Answer to Question #120992 in Differential Equations for Jhansi

Question

sove pde d²z/dx²+d²z/dxdy+dz/dy-Z=e^-x

Answer 1

$\frac{\partial^{2}z}{\partial^{2}x^{2}} + \frac{\partial^{2}z}{\partial x\partial y} + \frac{\partial z}{\partial y} -z = e^{-x}$

Equation can be written as:

$(D^{2} + DD' + D' - 1)z = e^{-x}$

$(D + 1)(D + D' -1)z = e^{-x}$

Comparing above equation with

$(D -m_1 D' - \alpha_1)(D -m_2 D' - \alpha_2) = F(x,y)$

so $m_1 = 0, m_2 = -1, \alpha_1 = -1, \alpha_2 = 2$

So C.F. of the equation is

$C.F. = e^{-x}\phi_1(y) + e^{x}\phi_2(y-x)$

then PI is calculated as

$P.I. = \frac{1}{(D+1)(D+D'-1)}e^{-x} = \frac{1}{(D+1)(D-1)}e^{-x}$ as exponential has no y.

$P.I. = \frac{1}{(D+1)(D-1)}e^{-x}$ $= e^{-x} \frac{1}{D(D-2)}1$ $= -\frac{xe^{-x}}{2}$

So solution to the problem is

$z(x,y) = e^{-x}\phi_1(y) + e^{x}\phi_2(y-x) - \frac{xe^{-x}}{2}$

Solution can be verified by taking partial derivatives and put them in differential equation.