Question #108832
1. Using u=XY solve the boundary valued problem

du/dx + 2du/dy = 4u

u(x,0) = 2e^3x + 4e^5x
1
Expert's answer
2020-04-15T08:38:26-0400

Given :

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

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

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

Using (2) in (1);

    XY+2XY=4XY\implies X'Y+2XY'=4XY

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

    (X/4X)=1(Y/2Y)\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)

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


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

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

    X(x)=Ae4kx\implies X(x)=Ae^{4kx} ; ---(3)

A is an arbitrary constant.

Again solving 1(Y/2Y)=k1-(Y'/2Y)=k ; we get;

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

    Y(y)=Be2(1k)y\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)=(Ae4kx)(Be2(1k)y)u(x,y)=X(x)Y(y)=(Ae^{4kx})(Be^{2(1-k)y})

    u(x,y)=Ce4kx+2(1k)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)=2e3xu(x,0)=2e^{3x} ; then

(5)     2e3x=Ce4kx    C=2;k=3/4\implies 2e^{3x}=Ce^{4kx} \implies C=2 ; k=3/4

Thus, the final solution is ;

    u(x,y)=2e3x+y/2\implies u(x,y)=2e^{3x+y/2} (Answer)


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