(d^2 y/dx^2 ) − 4(dy/dx) +11y = e^(−2x) .sin 4x


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Question Number 123107 by benjo_mathlover last updated on 23/Nov/20

$\frac{d^{2} y}{{d x}^{2}} - 4 \frac{d y}{d x} + 11 y = e^{- 2 x} . \sin 4 x$
	Answered by mathmax by abdo last updated on 23/Nov/20
	$h→y^(′′) −4y^′ +11y =0→r^2 −4r+11=0 Δ^′ =4−11=−7 ⇒r_1 =2+i(√7) and r_2 =2−i(√7) y_h =ae^((2+i(√7))x) +b e^((2−i(√7))x) =e^(2x) (αcos((√7)x)+β sin((√7)x)) =αu_1 +βu_2 W(u_1 ,u_2 ) = determinant (((e^(2x) cos((√7)x) e^(2x) sin((√7)x))),((e^(2x) (2cos((√7)x)−(√7)sin((√7))x e^(2x) (2sin((√7)x)+(√7)cos((√7)x)))) =e^(4x) {2cos((√7)x)sin((√7)x)+(√7)cos^2 ((√7)x)} −e^(4x) {2sin((√7)x)cos((√7)x)−(√7)sin^2 ((√7)x)} =(√7)e^(4x) ≠0 W_1 = determinant (((0 e^(2x) sin((√7)x))),((e^(−2x) sin(4x) e^(2x) (2sin((√7)x)+(√7)cos((√7)x)))) =−sin(4x)sin((√7)x) W_2 = determinant (((e^(2x) cos((√7)x) 0)),((e^(2x) (2cos((√7)x)−(√7)sin((√7)x) e^(−2x) sin(4x)))) =cos((√7)x)sin(4x) v_1 =∫ (w_1 /w)dx =−∫ ((sin(4x)sin((√7)x))/( (√7)e^(4x) ))dx =(1/( (√7)))∫ e^(−4x) sin(4x)sin((√7)x)dx (eazy to find by use sina.sinb=....) v_2 =∫ (w_2 /w)dx =∫ ((cos((√7)x)sin(4x))/( (√7)e^(4x) ))dx =(1/( (√7)))∫ e^(−4x) cos((√7)x)sin(4x)dx =.....(use trigo.formulae) ⇒y_p =u_1 v_1 +u_2 v_2 the general solution is y =y_h +y_p$
	$h \to y^{^{″}} - {4 y}^{^{'}} + 11 y = 0 \to r^{2} - 4 r + 11 = 0$ $Δ^{^{'}} = 4 - 11 = - 7 \Rightarrow r_{1} = 2 + i \sqrt{7} and r_{2} = 2 - i \sqrt{7}$ $y_{h} = {ae}^{(2 + i \sqrt{7}) x} + b e^{(2 - i \sqrt{7}) x} = e^{2 x} (α \cos (\sqrt{7} x) + β \sin (\sqrt{7} x)) = α u_{1} + β u_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{2 x} \cos (\sqrt{7} x) e^{2 x} \sin (\sqrt{7} x) \\ e^{2 x} (2 \cos (\sqrt{7} x) - \sqrt{7} \sin (\sqrt{7}) x e^{2 x} (2 \sin (\sqrt{7} x) + \sqrt{7} \cos (\sqrt{7} x) \end{matrix} \|$ $= e^{4 x} {2 \cos (\sqrt{7} x) \sin (\sqrt{7} x) + \sqrt{7} \cos^{2} (\sqrt{7} x)}$ $- e^{4 x} {2 \sin (\sqrt{7} x) \cos (\sqrt{7} x) - \sqrt{7} \sin^{2} (\sqrt{7} x)}$ $= \sqrt{7} e^{4 x} \neq 0$ $W_{1} = \| \begin{matrix} 0 e^{2 x} \sin (\sqrt{7} x) \\ e^{- 2 x} \sin (4 x) e^{2 x} (2 \sin (\sqrt{7} x) + \sqrt{7} \cos (\sqrt{7} x) \end{matrix} \|$ $= - \sin (4 x) \sin (\sqrt{7} x)$ $W_{2} = \| \begin{matrix} e^{2 x} \cos (\sqrt{7} x) 0 \\ e^{2 x} (2 \cos (\sqrt{7} x) - \sqrt{7} \sin (\sqrt{7} x) e^{- 2 x} \sin (4 x) \end{matrix} \|$ $= \cos (\sqrt{7} x) \sin (4 x)$ $v_{1} = \int \frac{w_{1}}{w} dx = - \int \frac{\sin (4 x) \sin (\sqrt{7} x)}{\sqrt{7} e^{4 x}} dx = \frac{1}{\sqrt{7}} \int e^{- 4 x} \sin (4 x) \sin (\sqrt{7} x) dx$ $(eazy to find by use sina . sinb = . . . .)$ $v_{2} = \int \frac{w_{2}}{w} dx = \int \frac{\cos (\sqrt{7} x) \sin (4 x)}{\sqrt{7} e^{4 x}} dx = \frac{1}{\sqrt{7}} \int e^{- 4 x} \cos (\sqrt{7} x) \sin (4 x) dx$ $= . . . . . (use trigo . formulae) \Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2}$ $the general solution is y = y_{h} + y_{p}$
	Commented by peter frank last updated on 24/Nov/20

	$thank you$
	Commented by talminator2856791 last updated on 08/Dec/20

	$how do you make big line like that$
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