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?