Q118 / 160 · AP EAMCET 2020 (21 Sep Shift 2) · Mathematics
Solve of the differential equation
\(\frac{d y}{d x}=\frac{1+y^2}{\left(\tan ^{-1} y\right)-x}\)
- A\(x e^{\tan ^{-1} y}=e^{-\tan ^{-1} y}\left(\left(\tan ^{-1} y\right)-1\right)+c\)
- Correct answerB\(x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\left(\tan ^{-1} y\right)-1\right)+c\)
- C\(x e^{\tan ^{-1} y}=e^{\tan ^{-1} y}\left(\left(\tan ^{-1} y\right)+1\right)+c\)
- D\(x e^{\tan ^{-1} y}=e^{-\tan ^{-1} y}\left(\left(\tan ^{-1} y\right)+1\right)+c\)
Solution
Given differential equation
\(\frac{d y}{d x}=\frac{1+y^2}{\tan ^{-1} y-x} \Rightarrow \frac{d x}{d y}=\frac{\tan ^{-1} y-x}{1+y^2}\)
\(\Rightarrow \frac{d x}{d y}+\frac{x}{1+y^2}=\frac{\tan ^{-1} y}{1+y^2}\) is a linear
differential equation, so.
\(\mathrm{IF}=e^{\int \frac{d y}{1+y^2}=e^{\tan ^{-1} y}}\)
So, the solution is
\(\begin{aligned}
& x e^{\tan ^{-1} y}=\int e^{\tan ^{-1} y} \frac{\tan ^{-1} y}{1+y^2} d y+c \\
\Rightarrow x e^{\tan ^{-1} y} & =\int t \cdot e^t d t+c, \quad\left\{\text {where } t=\tan ^{-1} y\right\} \\
\Rightarrow x e^{\tan ^{-1} y} & =t e^t-\int e^t+c \\
\Rightarrow x e^{\tan ^{-1} y} & =\left(\tan ^{-1} y\right) e^{\tan ^{-1} y}-e^{\tan ^{-1} y}+c \\
\Rightarrow x e^{\tan ^{-1} y} & =e^{\tan ^{-1} y}\left(\left(\tan ^{-1} y\right)-1\right)+c
\end{aligned}\)
Hence, option (b) is correct.
Hence, the correct answer is option: 'B'.