Answer to Question #205556 in Differential Equations for Devansh

Question

(D²-D')z=e^x-ysin(x+2y)

Answer 1

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

Let $D' = 1$

Then auxiliary equation will be, $m^2-1=0 \implies m = 1,-1$

CF of the equation, $z = f_1(y+x)+f_2(y-x)$

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

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

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

Now Multiplying by D in numerator and denominator,

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

Putting values, $D^2 = -1, DD' = -2$

$PI= e^{x-y}\frac{D}{-2+2+D}sin(x+2y) = e^{x-y}sin(x+2y)$

Hence, complete solution will be,

$z = f_1(y+x)+f_2(y-x) + e^{x-y}sin(x+2y)$