Answer to Question #206811 in Differential Equations for Rakul

Question

Solve (D²+2DD'+D'2)z=cos(x+2y)+e^x-2y.

Answer 1

Given, $(D^2 + 2DD'+D'^2)z = \cos(x+2y) + e^{x - 2y}$

The auxiliary equation is $m^2 + 2m +1 =0$. The roots of the auxiliary equation are $m = -1~(\text{twice})$. The complementary function is $C.F = f_1(y-x) + xf_2(y-x).$

The particular integral is

$\begin{aligned} P.I &= \dfrac{1}{D^2 + 2DD'+D'^2} \cos(x + 2y) + \dfrac{1}{D^2 + 2DD'+D'^2} e^{x - 2y}\\ &~~~~~~\text{Replacing $D^2 = -1, DD' =-2, D'^2 = -2^2$ for the first term and }\\ &~~~~~~\text{Replacing $D = 1, D' = -2$ for the second term, we get}\\ &= \dfrac{1}{-1^2+2*(-2)-2^2}\cos(x+2y) + \dfrac{1}{1+2*1*-2+(-2)^2}e^{x-2y}\\ &=\dfrac{-\cos(x+2y)}{9}+e^{x-2y} \end{aligned}$

Hence the solution is $z = \text{C.F + P.I} = f_{1}(y-x)+xf_{2}(y-x)+e^{x-2y}-\dfrac{\cos(x+2y)}{9}$