Answer to Question #177556 in Differential Equations for Amna

Question

Solve the differential equation by an appropriate method. y''+2y'=x^2+4e^2x

Answer 1

We have given the differential equation,

$y''+2y'=x^2+4e^{2x}$

$(D^2+2D)y = x^2 + 4e^{2x}$

Auxiliary equation : $m^2 + 2m = 0$

$m(m+2) = 0$

$m = 0, m = -2$

Hence, $C.F = C_1 + C_2e^{-2x}$

$P.I = \dfrac{1}{D^2+2D}4e^{2x} + \dfrac{1}{D^2+2D}x^2$

$= \dfrac{4e^{2x}}{2^2+2(2)} + \dfrac{1}{2D}[1+\dfrac{D}{2}]^{-1}x^2$

$= \dfrac{4e^{2x}}{8} + \dfrac{1}{2D}[1-\dfrac{D}{2}+\dfrac{D^2}{4}]x^2$

$= \dfrac{4e^{2x}}{8} + [\dfrac{1}{2D}-\dfrac{1}{4}+\dfrac{D}{8}]x^2$

$P.I = \dfrac{e^{2x}}{2} + \dfrac{x^3}{6} - \dfrac{x^2}{4} + \dfrac{x}{4}$

Hence, complete solution of the differential equation is

$y = C.F + P.I$

$y = C_1 + C_2e^{2x} + \dfrac{e^{2x}}{2} + \dfrac{x^3}{6} - \dfrac{x^2}{4} + \dfrac{x}{4}$