I don't have much experience in partial differential equations and i'm stuck. I need to find a solution to this : $$x\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}=0$$ with initial condition of $f(1,y)=e^{-2y}$ with the help of changing variables $$u=xe^y,v=xe^{-y}$$ my solution is this: $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}e^y+\frac{\partial f}{\partial v}e^{-y}$$ $$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}xe^y+\frac{\partial f}{\partial v}(-xe^{-y})$$ $$\frac{\partial f}{\partial u}2xe^y=0\implies\frac{\partial f}{\partial u}2u=0 $$ $$\int 2u \frac{\partial f}{\partial u} =\int 0$$ $$u^2= C(v)\implies xe^{2y}=C(xe^{-y})$$ I'm stuck from here i don't know if my integral evaluation is correct and how can i solve this with initial condition as i said earlier don't have much experience in partial differential equation, can somebody please help me

Thanks in advance


You have performed the integration without writing what variables you are integrating with respect to.

Normally, you would first 'separate' the equation $$\frac{\partial f}{\partial u}2u=0$$

Into something like

$$\partial f=\frac{0}{2u} \partial u=0 \;\partial u$$

And then integrate both sides

$$f(x,y)=\int df=\int 0 \; du=g(v)$$

where $g(v)$ is an arbitrary function in the single variable $v$.

If you plug this arbitrary $g(v)$ back into the original differential equation, you'll find that it is satisfied regardless of what $g$ is specifically.

So the general solution to your equation is $f(x,y)=g(xe^{-y})$ for any single-variable function $g$ (that obeys some basic conditions relating to differentiability, of course).

It is only once you add in the initial condition that you can finally get a single specific solution (which I'll leave for you to work out).

(feel free to comment or edit for any corrections or suggestions)

  • $\begingroup$ i think that i made some mistake, at the end it shouldn't be $\frac{\partial f}{\partial u}=0$ ? $\endgroup$
    – simon
    Sep 23 '20 at 7:49
  • $\begingroup$ @simon What makes you think you made some mistake? You converted the derivatives correctly, so $\frac{\partial f}{\partial u} 2u=0$ has to hold. If $\frac{\partial f}{\partial u}$ is not $0$ everywhere, then this would mean $u$ is $0$ (almost) everywhere, which is clearly impossible. $\endgroup$ Sep 23 '20 at 10:18

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.