Given :
∂u/∂x+2∂u/∂y=4u ---(1)
Now, let us assume that u(x,y)=X(x)Y(y)
Thus; ∂u/∂x=X′Y;∂u/∂y=XY′ ---(2)
Using (2) in (1);
⟹X′Y+2XY′=4XY
⟹(X′/4X)+(Y′/2Y)=1
⟹(X′/4X)=1−(Y′/2Y)
As LHS is a function of x alone and RHS is a function of y alone.
Then, LHS=RHS=constant=k(say)
⟹(X′/4X)=1−(Y′/2Y)=k
Solving (X′/4X)=k ; we get;
∫X′dx/X=∫4kdx⟹lnX=4kx+c ; where c is constant of integration.
⟹X(x)=Ae4kx ; ---(3)
A is an arbitrary constant.
Again solving 1−(Y′/2Y)=k ; we get;
∫Y′dy/Y=∫2(1−k)dy⟹lnY=2(1−k)y+c ; where c is constant of integration.
⟹Y(y)=Be2(1−k)y ---(4)
B is an arbitrary constant.
Using (3) and (4), we get;
u(x,y)=X(x)Y(y)=(Ae4kx)(Be2(1−k)y)
⟹u(x,y)=Ce4kx+2(1−k)y ; ---(5)
where C=AB is a arbitrary constant.
The boundary condition is incorrect, there cannot be two terms.
If u(x,0)=2e3x ; then
(5) ⟹2e3x=Ce4kx⟹C=2;k=3/4
Thus, the final solution is ;
⟹u(x,y)=2e3x+y/2 (Answer)
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