Answer to Question #254689 in Differential Equations for sanket

Question

(D³-2D²D')z=3x²y

Answer 1

Given partial differential equation is,

$(D^3-2D^2D')z=3x^2y$

Let $D' = 1$ then auxiliary equation,

$D^3-2D^2 = 0$

$m^3-2m^2 = 0 \implies m^2(m-2) = 0$

$m = 0, 0, 2$

Then

$C.F. , z = \phi(y)+x\phi(y)+\phi(y+2x)$

P.I. $\frac{1}{(D^3-2D^2D')}3x^2y = 3\frac{1}{(D^3-2D^2D')}x^2y = 3\frac{1}{D^3(1-\frac{2D'}{D})}x^2y$

Expanding biomomially,

$= \frac{3}{D^3}(1-\frac{2D'}{D})^{-1}x^2y = \frac{3}{D^3}(1+\frac{2D'}{D}+....)x^2y$

$= \frac{3}{D^3}[x^2y+\frac{2}{D}x^2]= \frac{3}{D^3}[x^2y+\frac{2x^3}{3}] = \frac{3x^2y}{D^3}+\frac{2x^3}{D^3}$

$= \frac{x^5y}{20}+\frac{x^6}{60}$

Complete solution will be,

$z = \phi(y)+x\phi(y)+\phi(y+2x) + \frac{x^5y}{20}+\frac{x^6}{60}$