solve y^(′′) −y^′ +2 =x^2 e^(−x) with y(0) =1 and y^′ (0) =−1


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Question Number 95465 by mathmax by abdo last updated on 25/May/20

$solve y^{^{″}} - y^{^{'}} + 2 = x^{2} e^{- x} with y (0) = 1 and y^{^{'}} (0) = - 1$
	Answered by mathmax by abdo last updated on 26/May/20
	$(he)→y^(′′) −y^′ +2 =0→r^2 −r+2 =0 Δ =−7 ⇒r_1 =((1+i(√7))/2) and r_2 =((1−i(√7))/2) ⇒ y(x) =a e^(((1+i(√7))/2)x) +b e^((1−i(√7))/2) =e^(x/2) (α cos((((√7)x)/2))+βsin((((√7)x)/2))) =α e^(x/2) cos(((√7)/2)x) +β e^(x/2) sin(((√7)/2)x) =αu_1 (x)+βu_2 (x) W(u_1 ,u_2 ) = determinant (((u_1 u_2 )),((u_1 ^′ u_2 ^′ ))) = determinant (((e^(x/2) cos((((√7)x)/2)) e^(x/2) sin((((√7)x)/2)))),(((1/2)e^(x/2) cos((((√7)x)/2))−((√7)/2)e^(x/2) sin((((√7)x)/2)) (1/2)e^(x/2) sin((((√7)x)/2))+((√7)/2)e^(x/2) cos((((√7)x)/2))))) =e^(x/2) cos((((√7)x)/2)){(1/2)e^(x/2) sin((((√7)x)/2))+((√7)/2)e^(x/2) cos((((√7)x)/2))−e^(x/2) sin((((√7)x)/2)){(1/2)e^(x/2) cos((((√7)x)/2))−((√7)/2)e^(x/2) sin((((√7)x)/2))}$
	$(he) \to y^{^{″}} - y^{^{'}} + 2 = 0 \to r^{2} - r + 2 = 0$ $Δ = - 7 \Rightarrow r_{1} = \frac{1 + i \sqrt{7}}{2} and r_{2} = \frac{1 - i \sqrt{7}}{2} \Rightarrow$ $y (x) = a e^{\frac{1 + i \sqrt{7}}{2} x} + b e^{\frac{1 - i \sqrt{7}}{2}} = e^{\frac{x}{2}} (α \cos (\frac{\sqrt{7} x}{2}) + β \sin (\frac{\sqrt{7} x}{2}))$ $= α e^{\frac{x}{2}} \cos (\frac{\sqrt{7}}{2} x) + β e^{\frac{x}{2}} \sin (\frac{\sqrt{7}}{2} x) = α u_{1} (x) + β u_{2} (x)$ $W (u_{1}, u_{2}) = \| \begin{matrix} u_{1} u_{2} \\ u_{1}^{^{'}} u_{2}^{^{'}} \end{matrix} \| = \| \begin{matrix} e^{\frac{x}{2}} \cos (\frac{\sqrt{7} x}{2}) e^{\frac{x}{2}} \sin (\frac{\sqrt{7} x}{2}) \\ \frac{1}{2} e^{\frac{x}{2}} \cos (\frac{\sqrt{7} x}{2}) - \frac{\sqrt{7}}{2} e^{\frac{x}{2}} \sin (\frac{\sqrt{7} x}{2}) \frac{1}{2} e^{\frac{x}{2}} \sin (\frac{\sqrt{7} x}{2}) + \frac{\sqrt{7}}{2} e^{\frac{x}{2}} \cos (\frac{\sqrt{7} x}{2}) \end{matrix} \|$ $= e^{\frac{x}{2}} \cos (\frac{\sqrt{7} x}{2}) {\frac{1}{2} e^{\frac{x}{2}} \sin (\frac{\sqrt{7} x}{2}) + \frac{\sqrt{7}}{2} e^{\frac{x}{2}} \cos (\frac{\sqrt{7} x}{2}) - e^{\frac{x}{2}} \sin (\frac{\sqrt{7} x}{2}) {\frac{1}{2} e^{\frac{x}{2}} \cos (\frac{\sqrt{7} x}{2}) - \frac{\sqrt{7}}{2} e^{\frac{x}{2}} \sin (\frac{\sqrt{7} x}{2})}$
	Commented by mathmax by abdo last updated on 26/May/20

	$W (u_{1}, u_{2}) = \frac{1}{2} e^{x} \cos (\frac{\sqrt{7} x}{2}) \sin (\frac{\sqrt{7} x}{2}) + \frac{\sqrt{7}}{2} e^{x} \cos^{2} (\frac{\sqrt{7} x}{2})$ $- \frac{1}{2} e^{x} \sin (\frac{\sqrt{7} x}{2}) \cos (\frac{\sqrt{7} x}{2}) + \frac{\sqrt{7}}{2} e^{x} \sin^{2} (\frac{\sqrt{7} x}{2})$ $= \frac{\sqrt{7}}{2} e^{x}$
	Commented by mathmax by abdo last updated on 26/May/20

	$W_{1} = \| \begin{matrix} 0 u_{2} \\ x^{2} e^{- x} u_{2}^{^{'}} \end{matrix} \| = - x^{2} e^{- x} e^{\frac{x}{2}} \sin (\frac{\sqrt{7} x}{2}) = - x^{2} e^{- \frac{x}{2}} \sin (\frac{\sqrt{7} x}{2})$ $W_{2} = \| \begin{matrix} u_{1} 0 \\ u_{1}^{^{'}} x^{2} e^{- x} \end{matrix} \| = x^{2} e^{- x} e^{\frac{x}{2}} \cos (\frac{\sqrt{7} x}{2}) = x^{2} e^{- \frac{x}{2}} \cos (\frac{\sqrt{7} x}{2})$ $v_{1} = \int \frac{w_{1}}{w} dx = - \int \frac{x^{2} e^{- \frac{x}{2}} \sin (\frac{\sqrt{7} x}{2})}{\frac{\sqrt{7}}{2} e^{x}} dx = - \frac{2}{\sqrt{7}} \int x^{2} e^{- \frac{3}{2} x} \sin (\frac{\sqrt{7} x}{2}) dx$ $v_{2} = \int \frac{w_{2}}{w} dx = \int \frac{x^{2} e^{- \frac{x}{2}} \cos (\frac{\sqrt{7} x}{2})}{\frac{\sqrt{7}}{2} e^{x}} dx = \frac{2}{\sqrt{7}} \int x^{2} e^{- \frac{3}{2} x} \cos (\frac{\sqrt{7} x}{2}) dx$ $y_{p} = u_{1} v_{1} + u_{2} v_{2} and y = y_{h} + y_{p} . . . .$
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