Differential equation (legacy) - Revision history

Rongzhou: Rongzhou moved page Differential equation to Differential equation (legacy) without leaving a redirect

2026-10-04T13:33:29Z

Rongzhou moved page Differential equation to Differential equation (legacy) without leaving a redirect

← Older revision		Revision as of 13:33, 4 October 2026
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Rongzhou: /* Method 2: linear equations with constant coefficients */

2026-10-04T13:26:10Z

Method 2: linear equations with constant coefficients

← Older revision		Revision as of 13:26, 4 October 2026
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	gives the characteristic equation $r^2+ar+b=0$, whose roots determine $y_h$:		gives the characteristic equation $r^2+ar+b=0$, whose roots determine $y_h$:

	* $r_1\neq r_2$ ~~real~~: $y_h=C_1e^{r_1x}+C_2e^{r_2x}$;		* $r_1\neq r_2, r_1,r_2 \in \mathbb{R}$: $y_h=C_1e^{r_1x}+C_2e^{r_2x}$;
	* $r_1=r_2=r$: $y_h=(C_1+C_2x)e^{rx}$;		* $r_1=r_2=r$: $y_h=(C_1+C_2x)e^{rx}$;
	* $r=\alpha\pm i\beta$: $y_h=e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)$.		* $r=\alpha\pm i\beta$: $y_h=e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)$.
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	'''Non-homogeneous case ($f\neq 0$).''' $y=y_h+y_p$, with $y_p$ found by undetermined coefficients as in Method 3.		'''Non-homogeneous case ($f\neq 0$).''' $y=y_h+y_p$, with $y_p$ found by undetermined coefficients as in Method 3.

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

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

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

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

Rongzhou: /* Method 1: direct integration */

2026-10-04T13:07:59Z

Method 1: direct integration

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	==== Method 1: direct integration ====		==== Method 1: direct integration ====

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

	$$y''=f(x)\;\Longrightarrow\;y'=\int f(x)\,dx+C_1\;\Longrightarrow\;y=\int\!\!\left(\int f(x)\,dx\right)dx+C_1x+C_2$$		$$y''=f(x)\;\Longrightarrow\;y'=\int f(x)\,dx+C_1\;\Longrightarrow\;y=\int\!\!\left(\int f(x)\,dx\right)dx+C_1x+C_2$$
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	and likewise $y^{(n)}=f(x)$ by $n$ integrations.		and likewise $y^{(n)}=f(x)$ by $n$ integrations.

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

	~~$$x''=-9.8\;\Longrightarrow\;\frac{dx}{dt}=-9.8t+v_0\;\Longrightarrow\;x(t)=-4.9t^2+v_0t+x_0$$~~		Integrate twice for the general solution:

	~~Numeric scenario~~: the ball is dropped from rest, $v_0=0$, at height $x_0=19.6\ \mathrm{m}$. It reaches the ground, $x=0$, when		$$x''=-9.8\;\Longrightarrow\;\frac{dx}{dt}=-9.8t+v_0\;\Longrightarrow\;x(t)=-4.9t^2+v_0t+C$$

			Fix the initial conditions for a particular solution:

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

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

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

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

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

2026-10-04T13:03:28Z

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

← Older revision		Revision as of 13:03, 4 October 2026
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	If you notice, the first term $y=\frac{1}{\mu}\int\mu\,q\,dx$$\frac{C}{\mu}$ provides one particular solution to the non-homogeneous equation, and the second term, $C/\mu=Ce^{-\int p\,dx}$, is a solution to the homogeneous counter par of the original equation: $y'+p(x)\,y=0$. It generalises the solution to the entire solution family.		If you notice, the first term $y=\frac{1}{\mu}\int\mu\,q\,dx$$\frac{C}{\mu}$ provides one particular solution to the non-homogeneous equation, and the second term, $C/\mu=Ce^{-\int p\,dx}$, is a solution to the homogeneous counter par of the original equation: $y'+p(x)\,y=0$. It generalises the solution to the entire solution family.

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

	To see the method in action, consider		To see the method in action, consider

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

2026-10-04T13:02:09Z

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

← Older revision		Revision as of 13:02, 4 October 2026
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	If you notice, the first term $y=\frac{1}{\mu}\int\mu\,q\,dx$$\frac{C}{\mu}$ provides one particular solution to the non-homogeneous equation, and the second term, $C/\mu=Ce^{-\int p\,dx}$, is a solution to the homogeneous counter par of the original equation: $y'+p(x)\,y=0$. It generalises the solution to the entire solution family.		If you notice, the first term $y=\frac{1}{\mu}\int\mu\,q\,dx$$\frac{C}{\mu}$ provides one particular solution to the non-homogeneous equation, and the second term, $C/\mu=Ce^{-\int p\,dx}$, is a solution to the homogeneous counter par of the original equation: $y'+p(x)\,y=0$. It generalises the solution to the entire solution family.

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

	To see the method in action, consider		To see the method in action, consider
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	Multiply the whole equation by $\mu(x)=x$:		Multiply the whole equation by $\mu(x)=x$:

	$$ x\frac{dy}{dx} + y = x^3~~. $$~~		$$ x\frac{dy}{dx} + y = x^3 \\
			\frac{d}{dx}\bigl(xy\bigr) = x^3. $$
	~~The left side is exactly the derivative of the product $x~~\~~,y$:~~

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

	Integrate with respect to $x$:		Integrate with respect to $x$:

	$$ xy = \int x^3\,dx = \frac{x^4}{4} + C. $$		$$ xy = \int x^3\,dx = \frac{x^4}{4} + C. \\
			y = \frac{x^3}{4} + \frac{C}{x}. $$
	~~Finally, divide by $x$ to obtain the general solution:~~

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

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

Rongzhou: /* Ordinary and partial */

2026-10-04T12:57:07Z

Ordinary and partial

← Older revision		Revision as of 12:57, 4 October 2026
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	=== Ordinary and partial ===		=== Ordinary and partial ===

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

			An example of a partial differential equation is the heat transfer equation describing the temperature $u$ of an insulated metal bar, according to position $x$ on the metal bar and time $t$

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

Rongzhou: /* Linearity and homogeneity */

2026-10-04T12:54:28Z

Linearity and homogeneity

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

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

			For example, if $y'=y^2$ has two solutions $y_1'=y_1^2$ and $y_2'=y_2^2$

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

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

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

Rongzhou: /* General and specific solution */

2026-10-03T20:53:30Z

General and specific solution

← Older revision		Revision as of 20:53, 3 October 2026
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	A surprising number of laws of nature can be described by differential equations: pendulums, cooling drinks, growing populations, and discharging capacitors.		A surprising number of laws of nature can be described by differential equations: pendulums, cooling drinks, growing populations, and discharging capacitors.

	== General and specific ~~solution~~ ==		== General and specific solutions ==

	~~The~~ simplest ~~differential equation prescribes the slope~~ of ~~a function $y(x)$:~~		Differential equations, including the simplest ones, often have an infinite number of solutions.

	~~$$\frac{dy}{dx}=2x$$~~		The entire solution set to a particular differential equation is known as its '''general solution'''.

	~~Integrating both sides gives~~		If an '''initial condition''' $x_1, y(x_1)$ is known, then the solution set can be further restricted to a single function, known as the '''particular solution'''.

	~~$$\int\frac{dy}{dx}\,dx=\int 2x\,dx\qquad\Longrightarrow\qquad y(x)=x^{2}+C$$~~		Take the example of the simplest differential equation in the world:

	~~Where~~ $C$ ~~is an unknown constant. Therefore, the solution is not one single function, but a family of functions, known as the '''general solution'''.~~		$$\frac{dy}{dx}=x$$

	~~If we know the value of $y$ at a particular $x$~~, ~~for instance, $y(0)=3$, then a '''particular solution''' function can be identified~~		To solve it, integrate both sides:

	$$~~3=0^2+C~~\~~qquad~~\~~Longrightarrow~~\~~qquad C~~=3\qquad\Longrightarrow\qquad y=x^2+3$$		$$\int\frac{dy}{dx}\,dx=\int x\,dx\qquad\Longrightarrow\qquad y(x)=\frac{1}{2}x^{2}+C$$

	~~A prescribed value such as~~ $y(a)=b$ ~~allowing us to pin down a~~ particular solution ~~is called an '''~~initial condition~~'''~~.		Where $C$ is an unknown constant. This means that any function in the family of $y=\frac{1}{2}x^2+C$, simply $y=\frac{1}{2}x^2$ or more complicatedly $y=\frac{1}{2}x^2+\pi$, is a solution to the original differential equation. So $y=\frac{1}{2}x^2+C$ is the general solution.

			If we know that $x=0, y=3$:

			$$3=\frac{1}{2}0^2+C\qquad\Longrightarrow\qquad C=3\qquad\Longrightarrow\qquad y=\frac{1}{2}x^2+3$$

			so $y=\frac{1}{2}x^2+3$ is the particular solution under the given initial condition.

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

Rongzhou at 11:56, 7 September 2026

2026-09-07T11:56:20Z

Show changes

RonzzWikiCowriterAI: AI-assisted (RonzzWikiCowriter): present Method 2 as the method of characteristics in general (general quasilinear first-order form, characteristic system, general algebraic solution u = F(ψ(x,t))), with the transport equation and the wave equation derived from it as examples rather than being the method itself. (via update-page on MediaWiki MCP Server)

2026-09-05T20:02:56Z

AI-assisted (RonzzWikiCowriter): present Method 2 as the method of characteristics in general (general quasilinear first-order form, characteristic system, general algebraic solution u = F(ψ(x,t))), with the transport equation and the wave equation derived from it as examples rather than being the method itself. (via update-page on MediaWiki MCP Server)

← Older revision		Revision as of 20:02, 5 September 2026
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	=== Partial differential equations ===		=== Partial differential equations ===

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

	==== Method 1: separation of variables ====		==== Method 1: separation of variables ====
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	so after one hour $u\approx 100e^{-0.82}\approx 44\,^{\circ}\mathrm{C}$, and $50\,^{\circ}\mathrm{C}$ is reached at $t=\ln 2/(2.3\times10^{-4})\approx 3050\ \text{s}\approx 51$ min.<ref>{{#cite:Q1579}}</ref>		so after one hour $u\approx 100e^{-0.82}\approx 44\,^{\circ}\mathrm{C}$, and $50\,^{\circ}\mathrm{C}$ is reached at $t=\ln 2/(2.3\times10^{-4})\approx 3050\ \text{s}\approx 51$ min.<ref>{{#cite:Q1579}}</ref>

	==== Method 2: ~~travelling waves (~~method of characteristics) ====		==== Method 2: the method of characteristics ====

	'''General form.''' The ~~transport equation for a quantity $u(x,t)$ carried~~ by ~~a flow at constant speed $c$ is~~ the first-order ~~linear PDE~~		'''General form.''' The method of characteristics solves first-order PDEs by tracing curves along which the PDE reduces to ordinary differential equations. In two independent variables the general quasilinear first-order equation is

	$$~~u_t~~+c\,~~u_x~~=0$$		$$A(x,t,u)\,u_x+B(x,t,u)\,u_t=C(x,t,u)$$

	~~'''General algebraic~~ solution~~.''' For any differentiable profile $f$, the travelling wave~~ $u(x,t)=f(x-ct)$ ~~satisfies the equation identically~~, ~~since~~ $~~u_t=-cf'~~(~~x-ct~~)$ ~~and~~ $u_x~~=f'(x-ct)~~$~~; conversely~~, ~~every~~ solution is of ~~this form~~, ~~with~~ $~~f$ fixed by~~ the ~~initial data:~~		A solution $u=u(x,t)$ is a surface in $(x,t,u)$-space. Its tangent plane at each point is spanned by $(1,0,u_x)$ and $(0,1,u_t)$, so a vector $(A,B,C)$ is tangent to the surface exactly when $C=A u_x+B u_t$, the condition expressed by the PDE itself. The solution surface is therefore swept out by the integral curves of the vector field $(A,B,C)$, the characteristic curves, which solve the characteristic system of ordinary differential equations

	$$u(x,t)=f(x~~-ct~~),\qquad u(x,~~0)=f(x~~)$$		$$\frac{dx}{ds}=A(x,t,u),\qquad \frac{dt}{ds}=B(x,t,u),\qquad \frac{du}{ds}=C(x,t,u)$$

	~~so $u$~~ is ~~constant on each~~ characteristic ~~line~~ $x~~-ct=\text{const}$~~, ~~and the initial profile is transported unchanged at speed $c$. For the forced equation $u_t+cu_x~~=s(x,t)$, the ~~source is accumulated along~~ the ~~characteristic:~~		Given data on a curve that is not itself characteristic, such as $u(x,0)=u_0(x)$, one characteristic issues from each point of the curve, and integrating the system carries the data across the region the characteristics cover. For the linear homogeneous case

	$$u(x,t)~~=f(x-ct)~~+~~\int_0^t s\bigl~~(x~~-c(~~t~~-\tau),\tau\bigr~~)\,~~d\tau~~$$		$$a(x,t)\,u_x+b(x,t)\,u_t=0$$

	'''Example: transport of a pollutant.''' A river flows steadily at speed $c=2\ \mathrm{m\,s^{-1}}$, and a factory releases a concentrated slug of pollutant at one point; as long as mixing and diffusion are negligible, the current simply carries the whole slug downstream without changing it. The concentration obeys ~~the transport equation~~ $u_t+2u_x=0$ with the Gaussian initial profile		the $x$- and $t$-equations do not involve $u$, and the third gives $du/ds=0$: the solution is constant along each characteristic. The characteristics form a one-parameter family; let $\psi(x,t)=\text{const}$ be a first integral, a function constant on each member of the family.

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

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

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

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

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

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

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

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

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

			'''Example: the wave equation.''' The wave equation

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

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

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

	~~'''Example:~~ the ~~wave equation.''' The second~~-~~order wave equation is the two~~-~~directional version of~~ the ~~same idea~~, and ~~its~~ general algebraic solution ~~is a pair of waves travelling left and right~~ (d'Alembert, 1747):		in which the operator becomes $u_{tt}-c^2u_{xx}=-4c^2u_{\xi\eta}$, so the equation reads $u_{\xi\eta}=0$. Hence $u_\xi$ depends on $\xi$ alone, and one further integration gives the general algebraic solution (d'Alembert, 1747):

	$$~~u_{tt}=c^2u_{xx}\;\Longrightarrow\;~~u(x,t)=f(x-ct)+g(x+ct)$$		$$u(x,t)=f(x-ct)+g(x+ct)$$

	~~The term $f(x-ct)$ obeys the transport equation with speed $c$ and $g(x+ct)$ with speed $-c$;~~ each ~~therefore satisfies the wave equation, and linearity superposes them into the general solution~~. The functions $f~~$ and $~~g$ are fixed by the initial displacement and velocity: for a string released from rest with initial displacement $\phi(x)$, the conditions $u(x,0)=\phi(x)$ and $u_t(x,0)=0$ give $f=g=\phi/2$, so		a superposition of two travelling waves, one in each direction. The functions $f,g$ are fixed by the initial displacement and velocity: for a string released from rest with initial displacement $\phi(x)$, the conditions $u(x,0)=\phi(x)$ and $u_t(x,0)=0$ give $f=g=\phi/2$, so

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