# Thread: 1st Order Semi Linear PDE xu_x+yu_y=u+3

1. ## 1st Order Semi Linear PDE xu_x+yu_y=u+3

Folks,

Could anyone check my attached work. Mathematica seems to output u(x,y)=-3+2x for the particular solution...

Thanks

2. If we parametrize x and y we have

$x(s) \text{ and } x(0)=1$

and

$y(s) \text{ and } y(0)=y_0$

Then

$u(x(s),y(s))$

Now if we take the derivative with respect to s we have

$\frac{d}{ds}u(x(s),y(s))=\frac{dx}{ds}u_x+\frac{dy }{ds}u_y=u+3$

This gives the system of equations

$\frac{dx}{ds}=x, \quad x(0)=1$

$\frac{dy}{ds}=y, \quad y(0)=y_0$

$\frac{du}{ds}=u+3$

Solving this gives

$x=e^{s} \quad y=y_0e^{s} \quad u=Ce^{s}-3$

Now if we sub out

$e^{s}$

we get

$u(x,y)=Cx-3 \implies u(1,y)=-1=C-3 \iff C=2$

$u(x,y)=2x-3$

3. Here's your mistake

First let $e^{f} = g$ so $u = x g\left(\dfrac{y}{x}\right) - 3$. If $u(1,y) = -1$ then $-1 = g(y) - 3$ so $g(y) = 2$ so $u = x \cdot 2 - 3$ (your Mathematica and TheEmptySet's solution)

4. Cheers Danny!

I like this alternative method, TheEmptySet. Thanks