$\lambda$ is typed as lambda, $\alpha$ as alpha.

Assume the PDE is separable by making the substitution $u = XY$:
=
Move the term with $Y^{\prime\prime}$ to the right hand side and divide both sides by $XY$ so that we get the separated ODE's:
= = $-\lambda$

Note: Use the prime notation for derivatives, so the derivative of $X$ is written as $X^\prime$. Do NOT use $X^\prime(x)$

Since these differential equations are independent of each other, they can be separated
DE in X: $= 0$
DE in Y: $= 0$

Now we solve the separate separated ODEs for the different cases in $\lambda$. In each case the general solution in X is written with constants a and b and the general solution in Y is written with constants c and d. Write the functions alphabetically, so that if the solutions involve cos and sin, your answer would be acos(x) + bsin(x).

For this set of differential equations we have two bifurcations, whether $\lambda = 0$ in the case of X or $\lambda = -1$ in the case of Y.

Case 1: $\lambda = 0$
$X(x) =$
$Y(y) =$
$u =$

Case 2: $\lambda = -1$
$X(x) =$
$Y(y) =$
$u =$

Case 3: $-1<\lambda < 0, \lambda = -\alpha^2$
$X(x) =$
$Y(y) =$
$u =$

Case 3: $\lambda < -1, \lambda = -\alpha^2$
$X(x) =$
$Y(y) =$
$u =$

Case 3: $\lambda > 0, \lambda = \alpha^2$
$X(x) =$
$Y(y) =$
$u =$

You can earn partial credit on this problem.