Q41 / 120 · VITEEE 2016 · Mathematics
The solution of the differential equation
$\left(1+y^{2}\right)+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0$
- A$(x-2)=k e^{-\tan ^{-1} y}$
- Correct answerB$2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+k$
- C$x e^{\tan ^{-1} y}=\tan ^{-1} y+k$
- D$x e^{2 \tan ^{-1} y}=e^{\tan ^{-1} y}+k$
Solution
$1+y^{2}+\left(x-e^{\tan ^{-1} y}\right) \frac{d y}{d x}=0$
$\Rightarrow\left(1+y^{2}\right) d x=\left(e^{\tan ^{-1} y}-x\right) d y$
$\Rightarrow \frac{d x}{d y}=\frac{e^{\tan ^{-1} y}-x}{1+y^{2}}$
$\Rightarrow \frac{d x}{d y}+\frac{1}{1+y^{2}} \cdot x=\frac{e^{\tan ^{-1} y}}{1+y^{2}}$
Which is the linear different equation of the form $\frac{d x}{d y}+R x=S$, where $R$ and $S$ are functions of $y$ or constant $(s)$
$$
\therefore \mathrm{I} . \mathrm{F}=\int_{e}^{\int \frac{1}{1+y^{2}} \cdot d y}=e^{\tan ^{-1} y}
$$
Hence required solution is
$$
\begin{array}{l}
x .(\mathrm{I} . \mathrm{F})=\int \frac{e^{\tan ^{-1} y}}{1+y^{2}}(\mathrm{I} . \mathrm{F}) d y \\
\Rightarrow x \cdot e^{\tan ^{-1} y}=\int \frac{e^{\tan ^{-1} y}}{1+y^{2}}\left(e^{\tan ^{-1} y}\right) d y \\
\Rightarrow x \cdot e^{\tan ^{-1} y}=\int \frac{e^{2 \tan ^{-1} y}}{1+y^{2}} d y
\end{array}
$$
Put $t=\tan ^{-1} y$
$$
\begin{array}{l}
\therefore \frac{d t}{d y}=\frac{1}{1+y^{2}} \Rightarrow d t=\frac{1}{1+y^{2}} \cdot d y \\
\therefore \int \frac{e^{2 \tan ^{-1} y}}{1+y^{2}} d y=\int e^{2 t} \cdot d t=\frac{e^{2 t}}{2}+K
\end{array}
$$
Hence equation (1) becomes,
$$
\begin{array}{l}
x e^{\tan ^{-1} y}=\frac{1}{2} e^{2 t}+K \\
\Rightarrow x e^{\tan ^{-1} y}=\frac{1}{2} e^{2 \tan ^{-1} y}+K \\
\Rightarrow 2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+K
\end{array}
$$
Hence, the correct answer is option: 'B'.