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Question Number 66420 by Rio Michael last updated on 14/Aug/19

$$solvethedifferentialequation$$$$2\frac{d^{2}y}{{dx}^2}+\frac{dy}{dx}-e^{-x}=4$$

Commented by mathmax by abdo last updated on 14/Aug/19

$$2y^{{}^{″}}+y^{{}^{\prime }}=4+e^{-x}let{y}^{{}^{\prime }}=zso\left(e\right)\Rightarrow 2z^{{}^{\prime }}+z=4+e^{-x}\left(e^{{}^{\prime }}\right)$$$$\left(he\right)\to 2z^{{}^{\prime }}+z=0\Rightarrow 2z^{{}^{\prime }}=-z\Rightarrow \frac{z^{{}^{\prime }}}{z}=-\frac{1}{2}\Rightarrow ln\mid z\mid =-\frac{x}{2}+\lambda \Rightarrow$$$$z\left(x\right)=K{e}^{-\frac{x}{2}}letusemvcmethod$$$$z^{{}^{\prime }}=K^{{}^{\prime }}{e}^{-\frac{x}{2}}-\frac{K}{2}{e}^{-\frac{x}{2}}$$$$\left(e^{{}^{\prime }}\right)\Rightarrow 2K^{{}^{\prime }}{e}^{-\frac{x}{2}}-K{e}^{-\frac{x}{2}}+K{e}^{-\frac{x}{2}}=4+e^{-x}\Rightarrow 2K^{{}^{\prime }}{e}^{-\frac{x}{2}}=4+e^{-x}\Rightarrow$$$$K^{{}^{\prime }}=\frac{1}{2}{e}^{\frac{x}{2}}(4+e^{-x})=2e^{\frac{x}{2}}+\frac{1}{2}{e}^{-\frac{x}{2}}\Rightarrow$$$$K\left(x\right)=\int (2e^{\frac{x}{2}}+\frac{1}{2}{e}^{-\frac{x}{2}})dx+c=4e^{\frac{x}{2}}-e^{-\frac{x}{2}}+c\Rightarrow$$$$z\left(x\right)=e^{-\frac{x}{2}}\{4e^{\frac{x}{2}}-e^{-\frac{x}{2}}+c\}=4-e^{-x}+c{e}^{-\frac{x}{2}}$$$$y^{{}^{\prime }}=z\Rightarrow y^{{}^{\prime }}=4-e^{-x}+c{e}^{-\frac{x}{2}}\Rightarrow y\left(x\right)=\int (4-e^{-x}+c{e}^{-\frac{x}{2}})dx+\lambda \Rightarrow$$$$y\left(x\right)=4x+e^{-x}-2c{e}^{-\frac{x}{2}}+\lambda$$$$$$

Commented by Rio Michael last updated on 15/Aug/19

$$thankyousir$$

Commented by mathmax by abdo last updated on 17/Aug/19

$$youarewelcome.$$

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