Question

) Solve the ODE y+4y+4y= e−2x using the method of variation of parameters x3

Answer 1

$$Solution$$

Given that:-

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

In operator form

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

Where $f(D)=(D^{2}+4D+4)$

$R=e^{-2x}$

The A.E is

$f(m)=0 \implies m^2+4m+4=0 \implies (m+2)^2=0\\ \implies m = -2, -2\\ \therefore y_c=C_1e^{-2x}+C_2xe^{-2x}\\ y_p=AU+BV\\ U=e^{-2x}\ ;\ U'=-2e^{-2x}\ ;\ \\ V=xe^{-2x}\ ;\ V'=-2xe^{-2x}+e^{-2x}\ \\ UV'+VU'=-2xe^{-4x}+e^{-4x}+2xe^{-4x}=e^{-4x}\\ \\ \therefore A=\intop \frac{-VR}{UV'-VU'}dx= - \intop \frac{xe^{-4x}}{e^{-4x}}dx = -\intop xdx\\ = -\intop xdx + \intop [\intop xdx]dx\\ = -\frac{x^2}{2}+\intop \frac{x^2}{2}dx = -\frac{x^2}{2} + \frac12 \intop x^2dx\\ \implies -\frac{x^2}{2} + \frac12 \cdot \frac{x^3}{3} = -\frac{x^2}{2} + \frac {x^3}{6}\\ \therefore B=\intop \frac{-UR}{UV'-VU'}dx= \intop \frac{e^{-2x} \cdot e^{-2x}}{e^{-4x}}dx= \intop dx\\ = \intop dx-\intop [\intop dx]dx\\ =x-\intop xdx = x-\frac{x^2}{2}\\ \implies y=y_c+y_{p}=C_1e^{-2x}+C_2xe^{-2x}+(x-\frac{x^2}{2})e^{-2x}+ (\frac{x^3}{6} +\frac{x^2}{2})e^{-2x}$