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linear partial differential eq.

Posted September 8th, 2007 by vic777
in

How can I find a solution for the equation with Cauchy's conditions ?
$ 3u_x+2u_y=0 $ with $ u(x,0)=\sin(x) $

‹ Linear partial diff. eqn. Linear partial differential eqn. ›

Comments

Great solution, thankyou

On September 8th, 2007 vic777 says:

Great solution, thankyou

As usually , first we have

On September 8th, 2007 Structure says:

As usually , first we have to find a solution for differential equation
$ \frac{dx}{3}=\frac{dy}{2}\; or\; dx-\frac{3}{2}dy=0 $
So, $ x-\frac{3}{2}y=c $
Then solution is of form $ u(x,y)=f(x-\frac{3}{2}y) $ where f is a funtion to be determined.
From u(x,0)=f(x)=sinx we deduce $ f(x)=\sin(x) $ and finally
$ u(x,y)=\sin(x)-\frac{3}{2}y $

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