Q68 / 160 · TS EAMCET 2017 · Mathematics
The solution of $\frac{d y}{d x}=\frac{x+y}{x-y}$ is
- Correct answerA$\tan ^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2+y^2}+C$
- B$\tan ^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2-y^2}+C$
- C$\sin ^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2+y^2}+C$
- D$\cos ^{-1}\left(\frac{y}{x}\right)=\log \sqrt{x^2-y^2}+C$
Solution
We have,
$$
\frac{d y}{d x}=\frac{x+y}{x-y}
$$
Put $y=v x$
$$
\begin{aligned}
& \frac{d y}{d x}=v+x \frac{d v}{d x} \\
& \therefore v+x \frac{d v}{d x}=\frac{x+v x}{x-v x}=\frac{1+v}{1-v} \\
& \Rightarrow \quad x \frac{d v}{d x}=\frac{1+v}{1-v}-v=\frac{1+v-v+v^2}{1-v}=\frac{1+v^2}{1-v} \\
& \Rightarrow \frac{1-v}{1+v^2} d v=\frac{d x}{x} \\
& \Rightarrow \int \frac{1}{1+v^2} d v-\frac{1}{2} \int \frac{2 v}{1+v^2} d v=\int \frac{d x}{x} \\
& \Rightarrow \tan ^{-1} v-\frac{1}{2} \log \left|1+v^2\right|=\log x+C \\
& \Rightarrow \quad \tan ^{-1} \frac{y}{x}=\log x+\frac{1}{2} \log \left(1+\frac{y^2}{x^2}\right)+C \\
& \Rightarrow \quad \tan ^{-1} \frac{y}{x}=\log \frac{x\left(\sqrt{x^2+y^2}\right)}{x}+C \\
& \tan ^{-1} \frac{y}{x}=\log \sqrt{x^2+y^2}+C \\
& \Rightarrow C^2
\end{aligned}
$$
Hence, the correct answer is option: 'A'.