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Question Number 165849 by daus last updated on 09/Feb/22

$$solvethedifferentialequation$$$$\frac{dy}{dx}+\frac{y}{x-1}=\frac{1}{x+1}$$

Commented by mkam last updated on 10/Feb/22

$$\mathit{p}\left(\mathit{x}\right)=\frac{1}{\mathit{x}-1},\mathit{Q}\left(\mathit{x}\right)=\frac{1}{\mathit{x}+1}$$$$$$$$(\mathit{I}.\mathit{f})={\mathit{e}}^{\int \mathit{P}\left(\mathit{x}\right)\mathit{d}\mathit{x}}=\mathit{e}{}^{\int \frac{1}{\mathit{x}-1}\mathit{d}\mathit{x}}={\mathit{e}}^{\mathit{l}\mathit{n}\mid \mathit{x}-1\mid }=\mathit{x}-1$$$$$$$$(\mathit{I}.\mathit{f})\mathit{y}=\int (\mathit{I}.\mathit{f}).\mathit{Q}\left(\mathit{x}\right)\mathit{d}\mathit{x}$$$$$$$$(\mathit{x}-1)\mathit{y}=\int \left(\frac{\mathit{x}-1}{\mathit{x}+1}\right)\mathit{d}\mathit{x}\Rightarrow (\mathit{x}-1)\mathit{y}=\int (1-\frac{2}{\mathit{x}+1})\mathit{d}\mathit{x}$$$$$$$$\therefore (\mathit{x}-1)\mathit{y}=\mathit{x}-2\mathit{l}\mathit{n}\mid \mathit{x}+1\mid +\mathit{C}$$$$$$$$\mathit{y}=\frac{\mathit{x}}{\mathit{x}-1}-\frac{2\mathit{l}\mathit{n}\mid \mathit{x}+1\mid }{\mathit{x}-1}+\frac{\mathit{C}}{\mathit{x}-1}$$$$$$$$◻\mathit{m}\mathit{o}\mathit{h}\mathit{a}\mathit{m}\mathit{m}\mathit{a}\mathit{d}\mathit{a}\mathit{l}\mathit{d}\mathit{o}\mathit{l}\mathit{a}\mathit{i}\mathit{m}\mathit{y}$$

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