Difficult partial differential equation 
Problem
  $$ \Omega\frac{\partial}{\partial t}A(y,t) +6\Lambda\Omega\left(y^2-y\right) \sin(t) = \frac{\partial^2}{\partial y^2}A(y,t) $$
  Boundary conditions 
  $$ \frac{\partial}{\partial y}A(t,0) = \frac{\partial}{\partial y}A(t,1) = 0 $$
  Solution 
  $$ A(y,t) =6\Lambda \cdot \operatorname{Im}\, \left\{ \left[\frac{i\sinh(\alpha y)}{\alpha}-\frac{i\big(1-\cosh(\alpha)\big)\cosh(\alpha y)}{\alpha\sinh(\alpha)}+i{y}^{2}-iy+2\Omega^{-1}\right]e^{it} \right\}$$
  where $$\alpha=\frac{1}{2}(1+i)\sqrt{2\Omega}$$

but I don't know how to get the solution. Maple didn't show anything. please help.
 A: The equation is the imaginary part of 
$$ \Omega\frac{\partial Z}{\partial t} + 6\lambda\Omega (y^2-y)e^{it} = \frac{\partial^2Z}{\partial y^2} $$
where $$ Z(y,t) = Y(y)e^{it} $$
Then the problem reduces to the ODE
$$ Y'' - i\Omega Y = 6\lambda\Omega(y^2-y) $$
This is linear and non-homogeneous, so we can apply the method of undetermined coefficients. Let $Y = Y_h + Y_p$, with homogeneous and particular solutions, respectively. 
For $Y_h$, the characteristic polynomial has roots $r = \pm \sqrt{i\Omega} = \pm\sqrt{\frac\Omega2}(1+i) = \pm \alpha$, therefore
$$ Y_h = E\cosh (\alpha y) + F\sinh(\alpha y) $$
For $Y_p$, we take the ansatz $Y_p = By^2 + Cy + D$. After substitution
$$ 2B - i\Omega(By^2+Cy+D) = 6\lambda\Omega(y^2-y) $$
Equating coefficients gives $B = -C = 6\lambda i$, $D = 12\lambda\Omega^{-1}$. Therefore your general solution takes the form
$$ Y(y) = 6\lambda \left[c_1\cosh(\alpha y) + c_2\sinh(\alpha y) + iy^2 - iy + \frac2\Omega \right] $$
Now you can use the BC $Y'(0) = Y'(1) = 0$ to determine the remaining constants, which are given in the solution as
$$ c_1 = -\frac{i(1-\cosh\alpha)}{\alpha\sinh\alpha}, \quad c_2 = \frac{i}{\alpha} $$
