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Question Number 97626 by mathmax by abdo last updated on 08/Jun/20

$$\mathrm{solve}{\mathrm{y}}^{{}^{″}}-{2\mathrm{y}}^{{}^{\prime }}+\mathrm{y}={\mathrm{x}}^2\mathrm{with}{\mathrm{y}}^{{}^{\prime }}\left(0\right)=\mathrm{y}\left(0\right)=-1$$

Commented by bemath last updated on 09/Jun/20

$${\lambda }^2-2\lambda +1=0$$$$\lambda =1,1\Rightarrow {\mathrm{y}}_{\mathrm{h}}={\mathrm{Ae}}^{\mathrm{x}}+{\mathrm{Bxe}}^{\mathrm{x}}$$$$\mathrm{particular}{\mathrm{y}}_{\mathrm{p}}={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}$$$${\mathrm{y}}^{\prime }=2\mathrm{ax}+\mathrm{b},{\mathrm{y}}^{″}=2\mathrm{a}$$$$\Rightarrow 2\mathrm{a}-4\mathrm{ax}-2\mathrm{b}+{\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}={\mathrm{x}}^2$$$$\Rightarrow \mathrm{a}=1,\mathrm{b}=4,\mathrm{c}=2\mathrm{b}-2\mathrm{a}=6$$$${\mathrm{y}}_{\mathrm{p}}={\mathrm{x}}^2+4\mathrm{x}+6$$$$\mathrm{generall}\mathrm{solution}$$$$\mathrm{y}={\mathrm{Ae}}^{\mathrm{x}}+{\mathrm{Bxe}}^{\mathrm{x}}+{\mathrm{x}}^2+4\mathrm{x}+6$$$$\mathrm{y}\left(0\right)=-1=\mathrm{A}+6,\mathrm{A}=-7$$$${\mathrm{y}}^{\prime }=-{7\mathrm{e}}^{\mathrm{x}}+{\mathrm{Be}}^{\mathrm{x}}+{\mathrm{Bxe}}^{\mathrm{x}}+2\mathrm{x}+4$$$${\mathrm{y}}^{\prime }\left(0\right)=-1=-7+\mathrm{B}+4$$$$\mathrm{B}=2\iff \therefore {\mathrm{y}}_{\mathrm{c}}=-{7\mathrm{e}}^{\mathrm{x}}+{2\mathrm{xe}}^{\mathrm{x}}+{\mathrm{x}}^2+4\mathrm{x}+6$$

Commented by bobhans last updated on 09/Jun/20

$$\mathrm{good}$$

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