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Question Number 133353 by Ahmed1hamouda last updated on 21/Feb/21

Commented by Ahmed1hamouda last updated on 21/Feb/21

$s o l v e t h e d i f f e r e n t i a l e q u a t i o n s$
	Answered by mathmax by abdo last updated on 21/Feb/21
	$2)y^(′′) +y^′ +y=(1+sinx)^2 h)→r^2 +r+1 =0 →Δ=−3 ⇒r_1 =((−1+i(√3))/2) and r_2 =((−1−i(√3))/2) ⇒y_h =a e^((((−1+i(√3))/2))x) +be^((((−1−i(√3))/2))x) =e^(−(x/2)) {αcos(((√3)/2)x)+βsin(((√3)/2)x)} =αu_1 +βu_2 W(u_1 ,u_2 )= determinant ((( e^(−(x/2)) cos(((√3)/2)x) e^(−(x/2)) sin(((√3)/2)x))),((e^(−(x/2)) (−(1/2)cos(((√3)/2)x)−((√3)/2)sin(((√3)/2)x)) e^(−(x/2)) (−(1/2)sin(((√3)/2)x)+((√3)/2)cos(((√3)/2)x)))) =e^(−x) {−(1/2)cos(((√3)/2)x)sin(((√3)/2)x)+((√3)/2)cos^2 (((√3)/2)x)} −e^(−x) {−(1/2)cos(((√3)/2)x)sin(((√3)/2)x)−((√3)/2)sin^2 (((√3)/2)x)} =((√3)/2)e^(−x) ≠0 W_1 = determinant (((o e^(−(x/2)) sin(((√3)/2)x))),(((1+sinx)^2 e^(−(x/2)) (−(1/2)sin(((√3)/2)x)+((√3)/2)cos(((√3)/2)x)))) =−e^(−(x/2)) (1+sinx)^2 sin(((√3)/2)x) W_2 = determinant (((e^(−(x/2)) cos(((√3)/2)x) 0)),((e^(−(x/2)) (....) (1+sinx)^2 ))) =e^(−(x/2)) (1+sinx)^2 cos(((√3)/2)x) V_1 =∫ (W_1 /W)dx =−∫ ((e^(−(x/2)) (1+sinx)^2 sin(((√3)/2)x))/(((√3)/2)e^(−x) )) =−(2/( (√3)))∫ e^(x/2) (1+sinx)^2 sin(((√3)/2)x) =−(2/( (√3)))∫ e^(x/2) (1+2sinx +((1−cos(2x))/2))sin(((√3)/2)x)dx =−(1/( (√3)))∫ e^(x/2) {3+4sinx −cos(2x))}sin(((√3)/2)x)dx =−(√3)∫ e^(x/2) sin(((√3)/2)x)dx−(4/( (√3)))∫ e^(x/2) sinx sin(((√3)/2)x)dx+(1/( (√3)))∫ e^(x/2) cos(2x)sin(((√3)/2)x)dx =.... V_2 =∫ (W_2 /W)dx =∫ ((e^(−(x/2)) (1+sinx)^2 cos(((√3)/2)x))/(((√3)/2)e^(−x) )) =(2/( (√3)))∫ e^(x/2) (1+2sinx +((1−cos(2x))/2))cos(((√3)/2)x)dx =(1/( (√3)))∫ e^(x/2) (3+4sinx−cos(2x))cos(((√3)/2)x)dx=..... ⇒y_p =u_1 v_1 +u_2 v_2 and general solution is y =y_p +y_h$
	$2) y^{^{″}} + y^{^{'}} + y = {(1 + sinx)}^{2}$ $h) \to r^{2} + r + 1 = 0 \to Δ = - 3 \Rightarrow r_{1} = \frac{- 1 + i \sqrt{3}}{2} and r_{2} = \frac{- 1 - i \sqrt{3}}{2}$ $\Rightarrow y_{h} = a e^{(\frac{- 1 + i \sqrt{3}}{2}) x} + {be}^{(\frac{- 1 - i \sqrt{3}}{2}) x} = e^{- \frac{x}{2}} {α \cos (\frac{\sqrt{3}}{2} x) + β \sin (\frac{\sqrt{3}}{2} x)}$ $= α u_{1} + β u_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{- \frac{x}{2}} \cos (\frac{\sqrt{3}}{2} x) e^{- \frac{x}{2}} \sin (\frac{\sqrt{3}}{2} x) \\ e^{- \frac{x}{2}} (- \frac{1}{2} \cos (\frac{\sqrt{3}}{2} x) - \frac{\sqrt{3}}{2} \sin (\frac{\sqrt{3}}{2} x)) e^{- \frac{x}{2}} (- \frac{1}{2} \sin (\frac{\sqrt{3}}{2} x) + \frac{\sqrt{3}}{2} \cos (\frac{\sqrt{3}}{2} x) \end{matrix} \|$ $= e^{- x} {- \frac{1}{2} \cos (\frac{\sqrt{3}}{2} x) \sin (\frac{\sqrt{3}}{2} x) + \frac{\sqrt{3}}{2} \cos^{2} (\frac{\sqrt{3}}{2} x)}$ $- e^{- x} {- \frac{1}{2} \cos (\frac{\sqrt{3}}{2} x) \sin (\frac{\sqrt{3}}{2} x) - \frac{\sqrt{3}}{2} \sin^{2} (\frac{\sqrt{3}}{2} x)}$ $= \frac{\sqrt{3}}{2} e^{- x} \neq 0$ $W_{1} = \| \begin{matrix} o e^{- \frac{x}{2}} \sin (\frac{\sqrt{3}}{2} x) \\ {(1 + sinx)}^{2} e^{- \frac{x}{2}} (- \frac{1}{2} \sin (\frac{\sqrt{3}}{2} x) + \frac{\sqrt{3}}{2} \cos (\frac{\sqrt{3}}{2} x) \end{matrix} \|$ $= - e^{- \frac{x}{2}} {(1 + sinx)}^{2} \sin (\frac{\sqrt{3}}{2} x)$ $W_{2} = \| \begin{matrix} e^{- \frac{x}{2}} \cos (\frac{\sqrt{3}}{2} x) 0 \\ e^{- \frac{x}{2}} (. . . .) {(1 + sinx)}^{2} \end{matrix} \|$ $= e^{- \frac{x}{2}} {(1 + sinx)}^{2} \cos (\frac{\sqrt{3}}{2} x)$ $V_{1} = \int \frac{W_{1}}{W} dx = - \int \frac{e^{- \frac{x}{2}} {(1 + sinx)}^{2} \sin (\frac{\sqrt{3}}{2} x)}{\frac{\sqrt{3}}{2} e^{- x}}$ $= - \frac{2}{\sqrt{3}} \int e^{\frac{x}{2}} {(1 + sinx)}^{2} \sin (\frac{\sqrt{3}}{2} x)$ $= - \frac{2}{\sqrt{3}} \int e^{\frac{x}{2}} (1 + 2 sinx + \frac{1 - \cos (2 x)}{2}) \sin (\frac{\sqrt{3}}{2} x) dx$ $= - \frac{1}{\sqrt{3}} \int e^{\frac{x}{2}} {3 + 4 sinx - \cos (2 x))} \sin (\frac{\sqrt{3}}{2} x) dx$ $= - \sqrt{3} \int e^{\frac{x}{2}} \sin (\frac{\sqrt{3}}{2} x) dx - \frac{4}{\sqrt{3}} \int e^{\frac{x}{2}} sinx \sin (\frac{\sqrt{3}}{2} x) dx + \frac{1}{\sqrt{3}} \int e^{\frac{x}{2}} \cos (2 x) \sin (\frac{\sqrt{3}}{2} x) dx$ $= . . . .$ $V_{2} = \int \frac{W_{2}}{W} dx = \int \frac{e^{- \frac{x}{2}} {(1 + sinx)}^{2} \cos (\frac{\sqrt{3}}{2} x)}{\frac{\sqrt{3}}{2} e^{- x}}$ $= \frac{2}{\sqrt{3}} \int e^{\frac{x}{2}} (1 + 2 sinx + \frac{1 - \cos (2 x)}{2}) \cos (\frac{\sqrt{3}}{2} x) dx$ $= \frac{1}{\sqrt{3}} \int e^{\frac{x}{2}} (3 + 4 sinx - \cos (2 x)) \cos (\frac{\sqrt{3}}{2} x) dx = . . . . .$ $\Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2} and general solution is$ $y = y_{p} + y_{h}$
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