Use method separation of variables: $\frac{{\partial u}}{{\partial y}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}} - 4u$

I did the dpe:
$$\frac{\partial u}{\partial y} = \frac{\partial ^2 u}{\partial x^2} - 4u$$

$0 < x < \pi$

With boundary conditions: $\begin{array}{l} u(0,y) = u(\pi ,y) = 0 \\ u(x,0) = {x^2} - \pi x \\ \end{array}$

I used method separation of variables to solve the problem, and i got the solution is:

$$u(x,y) = \sum\limits_{n = \text{even}} \frac{- 8}{\pi n^3} \sin (nx) e^{ - (4 + n^2)y}$$

However, I feel that the solution is not fit with the boundary conditions Could you give some hint??

Using separation of variables $$u = XY \\ u_y = XY' \\ u_{xx} = X''Y \\ XY' = X''Y - XY \\ \frac {Y'}Y = \frac {X''}X - 4$$ Since LHS and RHS are functions of different variables, they must be equal to some constant $$\frac {Y'}Y = \frac {X''}X - 4 = -\lambda^2-4 \\ \frac {Y'}Y = -\lambda^2-4 \\ Y = Y_0 e^{-(\lambda^2+4) y} \\ \frac {X''}X = -\lambda^2 \\ X = X_{01} \cos( \lambda x) + X_{02} \sin (\lambda x) \\ u = Y_0 e^{-(\lambda^2+4) y} \left (X_{01} \cos(\lambda x) + X_{02} \sin (\lambda x) \right )$$ Applying first and second BCs $$u(0, y) = Y_0 e^{-(\lambda^2+4) y} X_{01} = 0 \\ u(\pi, y) = Y_0 e^{-(\lambda^2+4) y} X_{02} \sin (\lambda \pi) = 0$$ From first equation $X_{01} = 0$, since if $Y_0$ one gets trivial solution. From second $$\lambda \pi = \pi n, \quad n \in \mathbb Z \\ \lambda = n\\ u = \sum_{n = 1}^\infty Y_0 e^{-(n^2+4) y} X_{01n} \sin nx$$ Coefficients $X_{01n}$ can absorb $Y_0$, so without loss of generality, one can say $$u = \sum_{n = 1}^\infty X_{01n} e^{-(n^2+4) y} \sin nx$$ Now apply third BC $$u(x, 0) = \sum_{n = 1}^\infty X_{01n} \sin nx = x^2 - \pi x$$ So, basically you need to find LHS expansion of $x^2-\pi x$. $$b_n = \frac 2\pi \int_0^\pi (x^2 - \pi x) \sin n x dx = -\frac 4 {n^3 \pi} (-1+\cos n \pi) = \frac 8{n^3 \pi},\quad n \text{ is odd}$$ so $$u = \frac 8{\pi^3} \sum_{n = odd} \frac {e^{-(n^2+4) y} \sin nx}{n^3}$$
• @Kaster: Thank you very much! Your solution and mine is:$u(x,y) = \sum\limits_{n = even} {\frac{{ - 8}}{{\pi {n^3}}}\sin (nx){e^{ - (4 + {n^2})y}}}$, Are they equal? – MacArthur Nguyen May 13 '13 at 14:28