-1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ This in an ODE, not a PDE. Unless you really mean: $$x^3 \frac {\partial u}{\partial x}+y\color {red}{\frac {\partial u}{\partial y}}=4+2x^2 u $$ $\endgroup$ – projectilemotion Dec 2 '17 at 8:19
0
$\begingroup$

$$x^3\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=4+2x^2u$$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.