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Question

# Solve the following differential equation: $\left({\mathrm{cot}}^{-1}y+x\right)dy=\left(1+{y}^{2}\right)dx$

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Solution

## The given differential equation is $\left({\mathrm{cot}}^{-1}y+x\right)dy=\left(1+{y}^{2}\right)dx$. This differential equation can be written as $\frac{dx}{dy}=\frac{{\mathrm{cot}}^{-1}y+x}{1+{y}^{2}}\phantom{\rule{0ex}{0ex}}⇒\frac{dx}{dy}+\left(-\frac{1}{1+{y}^{2}}\right)x=\frac{{\mathrm{cot}}^{-1}y}{1+{y}^{2}}$ This is a linear differential equation with $P=-\frac{1}{1+{y}^{2}}\mathrm{and}Q=\frac{{\mathrm{cot}}^{-1}y}{1+{y}^{2}}$. I.F. = ${\mathrm{e}}^{-\int \frac{1}{1+{y}^{2}}dy}={\mathrm{e}}^{{\mathrm{cot}}^{-1}y}$ Multiply the differential equation by integration factor (I.F.), we get $\frac{dx}{dy}{\mathrm{e}}^{{\mathrm{cot}}^{-1}y}-\frac{x}{\left(1+{y}^{2}\right)}{\mathrm{e}}^{{\mathrm{cot}}^{-1}y}=\frac{{\mathrm{cot}}^{-1}y}{\left(1+{y}^{2}\right)}{\mathrm{e}}^{{\mathrm{cot}}^{-1}y}\phantom{\rule{0ex}{0ex}}⇒\frac{d}{dy}\left(x{\mathrm{e}}^{{\mathrm{cot}}^{-1}\mathrm{y}}\right)=\frac{{\mathrm{cot}}^{-1}y}{\left(1+{y}^{2}\right)}{\mathrm{e}}^{{\mathrm{cot}}^{-1}y}$ Integrating both sides with respect y, we get $x{\mathrm{e}}^{{\mathrm{cot}}^{-1}y}=\int \frac{{\mathrm{cot}}^{-1}y}{\left(1+{y}^{2}\right)}{\mathrm{e}}^{{\mathrm{cot}}^{-1}y}dy+C$ Putting $t={\mathrm{cot}}^{-1}y$ and $dt=-\frac{1}{1+{y}^{2}}dy$, we get $x{\mathrm{e}}^{{\mathrm{cot}}^{-1}y}=-\int t{\mathrm{e}}^{t}dt+C\phantom{\rule{0ex}{0ex}}⇒x{\mathrm{e}}^{{\mathrm{cot}}^{-1}y}=-{\mathrm{e}}^{t}\left(t-1\right)+C\phantom{\rule{0ex}{0ex}}⇒x{e}^{{\mathrm{cot}}^{-1}\mathrm{y}}={\mathrm{e}}^{{\mathrm{cot}}^{-1}\mathrm{y}}\left(1-{\mathrm{cot}}^{-1}y\right)+C$

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