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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

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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)

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  • $\begingroup$ 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

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