# How can I solve these pde's?

Three different problem I got:

1.. $xu_x+2x^2u_y-u=x^2e^x$ and $u(x,x^2+x)=xe^x+x^2$

2.. $yu_{xx}+(x+y)u_{xy}+xu_{yy}=0, \quad x\neq y$

3.. $(y+xu)u_x+(x+yu)u_y=u^2-1$

Couldnt even start. Could you at least give me a hint? thanks.

edit1:

at 2. from $B^2-AC=\frac {(x-y)^2}4>0$ so the equation is hyperbolic. so we can write its cononical forms. is it true?

edit2: for the first one :

by taking $\eta=x^2-y$ and $\xi=x$ I found $u(x,y)=xe^x+x(-x+1-e^{-x})$ for the second one :

I took $\xi=\frac{x^2-y^2}{2}$ and $\eta=x-y$ and found $u(x,y)=\frac {f^*(\frac{x^2-y^2}{2})}{x-y}+g(x-y)$

edit 3:----------------------

1-) $\frac ba=\frac {2x^2}{x}=2x=\frac {dy}{dx}$

$x^2-y=c$, $\eta =x^2-y$ and $\xi =x$

$u_x=w_{\xi}+w_{\eta}2x$ and $u_y=-w_{\eta}$

$\xi(w_{\xi}+w_{\eta}2\xi)-2\xi ^2w_{\xi}-w=\xi^2e^\xi$

$w_{\xi}-\frac w \xi=e^\xi$ multiplying by $\frac 1\xi$

$\frac{d(w\frac 1\xi)}{d\xi}=e^\xi$

so we got $w=\xi e^\xi+g(\eta)$

$u(x,y)=xe^x+g(x^2-y)$

and $u(x,x^2+x)=xe^x+x^2 =xe^x+g(x^2-x^2-x)$

$g(-x)=x^2$, $g(x)=x^2$ so the $u(x,y)=xe^x+x^2$

//--------------------------------------------

3)$\frac {\frac{dx}{dt}}{y+ux}=\frac {\frac{dy}{dt}}{x+yu}=\frac {\frac{du}{dt}}{u^2-1}$

$\frac {\frac{dx}{dt}-\frac{dy}{dt}}{y+ux-x-yu}=\frac {\frac{du}{dt}}{u^2-1}$

$\frac {\frac{dx}{dt}-\frac{dy}{dt}}{(1-u)(y-x)}=\frac {\frac{du}{dt}}{u^2-1}$

$\frac {\frac{d(x-y)}{dt}}{y-x}=\frac {\frac{du}{dt}}{u+1}$

$ln(u+1)+ln(x-y)=c_1$------------------- (1) we got one independent solution we need another

$\frac {\frac{dx}{dt}+\frac{dy}{dt}}{y+ux+x+yu}=\frac {\frac{du}{dt}}{u^2-1}$

$\frac {\frac{dx}{dt}+\frac{dy}{dt}}{(1+u)(x+y)}=\frac {\frac{du}{dt}}{u^2-1}$

$\frac {\frac{d(x+y)}{dt}}{x+y}=\frac {\frac{du}{dt}}{u-1}$

$ln(u-1)-ln(x+y)=c_2$----------------------(2)

from (1) and (2) which form is $h(x,y,u)=c_1$ and $j(x,y,u)=c_2$

general solution would be in $j(x,y,u)=F(h(x,y,u))$ Where F is an arbitrary.

so $ln(u-1)-ln(x+y)=F(ln(u+1)+ln(x-y))$

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I notice you have asked 45 questions, and many people have given answers, but you have only voted up 7 answers. Be sure to show appreciation of people who help you by voting up their answers. Thanks! –  Matthew Conroy May 22 '14 at 23:24
@MatthewConroy I didnt notice :) I always accept answers as soon as possible but didnt know upvoting was important so my bad then –  lyme May 22 '14 at 23:30
Your first question is very odd. Why both $x$ and $x'$ exist? –  doraemonpaul May 23 '14 at 22:02
@doraemonpaul fixed. I was in a hurry. sorry –  lyme May 25 '14 at 21:01

Some hints:

• First problem: 1st order linear PDE, use the method of characteristics and then apply the "boundary" condition. Edit: note that your solution doesn't satisfy nor the equation nor the "boundary condition". The characteristic equations are given by: $$\frac{\mathrm{d} x}{x} = \frac{\mathrm{d} y}{2 x^2} = \frac{\mathrm{d} u}{u + x^2 e^x},$$ which from the first equality yields: $$y - x^2 = \text{constant} \equiv \eta ,$$ which is the characteristic of the PDE. The remaining equation, which comes up from the equality of the first and third terms is a 1st order linear ODE for $u$ as a function of $x$: $$\frac{\mathrm{d} u}{ \mathrm{d} x} = \frac{u}{x} + x e^x ,$$ whose solution (use an integrating factor) is given by $u(x,y) = x \, ( e^x + C)$, being $C$ a constant of integration. Put $C = C(\eta(x,y))$ in order to get the complete solution.

• Second problem: 2nd oder linear PDE, use the method of characteristics knowing that for $x - y \neq 0$ the equation is hyperbolic but parabolic when $x = y$ (I notice now that you are told $x \neq y$).

• Third problem: 1st order non-linear PDE, use the Lagrange-Charpit equations, which are a generalization of the characteristics. Edit: this method reads: $$\frac{\mathrm{d}x}{F_p} = \frac{\mathrm{d}y}{F_p} = \frac{\mathrm{d}u}{p F_p + q F_q} = - \frac{\mathrm{d} p}{ F_x + p F_u} = - \frac{\mathrm{d} q}{F_y + q F_u} ,$$ where $F(x,y,u;p,q) = (y+xu)u_x + (x+yu) u_y-u^2 -1$ and subindices denote partial differentiation with respect the subscript.

Cheers!

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thanks. I edited my answer but got problem with third. Could you give a link of example of this method –  lyme May 26 '14 at 21:31
Of course. Give me some minutes. –  Dmoreno May 26 '14 at 21:32
I fixed my answer. I guess first one is true now. I tried something at the third one which I'm not sure if it is true. Can you check again please –  lyme May 27 '14 at 10:56

1..

$xu_x+2x^2u_y-u=x^2e^x$

$u_x+2xu_y=\dfrac{u}{x}+xe^x$

$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$

$\dfrac{dy}{dt}=2x=2t$ , letting $y(0)=y_0$ , we have $y=t^2+y_0=x^2+y_0$

$\dfrac{du}{dt}=\dfrac{u}{x}+xe^x=\dfrac{u}{t}+te^t$ , we have $u=te^t+tf(y_0)=xe^x+xf(y-x^2)$

$u(x,x^2+x)=xe^x+x^2$ :

$xe^x+xf(x)=xe^x+x^2$

$f(x)=x$

$\therefore u(x,y)=xe^x+x(y-x^2)=xe^x-x^3+xy$

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Could you explain how did you find $u=te^t+tf(y_0)$ –  lyme Jun 1 '14 at 12:43
isnt it $u(t)=te^t+ct$ but $u(0)=0=f(y_0)$? –  lyme Jun 1 '14 at 15:21