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Question Number 156744 by joki last updated on 15/Oct/21

$$\mathrm{y}‘‘+{\mathrm{y}}^{\prime }={\mathrm{e}}^{\mathrm{x}}+3\mathrm{x}$$$$$$

Answered by qaz last updated on 15/Oct/21

$${\mathrm{y}}_{\mathrm{p}}=\frac{1}{{\mathrm{D}}^2+\mathrm{D}}({\mathrm{e}}^{\mathrm{x}}+3\mathrm{x})$$$$=\frac{1}{\mathrm{D}+1}({\mathrm{e}}^{\mathrm{x}}+\frac{3}{2}{\mathrm{x}}^2)$$$$=\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}+(1-\mathrm{D}+{\mathrm{D}}^2-...)\frac{3}{2}{\mathrm{x}}^2$$$$=\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}+\frac{3}{2}{\mathrm{x}}^2-3\mathrm{x}+3$$$$\Rightarrow \mathrm{y}={\mathrm{C}}_1+{\mathrm{C}}_2{\mathrm{e}}^{-\mathrm{x}}+\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}+\frac{3}{2}{\mathrm{x}}^2-3\mathrm{x}$$

Commented by puissant last updated on 15/Oct/21

$$whatistheformulalikethatsirQaz..?$$

Answered by mathmax by abdo last updated on 16/Oct/21

$${\mathrm{y}}^{\left(2\right)}+{\mathrm{y}}^{{}^{\prime }}={\mathrm{e}}^{\mathrm{x}}+3\mathrm{x}\mathrm{we}\mathrm{put}{\mathrm{y}}^{{}^{\prime }}=\mathrm{z}\Rightarrow {\mathrm{z}}^{{}^{\prime }}+\mathrm{z}={\mathrm{e}}^{\mathrm{x}}+3\mathrm{x}\left(\mathrm{e}\right)$$$$\mathrm{h}\to {\mathrm{z}}^{{}^{\prime }}+\mathrm{z}=0\to \frac{{\mathrm{z}}^{{}^{\prime }}}{\mathrm{z}}=-1\Rightarrow \ln \mid \mathrm{z}\mid =-\mathrm{x}+\mathrm{c}\Rightarrow \mathrm{z}={\mathrm{ke}}^{-\mathrm{x}}$$$$\mathrm{let}\mathrm{use}\mathrm{mvc}\mathrm{method}\to {\mathrm{z}}^{{}^{\prime }}={\mathrm{k}}^{{}^{\prime }}{\mathrm{e}}^{-\mathrm{x}}-{\mathrm{ke}}^{-\mathrm{x}}$$$$\left(\mathrm{e}\right)\Rightarrow {\mathrm{k}}^{{}^{\prime }}{\mathrm{e}}^{-\mathrm{x}}-{\mathrm{ke}}^{-\mathrm{x}}+{\mathrm{ke}}^{-\mathrm{x}}={\mathrm{e}}^{\mathrm{x}}+3\mathrm{x}\Rightarrow {\mathrm{k}}^{{}^{\prime }}{\mathrm{e}}^{-\mathrm{x}}={\mathrm{e}}^{\mathrm{x}}+3\mathrm{x}\Rightarrow$$$${\mathrm{k}}^{{}^{\prime }}={\mathrm{e}}^{2\mathrm{x}}+3\mathrm{x}{\mathrm{e}}^{\mathrm{x}}\Rightarrow \mathrm{k}=\int ({\mathrm{e}}^{2\mathrm{x}}+{3\mathrm{xe}}^{\mathrm{x}})\mathrm{dx}$$$$=\frac{1}{2}{\mathrm{e}}^{2\mathrm{x}}+3\int {\mathrm{xe}}^{\mathrm{x}}\mathrm{dx}\mathrm{we}\mathrm{have}$$$$\int {\mathrm{xe}}^{\mathrm{x}}\mathrm{dx}{=}_{\mathrm{by}\mathrm{parts}}{\mathrm{xe}}^{\mathrm{x}}-\int {\mathrm{e}}^{\mathrm{x}}\mathrm{dx}={\mathrm{xe}}^{\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}\Rightarrow$$$$\mathrm{k}=\frac{1}{2}{\mathrm{e}}^{2\mathrm{x}}+(3\mathrm{x}-3){\mathrm{e}}^{\mathrm{x}}+\mathrm{c}\mathrm{on}\mathrm{z}={\mathrm{ke}}^{-\mathrm{x}}\Rightarrow$$$$\mathrm{z}=(\frac{1}{2}{\mathrm{e}}^{2\mathrm{x}}+(3\mathrm{x}-3){\mathrm{e}}^{\mathrm{x}}+\mathrm{c}){\mathrm{e}}^{-\mathrm{x}}=\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}+3\mathrm{x}-3+{\mathrm{ce}}^{-\mathrm{x}}$$$${\mathrm{y}}^{{}^{\prime }}=\mathrm{z}\Rightarrow \mathrm{y}=\int \mathrm{zdx}=\int (\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}+3\mathrm{x}-3+\mathrm{c}{\mathrm{e}}^{-\mathrm{x}})\mathrm{dx}$$$$=\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}+\frac{3}{2}{\mathrm{x}}^2-3\mathrm{x}-{\mathrm{ce}}^{-\mathrm{x}}+\lambda$$

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