# Using method of characteristics to find general solution of PDE $x^3 u_x + y u_x = 4 + 2x^2u$

Use the method of characteristics to find the general solution $u(x; y)$ of the partial differential equation $$x^3 \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial x} = 4 + 2 x^2 u$$

• This in an ODE, not a PDE. Unless you really mean: $$x^3 \frac {\partial u}{\partial x}+y\color {red}{\frac {\partial u}{\partial y}}=4+2x^2 u$$ – projectilemotion Dec 2 '17 at 8:19

## 1 Answer

$$x^3\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=4+2x^2u$$ System of characteristic ODEs : $$\frac{dx}{x^3}=\frac{dy}{y}=\frac{du}{4+2x^2u}$$ First family of characteristic curves, from $\quad \frac{dx}{x^3}=\frac{dy}{y} \quad\to\quad y\:e^{\left(\frac{1}{2x^2}\right)}=c_1$

Second family of characteristic curves, from $\quad\frac{dx}{x^3}=\frac{du}{4+2x^2u}\quad\to\quad \frac{u}{x^2}+\frac{1}{x^4}=c_2$

General solution of the PDE expressed on the form of implicit equation : $$\Phi\left(y\:e^{\left(\frac{1}{2x^2}\right)} \:,\: \frac{u}{x^2}+\frac{1}{x^4}\right)=0$$ $\Phi$ is any differentiable function of two variables.

Equivalently, general solution of the PDE expressed on the form of explicit equation : $\quad \frac{u}{x^2}+\frac{1}{x^4}=F\left(y\:e^{\left(\frac{1}{2x^2}\right)} \right)$ $$u(x,y)=-\frac{1}{x^2}+x^2 F\left(y\:e^{\left(\frac{1}{2x^2}\right)} \right)$$ $F$ is any differentiable function.