Differential equations formulas

Linear means $y$ and every derivative appear only to the first power, are never multiplied together, and never sit inside another function. The coefficients may depend on $x$ only, and the equation has order $n$ wherever $a_0(x)\ne 0$. The equation is homogeneous when $r(x)=0$ and non-homogeneous otherwise.

Valid where $g(y)\ne 0$. Each root $y=y_0$ of $g(y)=0$ is a separate constant solution, and dividing by $g(y)$ can lose it (sometimes it reappears for a particular value of the constant). For $\frac{dy}{dx}=f(ax+by+c)$, substitute $z=ax+by+c$ to get the separable equation $\frac{dz}{dx}=a+bf(z)$.

Applies when $f(x,y)$ is unchanged by $x\to kx,y\to ky$, that is, when $M$ and $N$ are homogeneous functions of the same degree. It uses $\frac{dy}{dx}=v+x\frac{dv}{dx}$. Valid where $F(v)\ne v$; each root $v_0$ of $F(v)=v$ gives the straight-line solution $y=v_{0}x$.

If $a_1{b}_2-a_2{b}_1\ne 0$, put $x=X+h$, $y=Y+k$, where $(h,k)$ solves $a_{1}h+b_{1}k+c_1=0$ and $a_{2}h+b_{2}k+c_2=0$. The equation becomes $\frac{dY}{dX}=\frac{a_{1}X+b_{1}Y}{a_{2}X+b_{2}Y}$, which is homogeneous. If $a_1{b}_2-a_2{b}_1=0$, the two linear expressions are proportional apart from their constants, so the right-hand side is a function of one combination $ax+by$. Substitute $z=ax+by$ as in the separable case above.

The test holds on a simply connected region where $M$, $N$ and their first partial derivatives are continuous. The solution is $\varphi (x,y)=C$, where ${\varphi }_x=M$ and ${\varphi }_y=N$. The shortcut on the right needs $\int Mdx$ written without hidden $x$-free parts (write $\ln x$, not $\ln (xy)$). The safe route is $\varphi =\int Mdx+h(y)$ with $h^{\prime }(y)=N-\frac{\partial }{\partial y}\int Mdx$.

Multiply the whole equation by $\mu$ to make it exact. Two more rules: if $M$ and $N$ are homogeneous of the same degree and $Mx+Ny\ne 0$, then $\mu =\frac{1}{Mx+Ny}$; if $M=y{f}_1(xy)$ and $N=x{f}_2(xy)$ with $Mx-Ny\ne 0$, then $\mu =\frac{1}{Mx-Ny}$.

First divide by the coefficient of $\frac{dy}{dx}$ so that it becomes 1. $P$ keeps its sign. If the equation is linear in $x$ instead, $\frac{dx}{dy}+P(y)x=Q(y)$, use IF $=e^{\int Pdy}$ and $x\cdot \text{IF}=\int Q\cdot \text{IF}dy+C$.

Divide by $y^n$ first. The substitution turns the equation into a linear one in $z$. When $n>0$, $y=0$ is also a solution, which the division by $y^n$ loses.

Differentiating gives $\frac{dp}{dx}\left[x+f^{\prime }(p)\right]=0$. The factor $\frac{dp}{dx}=0$ gives the general solution, the family of straight lines $y=cx+f(c)$ (replace $p$ by $c$). The factor $x+f^{\prime }(p)=0$, with $p$ eliminated between it and the equation, gives the singular solution, which is the envelope of those lines. Example: $y=xp+\frac{1}{p}$ has general solution $y=cx+\frac{1}{c}$; $x-\frac{1}{p^2}=0$ gives the singular solution $y^2=4x$.

Form the family's ODE with the parameter eliminated, make the replacement, then solve. Example: $y=c{x}^2$ gives $\frac{dy}{dx}=\frac{2y}{x}$. The trajectories satisfy $-\frac{dx}{dy}=\frac{2y}{x}$, which gives the ellipses $x^2+2y^2=k$.

$y_c$ comes from the roots of $f(m)=0$ and is the general solution of $f(D)y=0$. $y_p$ is any one solution of the full equation.

For $n$ functions, use the $n\times n$ determinant whose rows are the functions and their derivatives up to order $n-1$. If $W\ne 0$ at some point of an interval, the functions are linearly independent on that interval.

Divide by the coefficient of $y^{\prime \prime }$ first so that it is 1. Holds for two solutions of a linear homogeneous ODE with $p,q$ continuous on an interval. There $W$ is either identically zero (the solutions are dependent) or never zero (they are independent and form a fundamental set).

Gives a second, independent solution on an interval where $y_1\ne 0$. It follows from Abel's formula, because ${\left(\frac{y_2}{y_1}\right)}^{\prime }=\frac{W}{y_1^2}$. Example: $x^2{y}^{\prime \prime }-3x{y}^{\prime }+4y=0$ ($x>0$) has $y_1=x^2$. Dividing by $x^2$ gives $p=-\frac{3}{x}$, so $e^{-\int pdx}=x^3$ and $y_2=x^2\int \frac{x^3}{x^4}dx=x^2\ln x$.

Replace $D$ by $a$. This also covers a constant ($a=0$) and $\cosh ax$ or $\sinh ax$ written as exponentials.

Differentiate $f$ with respect to $D$ $k$ times. For a simple root this gives $\frac{x{e}^{ax}}{f^{\prime }(a)}$, and for a double root $\frac{x^2{e}^{ax}}{f^{\prime \prime }(a)}$.

Replace $D^2$ by $-a^2$. Never replace $D$ itself. If odd powers of $D$ remain, as in $\frac{1}{pD+q}$, multiply numerator and denominator by $pD-q$ so that $D^2$ appears. Replace it by $-a^2$ again, then apply the $D$ now in the numerator as a derivative.

Used when $\pm ja$ are roots of the auxiliary equation, so that $\varphi (-a^2)=0$. Note the minus sign in the sine case.

Take out any factor $D^k$ first. Then write the rest as $c[1\pm \varphi (D)]$, expand in ascending powers of $D$, and stop at $D^m$, because higher derivatives of $x^m$ vanish. Example: $\frac{1}{1+D}{x}^2=(1-D+D^2)x^2=x^2-2x+2$.

Move $e^{ax}$ to the left and replace $D$ by $D+a$. Then handle $V$ (a polynomial, sine or cosine) with the rules above. This covers $x^n{e}^{ax}$ and $e^{ax}\sin bx$.

For $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=r(x)$ with the coefficient of $y^{\prime \prime }$ equal to 1, so divide through first. $y_1$ and $y_2$ are independent solutions of the homogeneous equation. It works for any $r(x)$ continuous on an interval where $p$ and $q$ are continuous.

$a$ and $b$ are constants. For $x<0$ use $x=-e^z$. This amounts to replacing $\ln x$ by $\ln |x|$ and $x^m$ by $|x{|}^m$.

Here $D=\frac{d}{dx}$ and $\theta =\frac{d}{dz}$. Rewrite the right-hand side in $z$ as well: $x^k=e^{kz}$ and $\ln x=z$.

Obtained by trying $y=x^m$ directly (equivalently, $\theta \to m$). The $x^2{y}^{\prime \prime }$ term contributes $m(m-1)$, not $m^2$.

For $\lambda \le 0$, only $y=0$ satisfies both conditions. With $y^{\prime }(0)=y^{\prime }(L)=0$ instead, ${\lambda }_n=(n\pi /L{)}^2$ and $y_n=\cos \frac{n\pi x}{L}$ for $n=0,1,2,\dots$ ($n=0$, that is $\lambda =0$, gives a constant). These are the spatial modes in the PDE solutions below.

One-sided (unilateral) transform, $Y(s)={\int }_0^{\infty }y(t)e^{-st}dt$, with $t\ge 0$ the independent variable. In general, $\mathcal{L}\{y^{(n)}\}=s^{n}Y(s)-s^{n-1}y(0)-\cdots -y^{(n-1)}(0)$. Transform the constant-coefficient ODE, solve for $Y(s)$ with the initial conditions already included, then invert by partial fractions. Example: $y^{\prime }+3y=0$, $y(0)=2$ gives $sY-2+3Y=0$, so $Y=\frac{2}{s+3}$ and $y=2e^{-3t}$.

For $A{u}_{xx}+B{u}_{xy}+C{u}_{yy}+\cdots =0$. Lower-order terms and the right-hand side do not affect the type.

For $u_t=c^2{u}_{xx}$, $0<x<L$, with $u(0,t)=u(L,t)=0$ and $u(x,0)=f(x)$. Every mode decays, and higher modes decay faster.

For $u_x(0,t)=u_x(L,t)=0$. As $t\to \infty$, $u\to a_0$, the average initial temperature.

For $u_{tt}=c^2{u}_{xx}$ with $u(0,t)=u(L,t)=0$, $u(x,0)=f(x)$ and $u_t(x,0)=g(x)$. Mode $n$ oscillates at ${\omega }_n=\frac{cn\pi }{L}$ rad/s, that is, $f_n=\frac{nc}{2L}$ Hz.

A string released from rest ($g=0$) has every $B_n^{\ast }=0$.

For the wave equation on an infinite string with $u(x,0)=f(x)$ and $u_t(x,0)=g(x)$. It shows two waves travelling at speed $c$ in opposite directions.

For $u_{xx}+u_{yy}=0$ on $0\le x\le a$, $0\le y\le b$, with $u=0$ on $x=0$, $x=a$ and $y=0$, and $u(x,b)=f(x)$. The sine in $x$ comes from the two zero side conditions, and $\sinh$ is the combination in $y$ that vanishes at $y=0$.