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Question Number 123593 by Bird last updated on 26/Nov/20

solve y^(′′) +2y^′ −y=xe^(−x^2 )

$${solve}\:{y}^{''} \:+\mathrm{2}{y}^{'} \:−{y}={xe}^{−{x}^{\mathrm{2}} } \\ $$

Answered by mathmax by abdo last updated on 27/Nov/20

h→r^2 +2r−1=0→Δ^′ =1+1=2 ⇒r_1 =−1+(√2) and r_2 =−1−(√2) y_h =ae^((−1+(√2))x) +b e^((−1−(√2))x ) =au_1 +bu_2 W(u_1 ,u_2 )= determinant (((e^((−1+(√2))x) e^((−1−(√2))x) )),(((−1+(√2))e^((−1+(√2))) (−1−(√2))e^((−1−(√2))) ))) =(−1−(√2))e^(−2x) +(1−(√2))e^(−2x) =−2(√2)e^(−2x) ≠0 W_1 = determinant (((o e^((−1−(√2))x) )),((xe^(−x^2 ) (−1−(√2))e^((−1−(√2))x) )))=xe^(−x^2 ) e^((−1−(√2))x) W_2 = determinant (((e^((−1+(√2))x) 0)),(((−1+(√2))e^((−1+(√2))x) xe^(−x^2 ) )))=xe^(−x^2 ) e^((−1+(√2))x) v_1 =∫ (w_1 /w)dx =∫ ((xe^(−x^2 ) e^((−1−(√2))x) )/(−2(√2)e^(−2x) ))dx =−(1/(2(√2)))∫ xe^(−x^2 ) e^((1−(√2))x) dx v_2 =∫ (w_2 /w)dx =∫ ((xe^(−x^2 ) e^((−1+(√2))x) )/(−2(√2)e^(−2x) ))dx =−(1/(2(√2)))∫ xe^(−x^2 ) e^((1+(√2))x) dx ⇒y_p =u_1 v_1 +u_2 v_2 and general soution is y =y_h +y_p

$$\mathrm{h}\rightarrow\mathrm{r}^{\mathrm{2}} \:+\mathrm{2r}−\mathrm{1}=\mathrm{0}\rightarrow\Delta^{'} \:=\mathrm{1}+\mathrm{1}=\mathrm{2}\:\Rightarrow\mathrm{r}_{\mathrm{1}} =−\mathrm{1}+\sqrt{\mathrm{2}}\:\mathrm{and}\:\mathrm{r}_{\mathrm{2}} =−\mathrm{1}−\sqrt{\mathrm{2}} \\ $$$$\mathrm{y}_{\mathrm{h}} =\mathrm{ae}^{\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)\mathrm{x}} \:+\mathrm{b}\:\mathrm{e}^{\left(−\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{x}\:} \:=\mathrm{au}_{\mathrm{1}} +\mathrm{bu}_{\mathrm{2}} \\ $$$$\mathrm{W}\left(\mathrm{u}_{\mathrm{1}} ,\mathrm{u}_{\mathrm{2}} \right)=\begin{vmatrix}{\mathrm{e}^{\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)\mathrm{x}} \:\:\:\:\:\:\:\:\:\:\mathrm{e}^{\left(−\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{x}} }\\{\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)\mathrm{e}^{\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:\:\:\:\left(−\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{e}^{\left(−\mathrm{1}−\sqrt{\mathrm{2}}\right)} }\end{vmatrix} \\ $$$$=\left(−\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{e}^{−\mathrm{2x}} \:+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{e}^{−\mathrm{2x}} \:=−\mathrm{2}\sqrt{\mathrm{2}}\mathrm{e}^{−\mathrm{2x}} \:\neq\mathrm{0} \\ $$$$\mathrm{W}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{o}\:\:\:\:\:\:\:\:\:\:\:\mathrm{e}^{\left(−\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{x}} }\\{\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \:\:\:\:\:\:\:\left(−\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{e}^{\left(−\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{x}} }\end{vmatrix}=\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{e}^{\left(−\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{x}} \\ $$$$\mathrm{W}_{\mathrm{2}} =\begin{vmatrix}{\mathrm{e}^{\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)\mathrm{x}} \:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)\mathrm{e}^{\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)\mathrm{x}} \:\:\:\:\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } }\end{vmatrix}=\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{e}^{\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)\mathrm{x}} \\ $$$$\mathrm{v}_{\mathrm{1}} =\int\:\frac{\mathrm{w}_{\mathrm{1}} }{\mathrm{w}}\mathrm{dx}\:=\int\:\:\frac{\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{e}^{\left(−\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{x}} }{−\mathrm{2}\sqrt{\mathrm{2}}\mathrm{e}^{−\mathrm{2x}} }\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\:\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{e}^{\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{x}} \mathrm{dx} \\ $$$$\mathrm{v}_{\mathrm{2}} =\int\:\frac{\mathrm{w}_{\mathrm{2}} }{\mathrm{w}}\mathrm{dx}\:=\int\:\:\frac{\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{e}^{\left(−\mathrm{1}+\sqrt{\mathrm{2}}\right)\mathrm{x}} }{−\mathrm{2}\sqrt{\mathrm{2}}\mathrm{e}^{−\mathrm{2x}} }\mathrm{dx}\:=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int\:\mathrm{xe}^{−\mathrm{x}^{\mathrm{2}} } \:\mathrm{e}^{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\mathrm{x}} \:\mathrm{dx} \\ $$$$\Rightarrow\mathrm{y}_{\mathrm{p}} =\mathrm{u}_{\mathrm{1}} \mathrm{v}_{\mathrm{1}} \:+\mathrm{u}_{\mathrm{2}} \mathrm{v}_{\mathrm{2}} \\ $$$$\mathrm{and}\:\mathrm{general}\:\mathrm{soution}\:\mathrm{is}\:\mathrm{y}\:=\mathrm{y}_{\mathrm{h}} +\mathrm{y}_{\mathrm{p}} \\ $$

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