Recipe

Solve a variable-coefficient linear first-order PDE

Use when: the equation is $a(x,y)u_x+b(x,y)u_y=0$ with $a,b$ depending on $x,y$ but not on $u$

Goal: the general solution $u(x,y)=f(\varphi(x,y))$ for the curve-invariant $\varphi$ , or a specific $u$ once initial data pins down $f$

Taught in Variable-coefficient characteristics

Quick recipe

Write the characteristic system
Solve for the characteristic curves
Write the general solution as an arbitrary function of the invariant
Match initial data to pin down $f$ , if given
Check the solution by substitution

1
From $a(x,y)u_x+b(x,y)u_y=0$ , write down $x'(t)=a(x,y)$ and $y'(t)=b(x,y)$ .

Example

For $xu_x+u_y=0$ : $x'(t)=x$ and $y'(t)=1$ .

Why: the characteristics are exactly the curves along which $u$ stays constant. This ODE system traces them out.
2
Find the invariant combination $\varphi(x,y)$ that stays fixed along each curve, by one of two routes.

Example

Parametrizing: $x'(t)=x\Rightarrow x=x_0e^t$ , and $y'(t)=1\Rightarrow y=t$ . Eliminating $t$ gives $x_0=xe^{-y}$ , a constant. For $2xu_x+u_y=0$ , eliminate directly: $dy/dx=1/(2x)\Rightarrow 2\,dy=dx/x\Rightarrow 2y=\ln|x|+C_1$ . So $xe^{-2y}$ is constant.
- you need $x(t)$ and $y(t)$ individually, such as once $u$ itself enters $a,b$ — parametrize by $t$ and solve the coupled ODEs directly.
- $a,b$ depend only on $x,y$ — eliminate $t$ by dividing the equations, giving $dy/dx=b/a$ , then solve that ODE.
⚡ Shortcut

Eliminating $t$ skips ever writing $x(t)$ explicitly. It's the faster path whenever $a,b$ depend only on $x,y$ . Use it when you only need the invariant, not the individual curves.

Why: either route produces the same curve-invariant $\varphi(x,y)$ . Which one takes less algebra depends on whether $dy/dx=b/a$ integrates easily on its own.
3
Once the curves are $\varphi(x,y)=\text{const}$ , the general solution is $u(x,y)=f(\varphi(x,y))$ for arbitrary $f$ .

Example

With $\varphi=xe^{-y}$ : $u(x,y)=f(xe^{-y})$ .

Why: $u$ is constant along each characteristic curve. So $u$ can only depend on $(x,y)$ through which curve it sits on.
4
Substitute the invariant's value along the curve where $u$ is specified. Then solve for $f$ explicitly.

Example

With $u(x,0)=x^3$ : setting $y=0$ in $u(x,y)=f(xe^{-y})$ gives $f(x)=x^3$ , so $f(s)=s^3$ and $u(x,y)=(xe^{-y})^3=x^3e^{-3y}$ .

Why: the characteristic method only determines $u$ up to an arbitrary function. Initial data is what removes that remaining freedom.
5
Differentiate the candidate $u$ with the chain rule. Confirm it satisfies the original PDE. Then check the initial condition too.

Example

$u_x=3x^2e^{-3y}$ , $u_y=-3x^3e^{-3y}$ , so $xu_x+u_y=3x^3e^{-3y}-3x^3e^{-3y}=0$ , and $u(x,0)=x^3e^0=x^3$ .

Why: this catches an algebra slip in solving the ODE, or in matching the initial data, before it's reported as the final answer.

Common mistakes

Integrating $1/x$ as $\ln x$ instead of $\ln|x|$ , then getting confused by a negative $x$ later.
Extending a solution found for $x>0$ across $x=0$ without separately checking where $a(x,y)$ vanishes.
Trusting the invariant found by eliminating $t$ without substituting it back into the original PDE to confirm it actually works.
Reporting $u(x,y)=f(\varphi(x,y))$ as the final answer when initial data was given, instead of solving for $f$ explicitly.