Answer to Question #291977 in Differential Equations for Omprakash

Question

(D^ 2 -40D' +4D'^2)Z=e^ (2x+y)

Answer 1

$(D^2-4DD'+4D'^2)z=e^{2x+y}\\ \text{The auxilliary equation is }\\ m^2-4m+4=0\\ (m-2)^2=0\\ m=2 \text{ twice}\\ \text{The complementary function is:}\\$

$$f_1(y+2x)+x \cdot f_2(y+2x)$$

To get the particular integral.

$P.I=\frac{1}{(D^2-4DD'+4D'^2)}e^{2x+y}\\ P.I= \frac{1}{(D-2D')^2}e^{2x+y}\\ P.I=\frac{x^2}{1^2\cdot \lfloor{2}}e^{2x+y}\\ P.I=\frac{x^2}{2}e^{2x+y}$

Hence, the complete solution is the sum of the complementary function and the particular integral.

$$=f_1(y+2x)+x \cdot f_2(y+2x)+\frac{x^2}{2}e^{2x+y}$$