2
$\begingroup$

How to solve this pde:

Let $u(x,y)=2f(y)\cos(x-2y)$ be a solution of the Initial Value Problem

$2u_x+u_y=u$; $u(x,0)=\cos(x)$.

Then find the value of $f(1)$.

By Lagrange's Auxiliary Equations $\dfrac{\operatorname{dx}}{2}=\dfrac{\operatorname{dy}}{1}=\dfrac{\operatorname{du}}{u}$

Hence the solutions are $x=2y+c_1;\ln u=y+c_2;x=\ln u^2+c_3$

But how can I find the required value from here.Please help.

$\endgroup$
2
$\begingroup$

If $u(x,y) = 2f(y)\cos(x-2y)$, then we have:

$$u(x,0) = 2f(0)\cos x$$

$$u_x = -2f(y)\sin(x-2y)$$

$$u_y = 2f'(y)\cos(x-2y)+4f(y)\sin(x-2y)$$

By using the given information, we get:

$$u(x,0) = \cos x$$

$$2f(0)\cos x = \cos x$$

$$f(0) = \dfrac{1}{2}$$

and

$$2u_x+u_y = u$$

$$2\left[-2f(y)\sin(x-2y)\right]+\left[2f'(y)\cos(x-2y)+4f(y)\sin(x-2y)\right] = 2f(y)\cos(x-2y)$$

$$f'(y) = f(y)$$

I'm sure you can solve for $f(y)$, and then get $f(1)$ from the above.

$\endgroup$
  • $\begingroup$ thank you very much sir ;I have got it to be $\frac{e}{2}$ $\endgroup$ – Learnmore Dec 9 '15 at 6:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.