Question

Solve the following differential equation:

$\left({\cot }^{-1}y+x\right)dy=\left(1+y^2\right)dx$

Answer 1

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