A question on PDE $xp-yq=u$ with $u(x,0)=sin(πx/4)$

Question

If $u(x,y)$ is a solution of the Partial differential equation $xp-yq=u$ with $u(x,0)=sin(πx/4)$ then $u(1/√2 ,1/√2)$ is

$(1/√2)e^{π/4}$
$(1/√2)e^{1/√2}$
$(π/4)e^{π/√2}$
$(π/4)e^{π/4}$. I tried to solve it using Lagrange's method and got $u(x,y)=f(x^2+y^2)e^{tan^{-1}y/x}$ as the solution but I do not know how can I fit this condition $u(x,0)=sin(πx/4)$ into my solution. Kindly help me out to solve it correctly or is there another method to hit such type of problem more efficiently. Thanks.

I am not seeing how you got that solution from Lagrange's method, can you show me? — Kushal Bhuyan, Nov 29 '15 at 03:35
Is this some competitive exam question? Where did you get this? Any reference will be helpful. — Kushal Bhuyan, Nov 29 '15 at 04:25
I am really sorry to accidentally post my comment in answer box .Can it not be deleted? — Nitin Uniyal, Nov 29 '15 at 06:15
You can delete it. Just look for the delete option below the sentence of your answer. — Kushal Bhuyan, Nov 29 '15 at 06:34

score 0 · Accepted Answer · answered Nov 29 '15 at 15:02

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

$\dfrac{dy}{dt}=-y$ , letting $y(0)=y_0$ , we have $y=y_0e^{-t}=\dfrac{y_0}{x}$

$\dfrac{du}{dt}=u$ , letting $u(0)=f(y_0)$ , we have $u(x,y)=e^tf(y_0)=xf(xy)$

$u(x,0)=\sin\dfrac{\pi x}{4}$ :

$xf(0)=\sin\dfrac{\pi x}{4}$ , which is impossible.

$\therefore$ There is no solution.

1 Answers1