Solve the differential equation; y′′+4y′+5y=xe^(−2x) sinx


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Question Number 105564 by Ar Brandon last updated on 30/Jul/20

$Solve the differential equation;$ $y^{″} + {4 y}^{'} + 5 y = {xe}^{- 2 x} sinx$
	Answered by bobhans last updated on 30/Jul/20
	$Homogenous equation υ^2 +4υ+5=0 →υ=((−4±2i)/2)=−2±i y_h =e^(−2x) {C_1 cos x+C_2 sin x} particular solution y_1 =e^(−2x) cos x⇒y_1 ′=−2e^(−2x) cos x−e^(−2x) sinx y_2 =e^(−2x) sin x⇒y_2 ′=−2e^(−2x) sin x+e^(−2x) cosx W= determinant ((( e^(−2x) cos x e^(−2x) sin x)),((e^(−2x) (−2cos x−sin x) e^(−2x) (−2sin x+cos x))) = e^(−4x) (−2sin xcos x+cos^2 x)−e^(−4x) (−2sinx cos x −sin^2 x) = e^(−4x) u_1 =−∫ ((e^(−2x) sin x(xe^(−2x) sin x)dx)/e^(−4x) ) u_1 =−∫ xsin^2 x dx = −∫x((1/2)−(1/2)cos2x)dx u_1 =−[(1/4)x^2 −((1/4)xcos 2x−(1/8)sin 2x)] u_1 =−(1/4)x^2 +(1/4)xcos 2x−(1/8)sin 2x u_2 =∫ ((e^(−2x) cos x(xe^(−2x) sin x)dx)/e^(−4x) ) u_2 = ∫(1/2)xsin 2x dx = −(1/4)xcos 2x+(1/(16))sin2x y_p = y_1 u_1 +y_2 u_2 y_p = e^(−2x) cos x(−(1/4)x^2 +(1/4)xcos 2x−(1/8)sin2x) + e^(−2x) sin x(−(1/4)xcos 2x+(1/(16))sin 2x) General solution y_G = y_b + y_p .★$
	$H o m o g e n o u s e q u a t i o n$ $υ^{2} + 4 υ + 5 = 0 \to υ = \frac{- 4 \pm 2 i}{2} = - 2 \pm i$ $y_{h} = e^{- 2 x} {C_{1} \cos x + C_{2} \sin x}$ $p a r t i c u l a r s o l u t i o n$ $y_{1} = e^{- 2 x} \cos x \Rightarrow y_{1}^{'} = - 2 e^{- 2 x} \cos x - e^{- 2 x} \sin x$ $y_{2} = e^{- 2 x} \sin x \Rightarrow y_{2}^{'} = - 2 e^{- 2 x} \sin x + e^{- 2 x} \cos x$ $W = \| \begin{matrix} e^{- 2 x} \cos x e^{- 2 x} \sin x \\ e^{- 2 x} (- 2 \cos x - \sin x) e^{- 2 x} (- 2 \sin x + \cos x \end{matrix} \|$ $= e^{- 4 x} (- 2 \sin x \cos x + \cos^{2} x) - e^{- 4 x} (- 2 \sin x$ $\cos x - \sin^{2} x) = e^{- 4 x}$ $u_{1} = - \int \frac{e^{- 2 x} \sin x ({x e}^{- 2 x} \sin x) d x}{e^{- 4 x}}$ $u_{1} = - \int x \sin^{2} x d x = - \int x (\frac{1}{2} - \frac{1}{2} \cos 2 x) d x$ $u_{1} = - [\frac{1}{4} x^{2} - (\frac{1}{4} x \cos 2 x - \frac{1}{8} \sin 2 x)]$ $u_{1} = - \frac{1}{4} x^{2} + \frac{1}{4} x \cos 2 x - \frac{1}{8} \sin 2 x$ $u_{2} = \int \frac{e^{- 2 x} \cos x ({x e}^{- 2 x} \sin x) d x}{e^{- 4 x}}$ $u_{2} = \int \frac{1}{2} x \sin 2 x d x = - \frac{1}{4} x \cos 2 x + \frac{1}{16} \sin 2 x$ $y_{p} = y_{1} u_{1} + y_{2} u_{2}$ $y_{p} = e^{- 2 x} \cos x (- \frac{1}{4} x^{2} + \frac{1}{4} x \cos 2 x - \frac{1}{8} \sin 2 x)$ $+ e^{- 2 x} \sin x (- \frac{1}{4} x \cos 2 x + \frac{1}{16} \sin 2 x)$ $G e n e r a l s o l u t i o n$ $y_{G} = y_{b} + y_{p} . ★$
	Commented by Ar Brandon last updated on 30/Jul/20
	Thanks, Sir Bobhans. Is this Wronskien's method ?��
	Commented by bobhans last updated on 30/Jul/20

	$y e s s i r .$
	Commented by Ar Brandon last updated on 30/Jul/20
	OK, gracias
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