The general solution of the differential equation $\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right)…
The general solution of the differential equation $\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0$ is
- $x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} x}\right)^{2}}{2}+c$
- $e^{\tan ^{-1} y}=\left(e^{\tan ^{-1} x}\right)^{2}+c$
- $x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{t a n^{-1} y}\right)^{2}}{2}+c$
- $e^{\tan ^{-1} y}=\left(e^{\tan ^{-1} y}\right)^{2}+c$
Solution
$\begin{array}{l}
\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0 \\
\therefore \quad\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=-\left(1+y^{2}\right) \Rightarrow \frac{d y}{d x}=\frac{-\left(1+y^{2}\right)}{x-e^{\tan ^{-1} y}} \\
\therefore \frac{d x}{d y}=\frac{\left(x-e^{\tan ^{-1} y}\right)}{-\left(1+y^{2}\right)} \Rightarrow \frac{d x}{d y}=\frac{-x}{\left(1+y^{2}\right)}+\frac{e^{\tan ^{-1} y}}{1+y^{2}} \\
\therefore \frac{d x}{d y}+\frac{x}{1+y^{2}}=\frac{e^{\tan ^{-1} y}}{1+y^{2}} \\
\quad \text { I.F. }=e^{\int \frac{1}{1+y^{2}} d y}=e^{\tan ^{-1} y}
\end{array}$
So, the general solution is
$\begin{array}{l}
x \cdot e^{\tan ^{-1} y}=\int \frac{e^{\tan ^{-1} y}}{1+y^{2}} \cdot e^{\tan ^{-1} y} d x \\
\text { Put } e^{\tan ^{-1} y}=t \Rightarrow \frac{e^{\tan ^{-1} y}}{1+y^{2}} d y=d t \\
\therefore x \cdot e^{\tan ^{-1} y}=\int t d t \\
\therefore x \cdot e^{\tan ^{-1} y}=\frac{t^{2}}{2}+c \Rightarrow x \cdot e^{\tan ^{-1} y}=\frac{\left(e^{\tan ^{-1} y}\right)^{2}}{2}+c
\end{array}$
Asked in: MHT CET 2020 (19 Oct Shift 2)
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