# Finding a solution to $xu_x+(y+1)u_y=u-1$ given an initial condition

Consider the following PDE: $$xu_x+(y+1)u_y=u-1.$$

Using this formula: $$\frac{dx}{x}=\frac{dy}{y+1}=\frac{du}{u-1}.$$

This yields $c_1=\frac{y+1}{x}$ and $c_2=\frac{u-1}{x}.$

We have: $$F\left(\frac{y+1}{x}\right)=\frac{u-1}{x}.$$

Given the following Cauchy condition - $u(x,2x-1)=e^x$. This yields $$xF(2)+1=e^x.$$

Am I on right right track? I'm a little confused because we have an $F(2)$.

• I don't follow the line "This yields..." Aren't you supposed to integrate $\frac{dx}{x}=\frac{dy}{y+1}=\frac{du}{u-1}$? – user147263 Oct 13 '14 at 17:59
• You integrate two at a time. For example, I did the left most and middle, and then the left most and the right most one. That's how I yielded $c_1$ and $c_2$. – emka Oct 13 '14 at 18:23
• But integration should produce something like $\ln x$, $\ln (y+1)$, $\ln (u-1)$... – user147263 Oct 13 '14 at 20:29

$\dfrac{dx}{dt}=x$ , letting $x(0)=1$ , we have $x=e^t$
$\dfrac{dy}{dt}=y+1$ , we have $y+1=y_0e^t=y_0x$
$\dfrac{du}{dt}=u-1$ , we have $u(x,y)=F(y_0)e^t+1=xF\left(\dfrac{y+1}{x}\right)+1$
$u(x,2x-1)=e^x$ :
$xF(2)+1=e^x$ , which is impossible.
$\therefore$ There is no solution.