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'''A differential equation''' is an equation in which the unknown is a function, and in which the derivatives (rates of change) of that function also appear. An ordinary equation such as $x^2 = 9$ has numbers as solutions, whereas a differential equation such as $2y+y'=0$ has functions $y(x)$ as solutions.
'''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 exact solution methods, each with a worked 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).
 
Most laws of nature are stated as rules about how quantities change, so differential equations appear throughout science and engineering: the swinging of a pendulum, the cooling of a hot drink, the growth of a population, and the discharge of a capacitor are all described by differential equations. This article explains what a differential equation is and how equations are classified, then works through the most common exact methods of solution, each illustrated by a concrete numerical example that needs no background knowledge beyond school algebra. It treats '''ordinary differential equations''' (ODEs), in which the unknown function depends on a single independent variable, in the most detail, and gives shorter accounts of '''partial differential equations''' (PDEs) and of the numerical, series, and qualitative methods used when no exact formula exists.


== A first example: slopes and a family of solutions ==
== A first example: slopes and a family of solutions ==


Take the simplest possible differential equation. Suppose the unknown is a function $y(x)$, and all we are told about it is how it changes:
The simplest differential equation prescribes the slope of a function $y(x)$:


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


Here $dy/dx$ is the slope of the graph of $y$. The equation says that whatever the solution is, its slope at the point $x$ must equal $2x$.
Integration inverts differentiation, so integrating both sides gives
 
To solve the equation is to find every function whose slope behaves this way. Integrating both sides with respect to $x$ undoes the differentiation on the left, so


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


where $C$ is an arbitrary constant, because differentiating $x^2 + C$ gives $2x$ for every value of $C$:
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'''.
 
$$\frac{d}{dx}\left(x^{2}+C\right)=2x$$
 
The solutions are therefore not one function but a whole family of parabolas $y = x^2 + C$, one for each choice of $C$, each a vertical shift of the others. This family is called the '''general solution'''.
 
How can we arrive at a '''particular solution''', that is, one specific function of the family? An extra piece of information is required.
 
Suppose we know that when $x = 0$, $y = 3$. Substituting those numbers in we have:
 
$$3=0^{2}+C\qquad\Longrightarrow\qquad C=3$$


Therefore
An extra condition picks out one member. If $y(0)=3$, then


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


Such an extra condition is called an '''initial condition'''.
Such a prescribed value is an '''initial condition'''.


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


A few labels do most of the work in deciding how to attack an equation. Three questions matter: how many derivatives appear (the ''order''), whether the unknown function enters linearly (''linearity''), and how many independent variables are involved (''ordinary or partial''). Equations that fit no convenient label are usually handled numerically or qualitatively, as described in the final section of this article.
Three features decide how to solve an equation: its '''order''', its '''linearity''', and how many independent variables it involves.


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


The '''order''' of a differential equation is the order of the highest derivative that appears in it. The equation $dy/dx = 2x$ of the previous section is first order.
The order is the order of the highest derivative present. $dy/dx=2x$ is first order; Newton's second law,
 
Newton's second law of motion is the standard second-order example: the acceleration of a body, the second derivative of its position, is proportional to the force acting on it:


{{#content:Q1583}}
{{#content:Q1583}}


where $x(t)$ is the position of a body of mass $m$ and $F$ is the net force.
is second order ($x(t)$ position of mass $m$, $F$ net force). An nth-order equation has $n$ arbitrary constants in its general solution, fixed by $n$ conditions, one constant appearing at each integration. Free fall shows the pattern: with only gravity $F_g=-mg$, Newton's law gives
 
In general, the general solution of an nth-order equation contains $n$ arbitrary constants, so $n$ extra conditions are needed to single out one particular solution. When the equation has the special form $y^{(n)}=f(x)$, the constants appear naturally, one at each of the $n$ integrations needed to undo the derivatives; the free-fall example below shows the pattern for $n = 2$.
 
Take a free-falling object, for example. Ignoring air resistance, the only force is gravity
 
$$F_g = -mg$$
 
By Newton's second law
 
$$m\frac{d^{2}x}{dt^{2}}=-mg\qquad\Longrightarrow\qquad\frac{d^{2}x}{dt^{2}}=-g$$
 
with $g \approx 9.8\ \mathrm{m\,s^{-2}}$ the acceleration of free fall. Integrating both sides once gives the velocity, and introduces the constant $v_{0}$, the speed at time $t = 0$:
 
$$\frac{dx}{dt}=-gt+v_{0}$$
 
Integrating again gives the height, and introduces a second constant, $x_{0}$, the height at $t = 0$:


$$x(t)=-\frac{g}{2}\,t^{2}+v_{0}t+x_{0}$$
$$m\frac{d^{2}x}{dt^{2}}=-mg\qquad\Longrightarrow\qquad\frac{d^{2}x}{dt^{2}}=-g\qquad(g\approx 9.8\ \mathrm{m\,s^{-2}})$$


Two conditions, the initial height and the initial velocity, are needed to fix both constants.
Integrate once (constant $v_0$, the speed at $t=0$), then again (constant $x_0$, the height at $t=0$):


A concrete check: a ball dropped from rest ($v_{0} = 0$) at a height of $19.6\ \mathrm{m}$ hits the ground when $x(t) = 0$:
$$\frac{dx}{dt}=-gt+v_{0}\qquad\Longrightarrow\qquad x(t)=-\frac{g}{2}t^{2}+v_{0}t+x_{0}$$


$$0=19.6-4.9\,t^{2}\qquad\Longrightarrow\qquad t=\sqrt{19.6/4.9}=2\ \text{seconds}$$
Two initial conditions are needed. A ball dropped from rest at height $19.6\ \mathrm{m}$ hits the ground ($x=0$) when


Notice how a differential equation can predict the future: the two initial conditions fix the whole trajectory.
$$0=19.6-4.9\,t^{2}\qquad\Longrightarrow\qquad t=\sqrt{19.6/4.9}=2\ \text{s}$$


=== Linearity and homogeneity ===
=== Linearity and homogeneity ===


A differential equation is '''linear''' when the unknown function and its derivatives appear only to the first power and are never multiplied together (multiplying by functions of the independent variable is allowed). A linear first-order equation can always be written as
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


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


Furthermore, a linear equation is called '''homogeneous''' when $q(x)=0$; it then takes the form
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 $\theta$).
 
$$\frac{dy}{dx}+p(x)\,y=0$$
 
The equations
 
$$\frac{dy}{dx}=y^{2},\qquad \frac{d^{2}\theta}{dt^{2}}+\sin\theta=0$$
 
are therefore nonlinear: the first contains the square of the unknown function, the second the sine of it.
 
Linearity and homogeneity matter because homogeneous linear equations obey the '''superposition principle'''. If $y_{1}$ and $y_{2}$ both solve a homogeneous linear equation, then any combination $c_{1}y_{1} + c_{2}y_{2}$ of those functions solves it too. This is why solutions of linear equations can be added together, a property used repeatedly later in this article (for example, to build the solution of the heat equation from simple building blocks).
 
Nonlinear equations do not obey the superposition principle. If $y_{1}$ and $y_{2}$ solve $dy/dx = y^{2}$, their sum does not.


In general, linear equations can be solved systematically, whereas most nonlinear equations cannot; the last section of this article describes what is done with those.
If $y_1,y_2$ solve a homogeneous linear equation, so does $c_1y_1+c_2y_2$ (the '''superposition principle'''). Nonlinear equations lack this property. Superposition underlies every linear method below.


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


An '''ordinary differential equation''' involves a function of a single independent variable, as in all the examples so far. A '''partial differential equation''' (PDE) involves a function of several independent variables, together with its partial derivatives. For example, the temperature $u(x, t)$ of a metal bar, which depends on the position $x$ along the bar and on the time $t$, satisfies the '''heat equation'''
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 obeys the heat equation


{{#content:Q1590}}
{{#content:Q1590}}


where the constant $\alpha$ measures how quickly heat spreads: a spot that is much warmer than its neighbours (large second derivative $\partial^2 u/\partial x^2$) warms or cools quickly.
where $\alpha$ is the thermal diffusivity. The equation says a spot cools fastest where the temperature profile is most curved ($\partial^2u/\partial x^2$ large).


== Slope fields: visualising solutions ==
== Slope fields ==


A first-order equation solved for its derivative has the general form
A first-order equation can be written


{{#content:Q1581}}
{{#content:Q1581}}


It assigns to every point $(x, y)$ of the plane the slope that any solution curve passing through that point must have there. Drawing a short line segment with exactly that slope at many points produces a '''direction field''' (or slope field) for the equation.
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.


[[File:Slope field of exponential growth.png|thumb|Direction field of $dy/dx = y$. Each short segment shows the slope that a solution must have there, and the drawn curves follow the field. Credit: jjbeard (public domain).]]
[[File:Slope field of exponential growth.png|thumb|Direction field of $dy/dx=y$. Credit: jjbeard (public domain).]]


A solution curve must be tangent to the field everywhere it passes, like a boat pushed by a current whose direction depends on where the boat is. The field therefore displays the whole solution family at a glance.<ref>{{#cite:Q1576}}</ref> Numerical methods, such as [[Euler's method]], work by following such arrows step by step: from a starting point, read the slope of the arrow there, take a small step in that direction, read the new arrow, and repeat.
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 single recipe that solves every differential equation. The practical approach is to recognise which family an equation belongs to, then apply that family's method. The subsections below work through the families that appear most often, each stated in general and then illustrated with numbers; the last subsection collects the routes taken when none of these formulas applies.


=== Separation of variables ===
=== Separation of variables ===


A first-order equation is '''separable''' when the right-hand side splits into a factor that depends only on $x$ and a factor that depends only on $y$:
A first-order equation is '''separable''' when the right-hand side factors into a function of $x$ times a function of $y$:


{{#content:Q1612}}
{{#content:Q1612}}


The arrow shows the whole method: divide both sides by $h(y)$ and integrate, so that all the $y$'s are on one side and all the $x$'s on the other. The two worked examples below carry this out completely, and both lead to the same conclusion: quantities whose rate of change is proportional to their own size change exponentially.
Divide by $h(y)$ and integrate; all $y$'s land on one side, all $x$'s on the other.


==== Worked example: exponential growth and decay ====
==== Exponential growth and decay ====


Many quantities change at a rate proportional to their own size. A population with unlimited food grows faster the larger it is, money in a bank earns interest in proportion to the balance, and the number of radioactive atoms left decreases in proportion to how many remain. Writing the constant of proportionality as $k$,
When a quantity changes at a rate proportional to its own size,


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


where $k$ is a constant. To solve it, divide both sides by $y$ and integrate both sides with respect to $t$:
divide by $y$ and integrate:
 
$$\frac{1}{y}\frac{dy}{dt}=k\qquad\Longrightarrow\qquad\int\frac{dy}{y}=\int k\,dt$$
 
The left-hand integral is $\ln|y|$, so
 
$$\ln|y| = kt + C$$
 
Exponentiating both sides (raising $e$ to the power of each side) removes the logarithm:


$$|y|=e^{kt+C}=e^{C}e^{kt}$$
$$\int\frac{dy}{y}=\int k\,dt\qquad\Longrightarrow\qquad \ln|y|=kt+C$$


The sign of $y$ never changes, so the absolute value can be dropped by absorbing the sign into the constant. Writing $y_{0}$ for the value at $t = 0$, we obtain
Exponentiating, $|y|=e^C e^{kt}$; the sign of $y$ never changes, so absorbing it into the constant and writing $y(0)=y_0$,


{{#content:Q1585}}
{{#content:Q1585}}


With numbers: suppose €1000 is deposited in an account paying 5% per year with interest added continuously, so $k = 0.05$ per year and $y_{0} = 1000$. The balance is $y(t) = 1000\,e^{0.05t}$, and after 10 years
With numbers: €1000 at 5% interest compounded continuously ($k=0.05\ \text{yr}^{-1}$) gives $y(t)=1000\,e^{0.05t}$, and
 
$$y(10)=1000\,e^{0.05\times 10}=1000\,e^{0.5}\approx 1648.7$$
 
so the deposit has grown to about €1649. To find when it doubles, set $y(t) = 2000$ and solve:


$$1000\,e^{0.05t}=2000\qquad\Longrightarrow\qquad e^{0.05t}=2\qquad\Longrightarrow\qquad t=\frac{\ln 2}{0.05}\approx 13.9\ \text{years}$$
$$y(10)=1000\,e^{0.5}\approx 1648.7$$


For $k < 0$ the same formula describes decay rather than growth. Radioactive substances are usually described by their '''half-life''', the time in which half the atoms decay; setting $y(t)=y_{0}/2$ gives $t_{1/2}=(\ln 2)/(-k)$.
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 $t_{1/2}=(\ln 2)/(-k)$.


==== Worked example: Newton's law of cooling ====
==== Newton's law of cooling ====


A hot object left in a cooler room cools at a rate proportional to how much hotter it is than the room. This is Newton's law of cooling,
A body hotter than its surroundings cools at a rate proportional to the temperature gap:


{{#content:Q1586}}
{{#content:Q1586}}


where $T(t)$ is the temperature of the object, $T_{a}$ is the constant room temperature, and the positive constant $k$ measures how easily heat escapes.
Separate and integrate:
 
Separating variables means bringing everything that involves $T$ to the left:
 
$$\frac{1}{T-T_{a}}\frac{dT}{dt}=-k\qquad\Longrightarrow\qquad\int\frac{dT}{T-T_{a}}=\int -k\,dt$$
 
Integrating both sides gives
 
$$\ln|T-T_{a}|=-kt+C$$
 
Exponentiating both sides,
 
$$|T-T_{a}|=e^{-kt+C}=e^{C}e^{-kt}$$
 
The difference $T - T_{a}$ keeps its sign: positive while the object cools, negative while it warms towards a warmer room. Absorbing the sign into the constant and fixing it with the initial temperature $T(0)=T_{0}$, we obtain
 
$$T(t)=T_{a}+\left(T_{0}-T_{a}\right)e^{-kt}$$


So the temperature does not decay exponentially, but the temperature gap $T - T_{a}$ does.
$$\int\frac{dT}{T-T_a}=\int-k\,dt\qquad\Longrightarrow\qquad \ln|T-T_a|=-kt+C$$


With numbers: suppose a drink at $80\,^{\circ}\mathrm{C}$ is left in a room at $20\,^{\circ}\mathrm{C}$, and measurement shows $k = 0.1$ per minute. Then
Exponentiating and folding the (constant-sign) factor $T-T_a$ into the constant, with $T(0)=T_0$:


$$T(t)=20+\left(80-20\right)e^{-0.1t}=20+60\,e^{-0.1t}$$
$$T(t)=T_a+(T_0-T_a)e^{-kt}$$


The gap starts at $60\,^{\circ}\mathrm{C}$ and shrinks by the factor $e^{-0.1t}$ each minute; it halves every $(\ln 2)/0.1 \approx 6.9$ minutes. To find when the drink reaches $40\,^{\circ}\mathrm{C}$, put $T(t)=40$:
The gap $T-T_a$ decays exponentially, not $T$ itself. Example: a drink at $80\,^{\circ}\mathrm{C}$ in a $20\,^{\circ}\mathrm{C}$ room, $k=0.1\ \text{min}^{-1}$:


$$40=20+60\,e^{-0.1t}\qquad\Longrightarrow\qquad e^{-0.1t}=\frac{1}{3}\qquad\Longrightarrow\qquad t=10\ln 3\approx 11\ \text{minutes}$$
$$T(t)=20+60\,e^{-0.1t}$$


In about 11 minutes the drink is two-thirds of the way from $80\,^{\circ}\mathrm{C}$ down to the room's $20\,^{\circ}\mathrm{C}$, and it approaches the room temperature without ever quite reaching it.<ref>{{#cite:Q1576}}</ref>
Reaches $40\,^{\circ}\mathrm{C}$ when $20+60e^{-0.1t}=40$, i.e. $t=10\ln 3\approx 11$ min.<ref>{{#cite:Q1576}}</ref>


=== First-order linear equations: the integrating factor ===
=== First-order linear equations: the integrating factor ===


The two equations just solved are separable as well as linear. Many first-order equations, however, are linear but not separable, and for those there is a general method based on the '''integrating factor'''.
For $y'+p(x)y=q(x)$ that is not separable, multiply by $\mu(x)$ chosen so the left side is a single derivative $(\mu y)'$. The product rule gives $(\mu y)'=\mu y'+\mu' y$, while multiplying the equation by $\mu$ gives $\mu y'+\mu p\,y$; matching coefficients requires $\mu'=p\mu$, whose solution is
 
Recall from the classification section that a linear first-order equation has the form
 
$$\frac{dy}{dx}+p(x)\,y=q(x)$$
 
The idea is to multiply both sides by a function $\mu(x)$, chosen so that the left-hand side becomes the derivative of a single product $\mu(x)\,y$, which can then be integrated in one stroke. Expanding the product rule,
 
$$\frac{d}{dx}\bigl(\mu(x)\,y\bigr)=\mu\,\frac{dy}{dx}+\frac{d\mu}{dx}\,y$$
 
Multiplying the equation by $\mu$ would give $\mu\,\frac{dy}{dx}+\mu p\,y$ on the left. For the two to agree, the coefficient of $y$ must match, so $\mu$ must satisfy $\mu' = p\,\mu$. This is itself a separable equation, solved exactly as in the previous subsection:


$$\frac{d\mu}{dx}=p(x)\,\mu\qquad\Longrightarrow\qquad\frac{d\mu}{\mu}=p(x)\,dx\qquad\Longrightarrow\qquad\ln\mu=\int p(x)\,dx\qquad\Longrightarrow\qquad\mu(x)=e^{\int p(x)\,dx}$$
$$\mu(x)=e^{\int p(x)\,dx}$$


(any constant of integration may be taken as zero, since multiplying $\mu$ by a constant changes nothing). Multiplying the original equation by this $\mu(x)$,
Multiplying the equation by $\mu$,


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{{#content:Q1613}}


and because the left-hand side is now a single derivative, both sides integrate immediately:
so both sides integrate directly:
 
$$\mu(x)\,y=\int \mu(x)\,q(x)\,dx + C$$
 
Worked example. Solve $y' + y = e^{-x}$. Here $p = 1$, so the integrating factor is
 
$$\mu=e^{\int 1\,dx}=e^{x}$$
 
Multiplying both sides by $e^{x}$,


$$e^{x}y'+e^{x}y=e^{x}e^{-x}=1$$
$$\mu(x)\,y=\int\mu(x)\,q(x)\,dx+C$$


The left-hand side is $(e^{x}y)'$, so
Example: $y'+y=e^{-x}$. Here $p=1$, $\mu=e^x$, and


$$\left(e^{x}y\right)'=1\qquad\Longrightarrow\qquad e^{x}y=x+C\qquad\Longrightarrow\qquad y=(x+C)e^{-x}$$
$$e^x y'+e^x y=1\qquad\Longrightarrow\qquad (e^x y)'=1\qquad\Longrightarrow\qquad e^x y=x+C\qquad\Longrightarrow\qquad y=(x+C)e^{-x}$$


{{#content:Q1607}}
{{#content:Q1607}}


An initial condition fixes the constant: if $y(0)=2$, then $2=(0+C)e^{0}$, so $C=2$ and $y=(x+2)e^{-x}$.<ref>{{#cite:Q1576}}</ref><ref>{{#cite:Q1577}}</ref>
The condition $y(0)=2$ gives $C=2$.<ref>{{#cite:Q1576}}</ref><ref>{{#cite:Q1577}}</ref>


=== Constant-coefficient linear equations of order two ===
=== Constant-coefficient linear equations of order two ===


The methods so far solve first-order equations. The most important second-order family is the linear equation with constant coefficients,
The equation $y''+a\,y'+b\,y=0$ (constant coefficients) models a mass on a spring, a small-angle pendulum, and an RLC circuit. It is solved by trying an exponential $y=e^{rx}$, since $y'=re^{rx}$ and $y''=r^2e^{rx}$:
 
$$y''+a\,y'+b\,y=0$$
 
which models systems pulled back towards a resting state: a mass on a spring, a pendulum swinging through small angles, and an electric circuit containing a capacitor and an inductor. The key idea is to try an exponential solution $y=e^{rx}$, because the derivative of an exponential is again a multiple of itself, so substituting turns the differential equation into an ordinary algebraic equation:


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{{#content:Q1644}}


To see where the arrow comes from, substitute $y=e^{rx}$ together with $y'=re^{rx}$ and $y''=r^{2}e^{rx}$ into the equation:
Substitution turns the equation into algebra: $(r^2+ar+b)e^{rx}=0$, and $e^{rx}\neq 0$, so
 
$$r^{2}e^{rx}+a\,r\,e^{rx}+b\,e^{rx}=\left(r^{2}+a\,r+b\right)e^{rx}=0$$
 
Since $e^{rx}$ is never zero, divide both sides by it:


$$r^{2}+a\,r+b=0$$
$$r^2+ar+b=0$$


This is the '''characteristic equation'''. It is an ordinary quadratic equation, and the form of the solution depends on the kind of roots it has:
This is the '''characteristic equation'''; its roots determine the solution:


* '''Two distinct real roots''', $r_{1} \neq r_{2}$: the general solution is $y=C_{1}e^{r_{1}x}+C_{2}e^{r_{2}x}$, a sum of two exponentials.
* distinct real roots $r_1\neq r_2$: $y=C_1e^{r_1x}+C_2e^{r_2x}$;
* '''One repeated real root''', $r$: the general solution is $y=\left(C_{1}+C_{2}x\right)e^{rx}$; the extra factor $x$ supplies the second arbitrary constant.
* one repeated root $r$: $y=(C_1+C_2x)e^{rx}$;
* '''A complex conjugate pair''', $r=\alpha \pm i\beta$: the general solution is $y=e^{\alpha x}\left(C_{1}\cos\beta x + C_{2}\sin\beta x\right)$, which oscillates, growing or fading in size according to the sign of $\alpha$.
* complex pair $r=\alpha\pm i\beta$: $y=e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)$.


A quick numerical example of the first case: solve $y''-3y'+2y=0$. Substituting $y=e^{rx}$ gives $r^{2}-3r+2=0=(r-1)(r-2)$, so the roots are $1$ and $2$, and
Example ($y''-3y'+2y=0$): $r^2-3r+2=(r-1)(r-2)$, so


{{#content:Q1608}}
{{#content:Q1608}}


Each term works, as can be checked directly: for $y=e^{x}$, one has $y''-3y'+2y=(1-3+2)e^{x}=0$, and similarly for $e^{2x}$.
Each term checks: for $y=e^x$, $y''-3y'+2y=(1-3+2)e^x=0$.


==== Example: the harmonic oscillator (a mass on a spring) ====
==== The harmonic oscillator (a mass on a spring) ====


Some quantities do not settle towards a level but swing back and forth. The complex-root case above is not an exotic exception: it is exactly what happens for the most common oscillating system of all. Consider a mass $m$ attached to a spring. If the spring is displaced a distance $x$ from its rest position, it pulls back with a force $-kx$ proportional to the displacement (Hooke's law), where the spring constant $k$ measures how stiff the spring is. Newton's second law therefore gives
A mass displaced $x$ from rest is pulled back by $-kx$ (Hooke's law), so Newton's second law gives


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


Writing $\omega_{0}^{2} = k/m$, this becomes the '''harmonic oscillator equation''':
With $\omega_0^2=k/m$ this is the '''harmonic oscillator equation'''


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


Its characteristic equation is $r^{2}+\omega_{0}^{2}=0$, with the purely imaginary roots $r=\pm i\omega_{0}$: this is the complex-conjugate case above with $\alpha=0$ and $\beta=\omega_{0}$. The equation asks for a function whose second derivative is a negative constant multiple of itself, and the sine and cosine have exactly this property. Differentiating $\cos(\omega_{0}t)$ twice brings out two factors of $\omega_{0}$ and a minus sign:
Its characteristic equation $r^2+\omega_0^2=0$ has roots $\pm i\omega_0$, the complex case above ($\alpha=0$). Since $\frac{d^2}{dt^2}\cos\omega_0 t=-\omega_0^2\cos\omega_0 t$, and likewise for sine, superposition gives
 
$$\frac{d}{dt}\cos(\omega_{0}t)=-\omega_{0}\sin(\omega_{0}t),\qquad \frac{d^{2}}{dt^{2}}\cos(\omega_{0}t)=-\omega_{0}^{2}\cos(\omega_{0}t)$$
 
so $x = \cos(\omega_{0}t)$ solves the equation, and so does $x = \sin(\omega_{0}t)$. Because the equation is linear and homogeneous, the superposition principle applies and the general solution is


$$x(t)=A\cos(\omega_{0}t)+B\sin(\omega_{0}t)$$
$$x(t)=A\cos\omega_0 t+B\sin\omega_0 t$$


The two constants $A$ and $B$ are fixed by the initial displacement and the initial velocity, exactly as the order of the equation requires.
with $A,B$ fixed by the initial position and velocity.


[[File:Simple harmonic motion animation.gif|thumb|A mass on a spring executes simple harmonic motion: the solution of the harmonic oscillator equation is a sinusoid of fixed amplitude and frequency. 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).]]


A concrete example. Take a mass of $2\ \mathrm{kg}$ on a spring with $k = 8\ \mathrm{N/m}$, so that $\omega_{0} = \sqrt{k/m} = \sqrt{4} = 2$ radians per second. Pull the mass $10\ \mathrm{cm}$ out and release it from rest: the initial velocity is zero, so $B = 0$, and $x(t) = 0.10\cos(2t)$ metres. The motion repeats after one period
Example: $m=2\ \mathrm{kg}$, $k=8\ \mathrm{N/m}$, 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


$$P=\frac{2\pi}{\omega_{0}}=\pi\ \text{seconds}\approx 3.14\ \text{s}$$
$$P=\frac{2\pi}{\omega_0}=\pi\approx 3.14\ \text{s}$$


so the mass returns to its starting point roughly every 3.14 seconds. One second after release, measuring angles in radians,
and after one second


$$x(1)=0.10\cos(2)\approx 0.10\times(-0.416)\approx -0.042\ \text{m}$$
$$x(1)=0.10\cos 2\approx 0.10(-0.416)\approx -0.042\ \text{m}$$


about 4 cm on the other side of the rest position. This kind of motion, a sinusoid of fixed amplitude, is called '''simple harmonic motion''', and the oscillator equation governs not only springs but pendulums (for small swings), electric circuits, and the vibrations of molecules.<ref>{{#cite:Q1577}}</ref>
Such fixed-amplitude sinusoidal motion is '''simple harmonic motion'''.<ref>{{#cite:Q1577}}</ref>


=== Partial differential equations: separating variables in the heat equation ===
=== Partial differential equations: separating variables in the heat equation ===


Exact formulas for partial differential equations exist only for a few simple, usually linear, problems, but these are precisely the problems at the heart of physics and engineering. The standard technique is again separation of variables, but now the variables to be separated are the independent variables themselves, here the position $x$ and the time $t$.
Solve the heat equation on a bar of length $L$, insulated sides, ends held at $0$:


Return to the metal bar of the classification section, this time of length $L$ with its sides insulated, so that heat flows only along the bar, and with both ends held at temperature $0$. Its temperature $u(x,t)$ satisfies the heat equation, reproduced here:
$$\frac{\partial u}{\partial t}=\alpha\frac{\partial^2u}{\partial x^2}$$


$$\frac{\partial u}{\partial t}=\alpha\,\frac{\partial^{2}u}{\partial x^{2}}$$
Seek a product solution $u(x,t)=X(x)T(t)$. Substitution gives $XT'=\alpha X''T$; dividing by $\alpha XT$,


Separation of variables looks for a solution that splits into a product of a function of $x$ alone and a function of $t$ alone:
{{#content:Q1622}}
 
$$u(x,t)=X(x)\,T(t)$$
 
Substituting into the heat equation gives $X\,T'=\alpha\,X''\,T$. Divide both sides by $\alpha X T$:


{{#content:Q1622}}
The left side depends only on $t$, the right only on $x$, so both equal one constant, $-\lambda$. This yields two ODEs,


The left-hand side now depends only on $t$ and the right-hand side only on $x$. Since $x$ and $t$ can vary independently, the only way two functions of different variables can be equal for all $x$ and $t$ is that both sides equal the same constant, written above as $-\lambda$. This splits the partial differential equation into two ordinary ones:
$$T'=-\alpha\lambda T\qquad\Longrightarrow\qquad T=e^{-\alpha\lambda t}$$


* $T'=-\alpha\lambda\,T$: exponential decay, the growth equation of this article with $k=-\alpha\lambda<0$, giving $T=e^{-\alpha\lambda t}$;
$$X''=-\lambda X\qquad\Longrightarrow\qquad X=A\cos(\sqrt{\lambda}\,x)+B\sin(\sqrt{\lambda}\,x)$$
* $X''=-\lambda X$: the function whose second derivative is a negative constant multiple of itself, giving sines and cosines of $\sqrt{\lambda}\,x$.


The ends of the bar are held at temperature $0$, so $u(0,t)=u(L,t)=0$, which forces $X(0)=X(L)=0$. The condition $X(0)=0$ removes the cosine, and $X(L)=0$ forces $\sin(\sqrt{\lambda}\,L)=0$, so $\sqrt{\lambda}\,L=n\pi$ for a positive whole number $n$. Thus only the special values $\lambda=(n\pi/L)^{2}$ are allowed, each giving one mode
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 $\lambda=(n\pi/L)^2$ gives one mode


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


a half-wave, or several half-waves, of a sine curve that sinks towards zero as heat leaks out of the ends. Because the heat equation is linear and homogeneous, superposition applies: any sum of modes is again a solution, and the general solution is
The equation is linear and homogeneous, so superposition applies, and the general solution is


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


The numbers $b_{n}$ are fixed by the initial temperature profile $u(x,0)$ at $t=0$; writing an arbitrary initial profile as such a sum of sine waves is exactly what Fourier's sine series does.
with the $b_n$ fixed by the initial profile $u(x,0)$ (a Fourier sine series). Modes with many wiggles (large $n$) decay fastest, since the decay rate $\alpha(n\pi/L)^2$ grows like $n^2$; soon only the $n=1$ mode remains.
 
Two features are worth noticing. First, every mode decays exponentially, so the bar does not oscillate: heat spreads and the temperature evens out. Second, modes with more wiggles decay faster, because their rate $-\alpha(n\pi/L)^{2}$ is larger in magnitude, so after a short time only the $n=1$ mode, one smooth half-sine across the whole bar, is left.
 
A concrete check with numbers: take a one-metre iron bar (iron's thermal diffusivity is $\alpha \approx 2.3\times10^{-5}\ \mathrm{m^{2}\,s^{-1}}$) whose initial temperature is $u(x,0)=100\sin(\pi x/L)$, that is, $0\,^{\circ}\mathrm{C}$ at the ends and $100\,^{\circ}\mathrm{C}$ in the middle. Only the $n=1$ mode is present, so


$$u(x,t)=100\,\sin\left(\frac{\pi x}{L}\right)e^{-\alpha\pi^{2}t/L^{2}}$$
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 $n=1$ is present:


With $L=1\ \mathrm{m}$, the decay rate is $\alpha\pi^{2}\approx 2.3\times10^{-5}\times9.87\approx 2.3\times10^{-4}$ per second. At the centre of the bar, where the sine equals $1$,
$$u(x,t)=100\sin\frac{\pi x}{L}\,e^{-\alpha\pi^2t/L^2}$$


$$u\!\left(\tfrac{1}{2},t\right)=100\,e^{-2.3\times10^{-4}\,t}$$
At the centre, with $L=1$ and $\alpha\pi^2\approx 2.3\times10^{-4}\ \text{s}^{-1}$,


After one hour ($t=3600\ \mathrm{s}$) the exponent is $2.3\times10^{-4}\times3600\approx0.82$, so the centre has cooled to $100\,e^{-0.82}\approx44\,^{\circ}\mathrm{C}$, and it reaches $50\,^{\circ}\mathrm{C}$ after $t=(\ln 2)/(2.3\times10^{-4})\approx 3050\ \mathrm{s}$, about 51 minutes. The heat has flowed out through the ends, exactly the phenomenon Fourier set out to describe.<ref>{{#cite:Q1579}}</ref>
$$u\!\left(\tfrac12,t\right)=100\,e^{-2.3\times10^{-4}t}$$


=== When no formula is possible: numerical, series, and qualitative methods ===
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>


The families above, separable and linear with convenient coefficients, cover a great many practical equations, but they are a small minority of all differential equations. There is no guarantee that a given equation, especially a nonlinear one, has a solution expressible in terms of the familiar functions at all. Such equations are studied in one of three ways:
=== When no formula exists ===


* '''Numerical approximation''', when numbers are what is needed: [[Euler's method]] steps forward along the slope field, replacing the exact solution by a polygonal path, and its refinements underlie most computer simulations;
Most equations, especially nonlinear ones, have no solution in terms of familiar functions. They are studied in one of three ways:<ref>{{#cite:Q1576}}</ref><ref>{{#cite:Q1578}}</ref>
* '''[[Qualitative methods]]''', when behaviour matters more than numbers: the solutions are studied through the slope field, equilibria, stability, and long-term behaviour, without ever writing a formula;
* '''Series and integral transforms''', for special but important cases: when an equation is linear, a solution can sometimes be written as an infinite series ([[Power series]]) or recovered from an algebraic equation by the [[Laplace transform]].


Choosing between these routes is part of the art of applying mathematics, and the diagram below summarises the decision process of this whole article.<ref>{{#cite:Q1576}}</ref><ref>{{#cite:Q1578}}</ref>
* '''Numerically''', when numbers suffice: [[Euler's method]] steps along the slope field;
* '''[[Qualitative methods]]''': equilibria, stability and long-term behaviour, without formulas;
* '''Series and transforms''', for linear cases: [[Power series]] or the [[Laplace transform]].


<uml type="uml">
<uml type="uml">
Line 355: Line 242:
start
start
:You have a differential equation;
:You have a differential equation;
if (Is it first order and separable?\ny' = g(x) h(y)?) then (yes)
if (First order and separable?\ny' = g(x) h(y)?) then (yes)
   :Separate and integrate:\n∫ dy/h(y) = ∫ g(x) dx;
   :Separate and integrate:\n∫ dy/h(y) = ∫ g(x) dx;
else (no)
else (no)
   if (Is it first order and linear?\ny' + p(x) y = q(x)?) then (yes)
   if (First order and linear?\ny' + p(x) y = q(x)?) then (yes)
     :Multiply by the integrating factor\nμ = e^{∫ p dx}, then integrate;
     :Integrating factor\nμ = e^{∫ p dx};
   else (no)
   else (no)
     if (Is it second order, linear, with constant coefficients?\ny'' + a y' + b y = 0?) then (yes)
     if (Second order, linear, constant coefficients?\ny'' + a y' + b y = 0?) then (yes)
       :Solve the characteristic equation\nr² + a r + b = 0;
       :Characteristic equation\nr² + a r + b = 0;
     else (no)
     else (no)
       if (Is it a linear PDE on a simple\nshape, e.g. the heat equation?) then (yes)
       if (Linear PDE on a simple shape,\ne.g. the heat equation?) then (yes)
         :Separate the variables:\nu(x,t) = X(x) T(t);
         :Separate variables\nu(x,t) = X(x) T(t);
       else (no)
       else (no)
         if (Are approximate numbers enough?) then (yes)
         if (Are approximate numbers enough?) then (yes)
           :Step forward numerically\n(article: Euler's method);
           :Numerical stepping\n(Euler's method);
         else (no)
         else (no)
           if (Is the equation linear?) then (yes)
           if (Linear?) then (yes)
             :Power series or Laplace transform\n(articles: Power series, Laplace transform);
             :Power series or Laplace transform;
           else (no)
           else (no)
             :Study behaviour without formulas:\nequilibria, stability, chaos\n(article: Qualitative methods);
             :Qualitative study:\nequilibria, stability, chaos;
           endif
           endif
         endif
         endif
Line 383: Line 270:
@enduml
@enduml
</uml>
</uml>
As the diagram shows, the exact methods occupy only the first few branches; most equations encountered in research are handled by the routes at the bottom of the diagram, each of which has its own article.


== 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>
Differential equations came with the calculus. Newton's laws in the ''Principia'' (1687) are differential equations; Leibniz's notation $dy/dx$ (1670s) is the one still used. In the mid-18th century [[Person:Leonhard Euler|Leonhard Euler]] made the subject systematic, contributing the exponential trial solution, series methods, and the first numerical scheme, Euler's method.<ref>{{#cite:Q1577}}</ref> Physics supplied the PDEs: [[Person:Jean le Rond d'Alembert|Jean le Rond d'Alembert]] solved the vibrating-string (wave) equation in 1747, and [[Person:Joseph Fourier|Joseph Fourier]] derived and solved the heat equation in 1822 by expanding initial data in sine series, founding Fourier analysis.<ref>{{#cite:Q1579}}</ref> When no formula exists, behaviour can still be studied: [[Person:Henri Poincaré|Henri Poincaré]] pioneered this qualitative view on the three-body problem, and in 1963 [[Person:Edward Lorenz|Edward Lorenz]] found chaos in a three-equation model of convection, ending hopes of long-term weather prediction.<ref>{{#cite:Q1578}}</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).]]
 
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>
 
[[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 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>
[[File:Isaac Newton portrait.jpg|thumb|left|Isaac Newton (after Godfrey Kneller, 1689). Credit: James Thronill (public domain).]]
[[File:Leonhard Euler portrait.jpg|thumb|Leonhard Euler (by Jakob Emanuel Handmann, 1753). Credit: Jakob Emanuel Handmann (public domain).]]


== References ==
== References ==

Revision as of 18:55, 5 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 $y(x)$. Laws of nature state how quantities change, so differential equations describe pendulums, cooling drinks, growing populations, and discharging capacitors. This article covers the standard exact solution methods, each with a worked 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 first example: slopes and a family of solutions

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

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

Integration inverts differentiation, so integrating both sides gives

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

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.

An extra condition picks out one member. If $y(0)=3$, then

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

Such a prescribed value is an initial condition.

Classifying differential equations

Three features decide how to solve an equation: its order, its linearity, and how many independent variables it involves.

Order

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

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

is second order ($x(t)$ position of mass $m$, $F$ net force). An nth-order equation has $n$ arbitrary constants in its general solution, fixed by $n$ conditions, one constant appearing at each integration. Free fall shows the pattern: with only gravity $F_g=-mg$, Newton's law gives

$$m\frac{d^{2}x}{dt^{2}}=-mg\qquad\Longrightarrow\qquad\frac{d^{2}x}{dt^{2}}=-g\qquad(g\approx 9.8\ \mathrm{m\,s^{-2}})$$

Integrate once (constant $v_0$, the speed at $t=0$), then again (constant $x_0$, the height at $t=0$):

$$\frac{dx}{dt}=-gt+v_{0}\qquad\Longrightarrow\qquad x(t)=-\frac{g}{2}t^{2}+v_{0}t+x_{0}$$

Two initial conditions are needed. A ball dropped from rest at height $19.6\ \mathrm{m}$ hits the ground ($x=0$) when

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

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

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

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 $\theta$).

If $y_1,y_2$ solve a homogeneous linear equation, so does $c_1y_1+c_2y_2$ (the superposition principle). Nonlinear equations lack this property. Superposition underlies every linear method below.

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 obeys the heat equation

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

where $\alpha$ is the thermal diffusivity. The equation says a spot cools fastest where the temperature profile is most curved ($\partial^2u/\partial x^2$ large).

Slope fields

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 run tangent to it.

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.[1]

Solving differential equations

Separation of variables

A first-order equation is separable when the right-hand side factors into a function of $x$ times a function of $y$:

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

Divide by $h(y)$ and integrate; all $y$'s land on one side, all $x$'s on the other.

Exponential growth and decay

When a quantity changes at a rate proportional to its own size,

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

divide by $y$ and integrate:

$$\int\frac{dy}{y}=\int k\,dt\qquad\Longrightarrow\qquad \ln|y|=kt+C$$

Exponentiating, $|y|=e^C e^{kt}$; the sign of $y$ never changes, so absorbing it into the constant and writing $y(0)=y_0$,

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

With numbers: €1000 at 5% interest compounded continuously ($k=0.05\ \text{yr}^{-1}$) gives $y(t)=1000\,e^{0.05t}$, and

$$y(10)=1000\,e^{0.5}\approx 1648.7$$

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 $t_{1/2}=(\ln 2)/(-k)$.

Newton's law of cooling

A body hotter than its surroundings cools at a rate proportional to the temperature gap:

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

Separate and integrate:

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

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

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

The gap $T-T_a$ decays exponentially, not $T$ itself. 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.[1]

First-order linear equations: the integrating factor

For $y'+p(x)y=q(x)$ that is not separable, multiply by $\mu(x)$ chosen so the left side is a single derivative $(\mu y)'$. The product rule gives $(\mu y)'=\mu y'+\mu' y$, while multiplying the equation by $\mu$ gives $\mu y'+\mu p\,y$; matching coefficients requires $\mu'=p\mu$, whose solution is

$$\mu(x)=e^{\int p(x)\,dx}$$

Multiplying the equation by $\mu$,

\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)

so both sides integrate directly:

$$\mu(x)\,y=\int\mu(x)\,q(x)\,dx+C$$

Example: $y'+y=e^{-x}$. Here $p=1$, $\mu=e^x$, and

$$e^x y'+e^x y=1\qquad\Longrightarrow\qquad (e^x y)'=1\qquad\Longrightarrow\qquad e^x y=x+C\qquad\Longrightarrow\qquad y=(x+C)e^{-x}$$

\frac{dy}{dx}+y=e^{-x}\qquad\Longrightarrow\qquad y=(x+C)e^{-x}

The condition $y(0)=2$ gives $C=2$.[1][2]

Constant-coefficient linear equations of order two

The equation $y''+a\,y'+b\,y=0$ (constant coefficients) models a mass on a spring, a small-angle pendulum, and an RLC circuit. It is solved by trying an exponential $y=e^{rx}$, since $y'=re^{rx}$ and $y''=r^2e^{rx}$:

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

Substitution turns the equation into algebra: $(r^2+ar+b)e^{rx}=0$, and $e^{rx}\neq 0$, so

$$r^2+ar+b=0$$

This is the characteristic equation; its roots determine the solution:

  • distinct real roots $r_1\neq r_2$: $y=C_1e^{r_1x}+C_2e^{r_2x}$;
  • one repeated root $r$: $y=(C_1+C_2x)e^{rx}$;
  • complex pair $r=\alpha\pm i\beta$: $y=e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)$.

Example ($y''-3y'+2y=0$): $r^2-3r+2=(r-1)(r-2)$, so

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

Each term checks: for $y=e^x$, $y''-3y'+2y=(1-3+2)e^x=0$.

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

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

With $\omega_0^2=k/m$ this is the harmonic oscillator equation

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

Its characteristic equation $r^2+\omega_0^2=0$ has roots $\pm i\omega_0$, the complex case above ($\alpha=0$). Since $\frac{d^2}{dt^2}\cos\omega_0 t=-\omega_0^2\cos\omega_0 t$, and likewise for sine, superposition gives

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

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).

Example: $m=2\ \mathrm{kg}$, $k=8\ \mathrm{N/m}$, 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

$$P=\frac{2\pi}{\omega_0}=\pi\approx 3.14\ \text{s}$$

and after one second

$$x(1)=0.10\cos 2\approx 0.10(-0.416)\approx -0.042\ \text{m}$$

Such fixed-amplitude sinusoidal motion is simple harmonic motion.[2]

Partial differential equations: separating variables in the heat equation

Solve the heat equation on a bar of length $L$, insulated sides, ends held at $0$:

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

Seek a product solution $u(x,t)=X(x)T(t)$. Substitution gives $XT'=\alpha X''T$; dividing by $\alpha XT$,

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

The left side depends only on $t$, the right only on $x$, so both equal one constant, $-\lambda$. This yields two ODEs,

$$T'=-\alpha\lambda T\qquad\Longrightarrow\qquad T=e^{-\alpha\lambda t}$$

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

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 $\lambda=(n\pi/L)^2$ gives one mode

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

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

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 the $b_n$ fixed by the initial profile $u(x,0)$ (a Fourier sine series). Modes with many wiggles (large $n$) decay fastest, since the decay rate $\alpha(n\pi/L)^2$ grows like $n^2$; soon only the $n=1$ mode remains.

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 $n=1$ is present:

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

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.[3]

When no formula exists

Most equations, especially nonlinear ones, have no solution in terms of familiar functions. They are studied in one of three ways:[1][4]

A short history

Differential equations came with the calculus. Newton's laws in the Principia (1687) are differential equations; Leibniz's notation $dy/dx$ (1670s) is the one still used. In the mid-18th century Leonhard Euler made the subject systematic, contributing the exponential trial solution, series methods, and the first numerical scheme, Euler's method.[2] Physics supplied the PDEs: Jean le Rond d'Alembert solved the vibrating-string (wave) equation in 1747, and Joseph Fourier derived and solved the heat equation in 1822 by expanding initial data in sine series, founding Fourier analysis.[3] When no formula exists, behaviour can still be studied: Henri Poincaré pioneered this qualitative view on the three-body problem, and in 1963 Edward Lorenz found chaos in a three-equation model of convection, ending hopes of long-term weather prediction.[4]

Isaac Newton (after Godfrey Kneller, 1689). Credit: James Thronill (public domain).
Leonhard Euler (by Jakob Emanuel Handmann, 1753). Credit: Jakob Emanuel Handmann (public domain).

References

  1. ↑ ↑ ↑ ↑ Boyce, W. E. (2012). Elementary Differential Equations and Boundary Value Problems (Book). In Elementary Differential Equations and Boundary Value Problems (Book). John Wiley & Sons.
  2. ↑ ↑ ↑ Tenenbaum, M. (1985). Ordinary Differential Equations (Book). In Ordinary Differential Equations (Book). Dover Publications.
  3. ↑ ↑ Strauss, W. A. (2008). Partial Differential Equations: An Introduction (Book). In Partial Differential Equations: An Introduction (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