Differential equation: Difference between revisions

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AI-assisted (RonzzWikiCowriter): restore the full history section and strip the narrator prose that restates what the demonstrations show; derivations and worked examples carry the explanation. (via update-page on MediaWiki MCP Server)
AI-assisted (RonzzWikiCowriter): reorganise the analytical-methods section by class — first-order ODE methods (separable, integrating factor), second-order ODE methods (direct integration, constant coefficients), PDE methods (separation of variables) — each with general case then concrete numeric example; add a falling-with-air-resistance example under the integrating factor. (via update-page on MediaWiki MCP Server)
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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 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 first example: slopes and a family of solutions ==
== A first example: slopes and a family of solutions ==
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{{#content:Q1583}}
{{#content:Q1583}}


is second order ($x(t)$ position of mass $m$, $F$ net force). Each integration introduces one arbitrary constant, so the general solution of an nth-order equation carries $n$ of them, fixed by $n$ initial conditions. For $n=2$, free fall with only gravity $F_g=-mg$ gives, by Newton's second law,
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 $n=2$.
 
$$m\frac{d^{2}x}{dt^{2}}=-mg\qquad\Longrightarrow\qquad\frac{d^{2}x}{dt^{2}}=-g\qquad(g\approx 9.8\ \mathrm{m\,s^{-2}})$$
 
Integrating once introduces $v_0$, the speed at $t=0$; a second integration introduces $x_0$, the height at $t=0$:
 
$$\frac{dx}{dt}=-gt+v_{0}\qquad\Longrightarrow\qquad x(t)=-\frac{g}{2}t^{2}+v_{0}t+x_{0}$$
 
A ball dropped from rest ($v_0=0$) at height $19.6\ \mathrm{m}$ reaches the ground, $x=0$, when
 
$$0=19.6-4.9\,t^{2}\qquad\Longrightarrow\qquad t=\sqrt{19.6/4.9}=2\ \text{s}$$


=== Linearity and homogeneity ===
=== Linearity and homogeneity ===
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== Solving differential equations ==
== Solving differential equations ==


=== Separation of variables ===
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; for second-order ODEs there are two more, equations reducible to direct integration and linear equations with constant coefficients; for linear PDEs on simple domains the standard tool is separation of variables. Each method below 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.


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


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


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


==== Exponential growth and decay ====
'''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 $g(t)=k$, $h(y)=y$:
 
When a quantity changes at a rate proportional to its own size,


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


divide by $y$ and integrate:
Divide by $y$ and integrate:


$$\int\frac{dy}{y}=\int k\,dt\qquad\Longrightarrow\qquad \ln|y|=kt+C$$
$$\int\frac{dy}{y}=\int k\,dt\qquad\Longrightarrow\qquad \ln|y|=kt+C$$
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{{#content:Q1585}}
{{#content:Q1585}}


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


$$y(10)=1000\,e^{0.5}\approx 1648.7$$
$$y(10)=1000\,e^{0.5}\approx 1648.7$$
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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 $t_{1/2}=(\ln 2)/(-k)$.
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 ====
'''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:
 
A body hotter than its surroundings cools at a rate proportional to the temperature gap:


{{#content:Q1586}}
{{#content:Q1586}}
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Reaches $40\,^{\circ}\mathrm{C}$ when $20+60e^{-0.1t}=40$, i.e. $t=10\ln 3\approx 11$ min.<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 ===
==== 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


If $y'+p(x)y=q(x)$ is not separable, multiply by $\mu(x)$ chosen so the left side is a single product 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
$$\frac{dy}{dx}+p(x)\,y=q(x)$$
 
and is solved by multiplying both sides by the '''integrating factor'''


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


Multiplying the equation by $\mu$,
so that the left-hand side collapses into a single derivative:


{{#content:Q1613}}
{{#content:Q1613}}


both sides integrate directly:
Both sides then integrate directly, giving the general solution


$$\mu(x)\,y=\int\mu(x)\,q(x)\,dx+C$$
$$y(x)=\frac{1}{\mu(x)}\left(\int \mu(x)\,q(x)\,dx + C\right)$$


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


$$e^x y'+e^x y=1\qquad\Longrightarrow\qquad (e^x y)'=1\qquad\Longrightarrow\qquad e^x y=x+C\qquad\Longrightarrow\qquad y=(x+C)e^{-x}$$
$$e^x y'+e^x y=1\qquad\Longrightarrow\qquad (e^x y)'=1\qquad\Longrightarrow\qquad e^x y=x+C\qquad\Longrightarrow\qquad y=(x+C)e^{-x}$$
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The condition $y(0)=2$ gives $C=2$.<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 ===
'''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
 
$$m\frac{dv}{dt}=mg-bv\qquad\Longrightarrow\qquad \frac{dv}{dt}+\frac{b}{m}\,v=g$$
 
Here $p=b/m$ and $q=g$, both constant, so $\mu=e^{(b/m)t}$ and
 
$$\frac{d}{dt}\left(e^{(b/m)t}v\right)=g\,e^{(b/m)t}\qquad\Longrightarrow\qquad e^{(b/m)t}v=\frac{mg}{b}e^{(b/m)t}+C$$
 
hence, with $v(0)=0$,
 
$$v(t)=\frac{mg}{b}\left(1-e^{-(b/m)t}\right)$$
 
As $t$ grows the exponential fades and the speed approaches the constant terminal velocity $mg/b$. 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>
 
=== Second-order ODEs ===
 
==== 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:
 
$$y''=f(x)\qquad\Longrightarrow\qquad y'=\int f(x)\,dx+C_1\qquad\Longrightarrow\qquad y=\int\!\!\left(\int f(x)\,dx\right)dx+C_1 x+C_2$$
 
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.
 
'''Example: free fall.''' With only gravity acting, the height $x(t)$ of a falling object obeys $x''=-g$. Integrating twice,
 
$$\frac{dx}{dt}=-gt+v_0\qquad\Longrightarrow\qquad x(t)=-\frac{g}{2}t^2+v_0 t+x_0$$
 
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
 
$$0=19.6-4.9\,t^{2}\qquad\Longrightarrow\qquad t=\sqrt{19.6/4.9}=2\ \text{s}$$
 
The two initial conditions have fixed the whole trajectory.
 
==== Method 2: linear equations with constant coefficients ====
 
'''General case.''' The equation
 
$$y''+a\,y'+b\,y=0$$


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 $y''=r^2e^{rx}$:
models a mass on a spring, a small-angle pendulum, and an RLC circuit. Its solutions are exponentials: try $y=e^{rx}$. Since $y'=re^{rx}$ and $y''=r^2e^{rx}$, substitution gives


{{#content:Q1644}}
{{#content:Q1644}}


Substituting gives $(r^2+ar+b)e^{rx}=0$, and $e^{rx}\neq 0$, so
The factor $e^{rx}$ is never zero, so the exponential solves the equation exactly when $r$ solves the algebraic '''characteristic equation'''


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


This is the '''characteristic equation'''; its roots determine the solution:
whose roots determine the general solution:


* distinct real roots $r_1\neq r_2$: $y=C_1e^{r_1x}+C_2e^{r_2x}$;
* distinct real roots $r_1\neq r_2$: $y=C_1e^{r_1x}+C_2e^{r_2x}$;
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* complex pair $r=\alpha\pm i\beta$: $y=e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)$.
* 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
If the equation has a nonzero right-hand side, $y''+ay'+by=f(x)$, the superposition principle gives $y=y_p+y_h$, where $y_h$ is the general homogeneous solution above and $y_p$ is any single solution of the full equation.
 
'''Worked demonstration:''' 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


{{#content:Q1608}}
{{#content:Q1608}}
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Check: for $y=e^x$, $y''-3y'+2y=(1-3+2)e^x=0$.
Check: for $y=e^x$, $y''-3y'+2y=(1-3+2)e^x=0$.


==== The harmonic oscillator (a mass on a spring) ====
'''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
 
A mass displaced $x$ from rest is pulled back by $-kx$ (Hooke's law), so Newton's second law gives


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


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


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


Its characteristic equation $r^2+\omega_0^2=0$ has roots $\pm i\omega_0$, the complex case above 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
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


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


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
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


$$P=\frac{2\pi}{\omega_0}=\pi\approx 3.14\ \text{s}$$
$$P=\frac{2\pi}{\omega_0}=\pi\approx 3.14\ \text{s}$$
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Such fixed-amplitude sinusoidal motion is '''simple harmonic motion'''.<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 ===


Solve the heat equation on a bar of length $L$, insulated sides, ends held at $0$:
==== Method 1: separation of variables (the heat equation) ====
 
'''General case.''' For a linear 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.
 
'''Application.''' Solve the heat equation on a bar of length $L$ with insulated sides and both ends held at $0$:


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


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


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


The left side depends only on $t$, the right only on $x$, so both equal one constant, $-\lambda$:
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:


$$T'=-\alpha\lambda T\qquad\Longrightarrow\qquad T=e^{-\alpha\lambda t}$$
$$T'=-\alpha\lambda T\qquad\Longrightarrow\qquad T=e^{-\alpha\lambda t}$$
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$$X''=-\lambda X\qquad\Longrightarrow\qquad X=A\cos(\sqrt{\lambda}\,x)+B\sin(\sqrt{\lambda}\,x)$$
$$X''=-\lambda X\qquad\Longrightarrow\qquad X=A\cos(\sqrt{\lambda}\,x)+B\sin(\sqrt{\lambda}\,x)$$


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
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


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


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


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


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:
'''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:


$$u(x,t)=100\sin\frac{\pi x}{L}\,e^{-\alpha\pi^2t/L^2}$$
$$u(x,t)=100\sin\frac{\pi x}{L}\,e^{-\alpha\pi^2t/L^2}$$
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=== When no formula exists ===
=== 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:<ref>{{#cite:Q1576}}</ref><ref>{{#cite:Q1578}}</ref>
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:<ref>{{#cite:Q1576}}</ref><ref>{{#cite:Q1578}}</ref>


* '''Numerically''', when numbers suffice: [[Euler's method]] steps along the slope field;
* '''Numerically''', when numbers suffice: [[Euler's method]] steps along the slope field;
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     :Integrating factor\nμ = e^{∫ p dx};
     :Integrating factor\nμ = e^{∫ p dx};
   else (no)
   else (no)
     if (Second order, linear, constant coefficients?\ny'' + a y' + b y = 0?) then (yes)
     if (Second order, of the form\ny'' = f(x)?) then (yes)
       :Characteristic equation\nr² + a r + b = 0;
       :Integrate twice;
     else (no)
     else (no)
       if (Linear PDE on a simple shape,\ne.g. the heat equation?) then (yes)
       if (Second order, linear, constant coefficients?\ny'' + a y' + b y = 0?) then (yes)
         :Separate variables\nu(x,t) = X(x) T(t);
         :Characteristic equation\nr² + a r + b = 0;
       else (no)
       else (no)
         if (Are approximate numbers enough?) then (yes)
         if (Linear PDE on a simple shape,\ne.g. the heat equation?) then (yes)
           :Numerical stepping\n(Euler's method);
           :Separate variables\nu(x,t) = X(x) T(t);
         else (no)
         else (no)
           if (Linear?) then (yes)
           if (Are approximate numbers enough?) then (yes)
             :Power series or Laplace transform;
             :Numerical stepping\n(Euler's method);
           else (no)
           else (no)
             :Qualitative study:\nequilibria, stability, chaos;
             if (Linear?) then (yes)
              :Power series or Laplace transform;
            else (no)
              :Qualitative study:\nequilibria, stability, chaos;
            endif
           endif
           endif
         endif
         endif
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@enduml
@enduml
</uml>
</uml>
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 ==
== A short history ==

Revision as of 19:13, 5 September 2026

Languages: English · français · Esperanto

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 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.

If $y(0)=3$, then

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

A prescribed value such as this is an initial condition.

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). 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 $n=2$.

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): 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

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

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. For example, the temperature $u(x,t)$ of an insulated metal bar, which depends on position $x$ and time $t$, obeys the heat equation

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

where $\alpha$ is the thermal diffusivity.

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

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; for second-order ODEs there are two more, equations reducible to direct integration and linear equations with constant coefficients; for linear PDEs on simple domains the standard tool is separation of variables. Each method below 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.

First-order ODEs

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:

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

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.

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 $g(t)=k$, $h(y)=y$:

\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}

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

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:

\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; $T$ 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.[1]

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

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

and is solved by multiplying both sides by the integrating factor

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

so that the left-hand side collapses 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)

Both sides then integrate directly, giving the general solution

$$y(x)=\frac{1}{\mu(x)}\left(\int \mu(x)\,q(x)\,dx + C\right)$$

Worked demonstration: solve $y'+y=e^{-x}$. 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]

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

$$m\frac{dv}{dt}=mg-bv\qquad\Longrightarrow\qquad \frac{dv}{dt}+\frac{b}{m}\,v=g$$

Here $p=b/m$ and $q=g$, both constant, so $\mu=e^{(b/m)t}$ and

$$\frac{d}{dt}\left(e^{(b/m)t}v\right)=g\,e^{(b/m)t}\qquad\Longrightarrow\qquad e^{(b/m)t}v=\frac{mg}{b}e^{(b/m)t}+C$$

hence, with $v(0)=0$,

$$v(t)=\frac{mg}{b}\left(1-e^{-(b/m)t}\right)$$

As $t$ grows the exponential fades and the speed approaches the constant terminal velocity $mg/b$. 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.[1]

Second-order ODEs

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:

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

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.

Example: free fall. With only gravity acting, the height $x(t)$ of a falling object obeys $x''=-g$. Integrating twice,

$$\frac{dx}{dt}=-gt+v_0\qquad\Longrightarrow\qquad x(t)=-\frac{g}{2}t^2+v_0 t+x_0$$

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

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

The two initial conditions have fixed the whole trajectory.

Method 2: linear equations with constant coefficients

General case. The equation

$$y''+a\,y'+b\,y=0$$

models a mass on a spring, a small-angle pendulum, and an RLC circuit. Its solutions are exponentials: try $y=e^{rx}$. Since $y'=re^{rx}$ and $y''=r^2e^{rx}$, substitution gives

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

The factor $e^{rx}$ is never zero, so the exponential solves the equation exactly when $r$ solves the algebraic characteristic equation

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

whose roots determine the general 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 $r=\alpha\pm i\beta$: $y=e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)$.

If the equation has a nonzero right-hand side, $y''+ay'+by=f(x)$, the superposition principle gives $y=y_p+y_h$, where $y_h$ is the general homogeneous solution above and $y_p$ is any single solution of the full equation.

Worked demonstration: 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

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

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

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

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

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

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

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).

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

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

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

Method 1: separation of variables (the heat equation)

General case. For a linear 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.

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

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

Seek $u(x,t)=X(x)T(t)$. Substituting 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 $t$ and the right only on $x$, so both must equal one and the same constant, written $-\lambda$. Each side is now an ODE:

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

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

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 value $\lambda=(n\pi/L)^2$ gives one basic solution, a 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). The decay rate $\alpha(n\pi/L)^2$ grows as $n^2$, so higher modes die out first.

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:

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

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, fit none of the classes above and have no solution in terms of familiar functions. They are studied in one of three ways:[1][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

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.[2]

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 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.[2]

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 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. 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.[3]

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, 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 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.[4]

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