Answer to Question #146714 in Differential Equations for Harsh mishra

Question

(D-D'+2)(D+D'-1)=e^x-y -x^2y

Answer 1

Given differential equation is $(D-D'+2)(D+D-1) = e^{x-y}-x^2y$

C.F. of the equation of type $(D-mD'+k)z = 0$ is $u = e^{kx}(y+mx)$

So the C.F. of the equation is, $C.F. = e^{-2x}(y+x) +e^x(y-x)$

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

For exponential part,

$\frac{1}{(D-D'+2)(D+D-1) } e^{x-y} = \frac{1}{(1+1+2)(1-1-1) } e^{x-y} = \frac{-1}{4}e^{x-y}$

For polynomial part,

$\frac{1}{(D-D'+2)(D+D-1) }( x^2y) = \frac{1}{D^2-D'^2+2D-2}x^2y$

$= -2(1 - \frac{D^2-D'^2+2D}{2})^{-1} x^2y$

$=-2 (1+(\frac{D^2-D'^2+2D}{2})+(\frac{D^2-D'^2+2D}{2})^2 + .....)x^2y$

Removing the terms which will give me zero, $i.e. D^3, D'^2.......$

$= -2(1+(\frac{D^2-D'^2+2D}{2}) +D^2)x^2y$

$= -(x^2y+\frac{1}{2}(2y+4xy)+\frac{1}{4}\times 8 ) = -(4+2y+4xy+2x^2y)$

So, Complete solution is,

$u(z) = e^{-2x}(y+x) +e^x(y-x)-\frac{1}{4}e^{x-y}-(4+2y+4xy+2x^2y)$