Q95 / 180 · COMEDK 2014 · Mathematics
The solution of
$(x+y)^{2}\left(x \frac{d y}{d x}+y\right)=x y\left(1+\frac{d y}{d x}\right)$ is
- Correct answerA$\log (x y)=-\frac{1}{x+y}+C$
- B$\log \left(\frac{x}{y}\right)=-\frac{1}{x+y}+C$
- C$\log (x y)=\frac{1}{x-y}+C$
- DNone of the above
Solution
Given, differential equation is
$$
\begin{aligned}
&\Rightarrow \quad(x y)^{-1}\left(x \frac{d y}{d x}+y\right)=(x+y)^{-2}\left(1+\frac{d y}{d x}\right) \\
&\Rightarrow \int(x y)^{-1}\left(x \frac{d y}{d x}+y\right) d x=\int(x+y)^{-2}\left(1+\frac{d y}{d x}\right) d x \ldots(\mathrm{i})
\end{aligned}
$$
Using integral,
$$
\int(f(x))^{n} f^{\prime}(x) d x=\frac{\left.(f(x))^{n+1}\right)}{n+1}
$$
and
$$
\begin{array}{ll}
\text { From Eq. (i) } & \log (x y)=\frac{(x+y)^{-1}}{-1}+C \\
\Rightarrow \quad & \log (x y)=\frac{-1}{x+y}+C
\end{array}
$$
Hence, the correct answer is option: 'A'.