Answer to Question #108832 in Differential Equations for harry

Question

1. Using u=XY solve the boundary valued problem

du/dx + 2du/dy = 4u

u(x,0) = 2e^3x + 4e^5x

Answer 1

Given :

$\partial u/\partial x+2\partial u/\partial y=4u$ ---(1)

Now, let us assume that $u(x,y)=X(x)Y(y)$

Thus; $\partial u/\partial x=X'Y; \partial u/\partial y=XY'$ ---(2)

Using (2) in (1);

$\implies X'Y+2XY'=4XY$

$\implies (X'/4X)+(Y'/2Y)=1$

$\implies (X'/4X)=1-(Y'/2Y)$

As LHS is a function of x alone and RHS is a function of y alone.

Then, LHS=RHS=constant=k(say)

$\implies (X'/4X)=1-(Y'/2Y)=k$

Solving $(X'/4X)=k$ ; we get;

$\int X'dx/X=\int 4kdx \implies lnX=4kx+c$ ; where c is constant of integration.

$\implies X(x)=Ae^{4kx}$ ; ---(3)

A is an arbitrary constant.

Again solving $1-(Y'/2Y)=k$ ; we get;

$\int Y'dy/Y=\int 2(1-k)dy \implies lnY=2(1-k)y+c$ ; where c is constant of integration.

$\implies Y(y)=Be^{2(1-k)y}$ ---(4)

B is an arbitrary constant.

Using (3) and (4), we get;

$u(x,y)=X(x)Y(y)=(Ae^{4kx})(Be^{2(1-k)y})$

$\implies u(x,y)=Ce^{4kx+2(1-k)y}$ ; ---(5)

where C=AB is a arbitrary constant.

The boundary condition is incorrect, there cannot be two terms.

If $u(x,0)=2e^{3x}$ ; then

(5) $\implies 2e^{3x}=Ce^{4kx} \implies C=2 ; k=3/4$

Thus, the final solution is ;

$\implies u(x,y)=2e^{3x+y/2}$ (Answer)