Question

Solve : $\left(1+y^2\right)dx=\left({\tan }^{-1}y-x\right)dy$.

Answer 1

Step1. Find the solution of the differential equation :

The solution of the given differential equation $\frac{dy}{dx}+px=Q$ is given by:

$x\left(\mathrm{IF}\right)=\int Q\left(\mathrm{IF}\right)dy+C$

Where,$\mathrm{IF}=e^{\int pdy}$

Step 2. Reduce the equation in the form of $\frac{dy}{dx}+px=Q$ :

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

$\begin{array}{l}\left(1+y^2\right)dx=\left({\tan }^{-1}y-x\right)dy\\ \Rightarrow \frac{dx}{dy}=\frac{{\tan }^{-1}y-x}{1+y^2}\\ \Rightarrow \frac{dx}{dy}=\frac{{\tan }^{-1}y}{1+y^2}-\frac{x}{1+y^2}\\ \Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{{\tan }^{-1}y}{1+y^2}\left[\because \frac{dy}{dx}+px=Q\right]\end{array}$

Here, $P=\frac{1}{1+y^2}$ and $Q=\frac{{\tan }^{-1}y}{1+y^2}$.

Step 3. Find the integrating factor (IF):

$\begin{array}{c}\mathrm{IF}=e^{\int \frac{1}{1+y^2}dy}\\ =e^{{\tan }^{-1}y}\end{array}$

Step 4. Find the solution of the given differential equation :

$$\begin{gathered} x\left(\mathrm{IF}\right)=\int Q\left(\mathrm{IF}\right)dy+C \\ \Rightarrow x{e}^{{\tan }^{-1}y}=\int \frac{{\tan }^{-1}y}{1+y^2}{e}^{{\tan }^{-1}y}dy+C \end{gathered}$$

Let, ${\tan }^{-1}y=t$

Therefore ,$\frac{1}{1+y^2}dy=dt$

Now,

$\begin{array}{c}x{e}^t=\int t{e}^{t}dt+C\\ \Rightarrow x{e}^t=t{e}^t-\int e^{t}dt+C\\ \Rightarrow x{e}^t=t{e}^t-e^t+C\\ \Rightarrow x{e}^t=e^t\left(t-1\right)+C\\ \Rightarrow x=\left(t-1\right)+\frac{C}{e^t}\\ \Rightarrow x={\tan }^{-1}y-1+C{e}^{-{\tan }^{-1}y}\end{array}$

Hence ,the solution of the given differential equation is $x={\tan }^{-1}y-1+C{e}^{-{\tan }^{-1}y}$.