help me with solving this..

Ive got this DE: 8(du/dx)+4(du/dy)=9u

I separate it with u=X(x)Y(y)

Let me give you my thoughts:

8*X' + k^2*X = 0

4*Y' - (k^2 + 9)*Y = 0

These yields:

solving the equations above for k^2 >0 , =0 or <0:

k^2 >0: yields:

X(x)= A1*e^-(( (K^2)/8)x) and Y(y)= B1*e^-(( (k^2 +9)/4)y)

k^2 =0: yields:

X(x)=A2 and Y(y)=B2*e^((9/4)y)

k^2 <0: yields:

X(x)= A1*e^(( (K^2)/8)x) and Y(y)= B1*e^(( (k^2 +9)/4)y)

Then U(x,t) is X(x)Y(y) for each of the threee cases:

Then I can write (by laws of superposition) U(x,t)= U(x,t)1 +U(x,t)2 +U(x,t)3

where U(x,t)1 is U(x,t) for case k^2 >0 etc...

Then I have my boundary condition:

u(0,y) = 33*exp(-y) +50*exp(4y)

How do I solve the rest?