ does not solve $y'=y^2$.

Ordinary and partial

An ordinary differential equation (ODE) has one independent variable. A partial differential equation (PDE) has several, with partial derivatives. All the examples we have seen above are ordinary differential equations. For an example of a partial differential equation, we can take the heat equation describing the temperature $u$ of an insulated metal bar, according to position $x$ on the metal bar and time $t$

\frac{\partial u}{\partial t}=\alpha\frac{\partial^{2}u}{\partial x^{2}}

where $\alpha$ is the thermal diffusivity.

Direction field

A first-order equation can be written

\frac{dy}{dx}=f(x,y)

assigning to each point $(x,y)$ the slope $f(x,y)$ a solution must have there. Drawing short segments of that slope gives a direction field. Solution curves must run tangent to the direction field for every point they pass through.

Direction field of $dy/dx=y$. Credit: jjbeard (public domain).

Solving differential equations

Just like there is no general formula solving all algebraic equations, there is no general method permitting the solution of all differential equations. However, some standard methods exist for solving a restricted class of simple differential equations.

First-order ODEs

Method 1: separable equations

General case. A first-order equation is separable when it can be rewritten as the equality of the derivative to the product of two functions, one containing only $x$, one containing only $y$.

\frac{dy}{dx}=g(x)\,h(y)\qquad\Longrightarrow\qquad\int\frac{dy}{h(y)}=\int g(x)\,dx

after which both integrals can be evaluated directly.

Example: exponential growth and decay. A quantity whose rate of change is proportional to its own size, such as an unchecked population or a radioactive sample, obeys

\frac{dy}{dt}=k\,y

Separating variables and integrating,

$$\int\frac{dy}{y}=\int k\,dt\;\Longrightarrow\;\ln|y|=kt+C_1\;\Longrightarrow\;y=Ce^{kt}$$

The initial condition $y(0)=y_0$ fixes $C=y_0$, giving

y(t)=y_{0}\,e^{kt} (1)

Observing (1) we notice this is very well an exponential growth (k>0)/decay(k<0).

Example: Newton's law of cooling. A hot object in a cooler room loses heat through its surface, and the larger the temperature gap, the faster it cools: The temperature of the object is modelled by

\frac{dT}{dt}=-k\bigl(T-T_{a}\bigr)

Separating variables and integrating,

$$\int\frac{dT}{T-T_a}=-k\int dt\;\Longrightarrow\;\ln|T-T_a|=-kt+C\;\Longrightarrow\;T-T_a=Ce^{-kt}$$

so with $T(0)=T_0$, hence $C=T_0-T_a$,

$$T(t)=T_a+(T_0-T_a)e^{-kt}$$

Method 2: linear first-order equations (integrating factor)

General case. For the linear equation

$$y'+p(x)\,y=q(x)$$

introduce the integrating factor $\mu=e^{\int p\,dx}$, chosen so that $\mu'=p\mu$; multiplying by $\mu$ collapses the left-hand side into a single derivative:

\frac{dy}{dx}+p(x)\,y=q(x),\text{multiply both sides by } \mu(x)=e^{\int p(x)\,dx}\;\Longrightarrow\;\frac{d}{dx}\bigl(\mu(x)\,y\bigr)=\mu(x)\,q(x)

Integrating both sides,

$$\mu y=\int\mu\,q\,dx+C\;\Longrightarrow\;y=\frac{1}{\mu}\int\mu\,q\,dx+\frac{C}{\mu}$$

If you notice, the first term $y=\frac{1}{\mu}\int\mu\,q\,dx$$\frac{C}{\mu}$ provides one particular solution to the non-homogeneous equation, and the second term, $C/\mu=Ce^{-\int p\,dx}$, is a solution to the homogeneous counter par of the original equation: $y'+p(x)\,y=0$. It generalises the solution to the entire solution family.

This is in fact a general principle: $y=y_p+y_h$, i.e., the solution family of a non-homogeneous equation is the sum of one particular solution plus the solution to its homogeneous counterpart.

To see the method in action, consider

$$ \frac{dy}{dx} + \frac{1}{x}\,y = x^2, \qquad x>0. $$

Here $P(x)=1/x$, so

$$ \mu(x)=e^{\int \frac{1}{x}\,dx}=e^{\ln x}=x. $$

Multiply the whole equation by $\mu(x)=x$:

$$ x\frac{dy}{dx} + y = x^3. $$

The left side is exactly the derivative of the product $x\,y$:

$$ \frac{d}{dx}\bigl(xy\bigr) = x^3. $$

Integrate with respect to $x$:

$$ xy = \int x^3\,dx = \frac{x^4}{4} + C. $$

Finally, divide by $x$ to obtain the general solution:

$$ y = \frac{x^3}{4} + \frac{C}{x}. $$

A quick substitution verifies that this function family satisfies the original equation. The arbitrary constant $C$ can be determined later if an initial condition $y(x_0)=y_0$ is given.

Method 3: constant-coefficient linear equations (trial solutions)

Second-order ODEs

Method 1: direct integration

General case. For $y''=f(x)$,

$$y''=f(x)\;\Longrightarrow\;y'=\int f(x)\,dx+C_1\;\Longrightarrow\;y=\int\!\!\left(\int f(x)\,dx\right)dx+C_1x+C_2$$

and likewise $y^{(n)}=f(x)$ by $n$ integrations.

Example: free fall. A ball released above the ground falls under gravity alone, which accelerates it downward at the constant rate $g\approx 9.8\ \mathrm{m\,s^{-2}}$; Newton's second law gives the second-order equation $x''=-g$, of the form above with $f(x)=-g$. Integrating twice,

$$x''=-9.8\;\Longrightarrow\;\frac{dx}{dt}=-9.8t+v_0\;\Longrightarrow\;x(t)=-4.9t^2+v_0t+x_0$$

Numeric scenario: the ball is dropped from rest, $v_0=0$, at height $x_0=19.6\ \mathrm{m}$. It reaches the ground, $x=0$, when

$$0=19.6-4.9t^2\;\Longrightarrow\;t=\sqrt{19.6/4.9}=2\ \text{s}$$

so the two initial conditions have pinned down the whole trajectory.

Method 2: linear equations with constant coefficients

General case.

$$y''+a\,y'+b\,y=f(x)$$

Homogeneous case ($f=0$). The exponential trial $y=e^{rx}$,

y''+a\,y'+b\,y=0,\qquad y=e^{rx}\ \Rightarrow\ r^{2}+a\,r+b=0

gives the characteristic equation $r^2+ar+b=0$, whose roots determine $y_h$:

  • $r_1\neq r_2$ real: $y_h=C_1e^{r_1x}+C_2e^{r_2x}$;
  • $r_1=r_2=r$: $y_h=(C_1+C_2x)e^{rx}$;
  • $r=\alpha\pm i\beta$: $y_h=e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)$.

Non-homogeneous case ($f\neq 0$). $y=y_h+y_p$, with $y_p$ found by undetermined coefficients as in Method 3.

Worked demonstration (homogeneous). $y''-3y'+2y=0$: $r^2-3r+2=(r-1)(r-2)=0$,

y''-3y'+2y=0\qquad\Longrightarrow\qquad y=C_{1}e^{x}+C_{2}e^{2x}

Check: $e^x$ gives $(1-3+2)e^x=0$.

Worked demonstration (non-homogeneous). $y''-3y'+2y=2e^{3x}$: keep $y_h$ above, try $y_p=Ae^{3x}$:

$$y_p''-3y_p'+2y_p=(9-9+2)Ae^{3x}=2Ae^{3x}\;\Longrightarrow\;A=1$$

$$y=C_1e^x+C_2e^{2x}+e^{3x}$$

Example: the harmonic oscillator (a mass on a spring). A mass attached to a spring is pulled back towards its rest position by a force $-kx$ proportional to the displacement (Hooke's law); once released it oscillates. Newton's second law models the motion,

$$m\frac{d^2x}{dt^2}=-kx\;\Longrightarrow\;x''+\omega_0^2x=0,\qquad \omega_0=\sqrt{\frac{k}{m}}$$

\frac{d^{2}x}{dt^{2}}+\omega_{0}^{2}x=0

The trial $x=e^{rt}$ gives the characteristic equation $r^2+\omega_0^2=0$ with roots $r=\pm i\omega_0$, the complex-pair case with $\alpha=0$, hence

$$x(t)=A\cos\omega_0t+B\sin\omega_0t$$

with $A,B$ fixed by the initial position and velocity.

A mass on a spring: the harmonic oscillator solution is a sinusoid of fixed amplitude. Credit: Evil saltine (public domain).

Numeric scenario: a mass $m=2\ \mathrm{kg}$ hangs on a spring with $k=8\ \mathrm{N\,m^{-1}}$, so $\omega_0=\sqrt{8/2}=2\ \mathrm{rad\,s^{-1}}$ and the displacement obeys $x''+4x=0$. Pulled $0.10\ \mathrm{m}$ from rest and released, the conditions $x(0)=0.10$, $x'(0)=0$ give $A=0.10$, $B=0$:

$$x(t)=0.10\cos 2t\ \mathrm{m},\qquad P=\frac{2\pi}{\omega_0}=\pi\approx 3.14\ \text{s}$$

After one second $x(1)=0.10\cos 2\approx -0.042\ \mathrm{m}$, and the motion is simple harmonic motion.[1]

Partial differential equations

Two standard techniques give closed-form solutions of linear PDEs: separation of variables, for separable problems on bounded domains, and the method of characteristics, for first-order equations. The wave equation is also solved by the second technique, because its operator factors into two first-order parts. Both methods below are stated in general and then applied to a concrete equation.

Method 1: separation of variables

General form. For a linear homogeneous PDE in two variables on a bounded domain with homogeneous boundary conditions, seek a solution of the separated form

$$u(x,t)=X(x)\,T(t)$$

Substituting into the PDE and dividing by $XT$ separates the variables into one ordinary differential equation in $x$ and one in $t$. Since the two sides are functions of different variables, they can be identically equal only if each equals the same constant, the separation constant $-\lambda$. The $x$-equation together with the boundary conditions is an eigenvalue problem: only a discrete sequence of constants $\lambda_n$, with eigenfunctions $X_n(x)$, is admissible. The $t$-equation then has a solution $T_n(t)$ for each $n$, and every product $X_nT_n$ solves the PDE.

General algebraic solution. The PDE is linear and homogeneous, so the separated modes superimpose:

$$u(x,t)=\sum_n c_n\,X_n(x)\,T_n(t)$$

with the coefficients $c_n$ chosen so that the series equals the initial profile $u(x,0)$; orthogonality of the eigenfunctions $X_n$ determines them.

Example: the heat equation. The temperature of a bar of length $L$ with insulated sides and both ends held at $0$ obeys

$$\frac{\partial u}{\partial t}=\alpha\frac{\partial^2u}{\partial x^2},\qquad u(0,t)=u(L,t)=0$$

Substituting $u=X(x)T(t)$ gives $XT'=\alpha X''T$, and dividing by $\alpha XT$,

u=X(x)\,T(t)\ \Rightarrow\ \frac{X''}{X}=\frac{T'}{\alpha\,T}=-\lambda

The $t$-equation $T'=-\alpha\lambda T$ has solution $T=e^{-\alpha\lambda t}$, and the $x$-equation

$$X''=-\lambda X\;\Longrightarrow\;X=A\cos(\sqrt\lambda\,x)+B\sin(\sqrt\lambda\,x)$$

together with the boundary conditions forces $X(0)=X(L)=0$: hence $A=0$ and $\sin(\sqrt\lambda\,L)=0$, so $\sqrt\lambda\,L=n\pi$, $n=1,2,\dots$. Each $\lambda=(n\pi/L)^2$ gives one mode

$$u_n(x,t)=\sin\frac{n\pi x}{L}\,e^{-\alpha(n\pi/L)^2t}$$

and the general algebraic solution above becomes

u(x,t)=\sum_{n=1}^{\infty}b_{n}\sin\Bigl(\frac{n\pi x}{L}\Bigr)\,e^{-\alpha (n\pi/L)^{2}t}

with $b_n$ determined by the Fourier sine series of the initial profile $u(x,0)$; the decay rate $\alpha(n\pi/L)^2$ grows as $n^2$.

Numeric scenario: a $1\ \mathrm{m}$ iron bar, heated so that its centre is at $100\,^{\circ}\mathrm{C}$ while both ends are held at $0\,^{\circ}\mathrm{C}$, cools by conduction with iron's diffusivity $\alpha\approx 2.3\times10^{-5}\ \mathrm{m^2s^{-1}}$. The initial profile $u(x,0)=100\sin(\pi x/L)$ is exactly the first mode, so only $n=1$ contributes and

$$u(x,t)=100\sin\frac{\pi x}{L}\,e^{-\alpha\pi^2t/L^2}$$

At the centre, with $L=1$ and $\alpha\pi^2\approx 2.3\times10^{-4}\ \text{s}^{-1}$,

$$u\!\left(\tfrac12,t\right)=100\,e^{-2.3\times10^{-4}t}$$

so after one hour $u\approx 100e^{-0.82}\approx 44\,^{\circ}\mathrm{C}$, and $50\,^{\circ}\mathrm{C}$ is reached at $t=\ln 2/(2.3\times10^{-4})\approx 3050\ \text{s}\approx 51$ min.[2]

Method 2: the method of characteristics

General form. The method of characteristics solves first-order PDEs by tracing curves along which the PDE reduces to ordinary differential equations. In two independent variables the general quasilinear first-order equation is

$$A(x,t,u)\,u_x+B(x,t,u)\,u_t=C(x,t,u)$$

A solution $u=u(x,t)$ is a surface in $(x,t,u)$-space. Its tangent plane at each point is spanned by $(1,0,u_x)$ and $(0,1,u_t)$, so a vector $(A,B,C)$ is tangent to the surface exactly when $C=A u_x+B u_t$, the condition expressed by the PDE itself. The solution surface is therefore swept out by the integral curves of the vector field $(A,B,C)$, the characteristic curves, which solve the characteristic system of ordinary differential equations

$$\frac{dx}{ds}=A(x,t,u),\qquad \frac{dt}{ds}=B(x,t,u),\qquad \frac{du}{ds}=C(x,t,u)$$

Given data on a curve that is not itself characteristic, such as $u(x,0)=u_0(x)$, one characteristic issues from each point of the curve, and integrating the system carries the data across the region the characteristics cover. For the linear homogeneous case

$$a(x,t)\,u_x+b(x,t)\,u_t=0$$

the $x$- and $t$-equations do not involve $u$, and the third gives $du/ds=0$: the solution is constant along each characteristic. The characteristics form a one-parameter family; let $\psi(x,t)=\text{const}$ be a first integral, a function constant on each member of the family.

General algebraic solution. Since $u$ is constant on every characteristic and the characteristics are the level sets of $\psi$, the general solution is an arbitrary function of the first integral,

$$u(x,t)=F\bigl(\psi(x,t)\bigr)$$

with $F$ fixed by the initial data. When the right-hand side of the PDE is nonzero, $u$ changes along a characteristic at the rate $C$ (or of the given source term), so the general solution acquires an integral of that term along the curve.

Example: transport of a pollutant. For constant coefficients $c$ the equation $u_t+c\,u_x=0$ has characteristics $dx/dt=c$, the straight lines $x-ct=\text{const}$; hence $\psi=x-ct$, and the general algebraic solution is the travelling wave

$$u(x,t)=F(x-ct),\qquad u(x,0)=F(x)$$

A river flows steadily at speed $c=2\ \mathrm{m\,s^{-1}}$, and a factory releases a concentrated slug of pollutant at one point; as long as mixing and diffusion are negligible, the current simply carries the whole slug downstream without changing it. The concentration obeys $u_t+2u_x=0$ with the Gaussian initial profile

$$u(x,0)=50\,e^{-(x/10)^2}\ \mathrm{mg\,L^{-1}}$$

(peak $50\ \mathrm{mg\,L^{-1}}$ at the release point, falling by $e^{-1}$ ten metres away). The solution above gives

$$u(x,t)=50\,e^{-((x-2t)/10)^2}\ \mathrm{mg\,L^{-1}}$$

After one minute the peak has moved from $x=0$ to $x=ct=120\ \mathrm{m}$, still reading $50\ \mathrm{mg\,L^{-1}}$; pure transport does not spread the slug, which would require the second-order term $\alpha u_{xx}$ of the heat equation. With a source $q(x,t)$, the value accumulates along each characteristic:

$$u(x,t)=F(x-ct)+\int_0^t q\bigl(x-c(t-\tau),\tau\bigr)\,d\tau$$

Example: the wave equation. The wave equation

$$u_{tt}=c^2u_{xx}$$

is second order, yet its operator factors into two first-order transport operators, so the method of characteristics still applies. Introduce the characteristic coordinates

$$\xi=x-ct,\qquad \eta=x+ct$$

in which the operator becomes $u_{tt}-c^2u_{xx}=-4c^2u_{\xi\eta}$, so the equation reads $u_{\xi\eta}=0$. Hence $u_\xi$ depends on $\xi$ alone, and one further integration gives the general algebraic solution (d'Alembert, 1747):

$$u(x,t)=f(x-ct)+g(x+ct)$$

a superposition of two travelling waves, one in each direction. The functions $f,g$ are fixed by the initial displacement and velocity: for a string released from rest with initial displacement $\phi(x)$, the conditions $u(x,0)=\phi(x)$ and $u_t(x,0)=0$ give $f=g=\phi/2$, so

$$u(x,t)=\frac{\phi(x-ct)+\phi(x+ct)}{2}$$

and the initial hump separates into two half-size copies travelling apart at speed $c$.[2]

When no formula exists

Most equations, especially nonlinear ones, fit none of the classes above and have no solution in terms of familiar functions. They are studied in one of three ways:[3][4]

The exact methods occupy the branches on the left; most equations encountered in research fall through to the routes on the right, each treated in its own article.

A short history

The origins of differential equations coincide with those of the calculus, since the calculus supplies the language in which rates of change are expressed and inverted. Newton's laws of motion and of universal gravitation, published in the Philosophiae Naturalis Principia Mathematica (1687), are differential equations; Newton treated them by the geometrical and infinite-series methods of his fluxional calculus. Although Newton developed a notation for fluxions, the differential notation $dy/dx$ introduced by Leibniz in the 1670s proved the more enduring: it exhibits the structure of the equation directly and is the notation adopted in this article.[1]

Isaac Newton (portrait after Godfrey Kneller, 1689). Newton's laws of motion and of gravitation (Principia, 1687) are differential equations. Credit: James Thronill after Godfrey Kneller (public domain).

The consolidation of these techniques into a systematic theory is due in large measure to Leonhard Euler, whose work in the middle decades of the eighteenth century established the principal exact methods. Euler showed that linear equations with constant coefficients are solved by the substitution $y=e^{rx}$, which reduces the problem to an algebraic equation, and he advanced the theory of series solutions. For equations that admitted no closed-form solution, he introduced the step-by-step numerical procedure, described above as Euler's method, that bears his name. The exact methods presented in this article derive, in large part, from his work.[1]

Leonhard Euler (portrait by Jakob Emanuel Handmann, 1753). Credit: Jakob Emanuel Handmann (public domain).

The theory of partial differential equations arose from the demands of eighteenth-century physics. In 1747, Jean le Rond d'Alembert derived the wave equation for the vibrating string and established that its general solution consists of two waves propagating in opposite directions. The problem of heat conduction proved more demanding, because the initial temperature distribution of a conducting body is arbitrary. In his Théorie analytique de la chaleur (1822), Joseph Fourier derived the heat equation from the physical principles of conduction and solved it by expanding the initial data into a trigonometric series. This work established separation of variables as a standard technique of mathematical physics, and the Fourier series introduced for the purpose has since become fundamental to the analysis of periodic phenomena, from acoustics to signal processing.[2]

The limits of closed-form methods became apparent towards the end of the nineteenth century, and the later history of the subject is concerned principally with equations for which elementary solutions do not exist. In his investigation of the three-body problem of celestial mechanics, Henri Poincaré demonstrated that qualitative properties of the motion, such as its equilibria, stability, and long-term behaviour, can be characterised without solving the equations, thereby founding the qualitative theory of dynamical systems. The subsequent development of electronic computing made numerical approximation, of which Euler's method is the simplest instance, a routine and general technique. The two strands converged in 1963, when Edward Lorenz, studying a simplified system of three ordinary differential equations that models atmospheric convection, established the phenomenon of deterministic chaos: although the equations are deterministic, their solutions are aperiodic and depend so sensitively on initial conditions that long-term weather prediction is not feasible in practice. These later approaches, qualitative analysis, numerical approximation, and series and transform methods, are treated in dedicated articles.[4]

References

  1. ↑ ↑ ↑ Tenenbaum, M. (1985). Ordinary Differential Equations (Book). In Ordinary Differential Equations (Book). Dover Publications.
  2. ↑ ↑ ↑ Strauss, W. A. (2008). Partial Differential Equations: An Introduction (Book). In Partial Differential Equations: An Introduction (Book). John Wiley & Sons.
  3. ↑ Boyce, W. E. (2012). Elementary Differential Equations and Boundary Value Problems (Book). In Elementary Differential Equations and Boundary Value Problems (Book). John Wiley & Sons.
  4. ↑ ↑ Strogatz, S. H. (2015). Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering (Book). In Nonlinear Dynamics and Chaos: With Applications to Physics, Biology, Chemistry, and Engineering (Book). Westview Press.

Further reading