Question

(1 + y²) dx = (tan⁻¹ y − x) dy

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(1+y^2\right)dx=\left({\tan }^{-1}y-x\right)dy \\ \Rightarrow \frac{dx}{dy}=\frac{{\tan }^{-1}y-x}{1+y^2} \\ \Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{{\tan }^{-1}y}{1+y^2} \end{gathered}$$
$$\begin{gathered} \mathrm{Comparing}\mathrm{with}\frac{dx}{dy}+Px=Q,\mathrm{we}\mathrm{get} \\ P=\frac{1}{1+y^2} \\ Q=\frac{{\tan }^{-1}y}{1+y^2} \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int \frac{1}{1+y^2}dy}=e^{{\tan }^{-1}y} \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ x\times {\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}=\int \frac{{\tan }^{-1}y}{1+y^2}\times {\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}dy+C \\ \Rightarrow x{\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}=I+C.....\left(1\right) \\ \mathrm{Now}, \\ I=\int \frac{{\tan }^{-1}y}{1+y^2}\times {\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}dy \\ \mathrm{Putting}t={\tan }^{-1}y,\mathrm{we}\mathrm{get} \\ dt=\frac{1}{1+y^2}dy \\ \therefore I=\int \underset{\mathrm{I}}{t}\times \underset{\mathrm{II}}{{\mathrm{e}}^{\mathrm{t}}}dt \\ =t\times \int e^{t}dt-\int \left(\frac{dt}{dt}\times \int e^{t}dt\right)dt \\ =t{e}^t-\int e^{t}dt \\ =t{e}^t-e^t \\ \therefore I={\tan }^{-1}y{e}^{{\tan }^{-1}y}-e^{{\tan }^{-1}y} \\ =e^{{\tan }^{-1}y}\left({\tan }^{-1}y-1\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}I\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ x{\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}=e^{{\tan }^{-1}y}\left({\tan }^{-1}y-1\right)+C \end{gathered}$$