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.


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:

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

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

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


$$2u_x+u_y = u$$

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

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

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

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

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