Q97 / 150 · MHT CET 2021 (21 Sep Shift 1) · Mathematics
The general solution of the differential equation $\left(3 x y+y^2\right) d x$ $+\left(x^2+x y\right) d y=0$ is
- A$x^2\left(2 x y-y^2\right)=c$
- B$x^2\left(y^2-2 x y\right)=c$
- C$x\left(2 x y+y^2\right)=c$
- Correct answerD$\mathrm{x}^2\left(2 \mathrm{xy}+\mathrm{y}^2\right)=\mathrm{c}$
Solution
$$
\begin{aligned}
& \left(3 x y+y^2\right) d x+\left(x^2+x y\right) d y=0 \\
& \therefore \frac{d y}{d x}=-\frac{\left(3 x y+y^2\right)}{\left(x^2+x y\right)}
\end{aligned}
$$
Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$$
\begin{aligned}
& \therefore v+x \frac{d v}{d x}=-\frac{\left(3 v x^2+v^2 x^2\right)}{\left(x^2+v x^2\right)}=-\frac{3 v+v^2}{1+v} \\
& \therefore x \frac{d v}{d x}=-\frac{3 v+v^2}{1+v}-v=-\frac{\left(2 v^2+4 v\right)}{1+v}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \int \frac{1+v}{\left(2 v^2+4 v\right)} d v=\int \frac{-1}{x} d x \\
& \therefore \frac{1}{2} \int \frac{v+1}{v^2+2 v} d v=-\int \frac{d x}{x} \Rightarrow \frac{1}{4} \int \frac{2(v+1)}{v^2+2 v} d v=-\log x+\log c_1
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \frac{1}{4} \log \left(\mathrm{v}^2+2 \mathrm{v}\right)+\log \mathrm{x}=\log \mathrm{c}_1 \Rightarrow \frac{1}{4} \log \left(\frac{\mathrm{y}^2}{\mathrm{x}^2}+\frac{2 \mathrm{y}}{\mathrm{x}}\right)+\log \mathrm{x}=\log \mathrm{c}_1 \\
& \therefore \frac{1}{4} \log \left(\frac{\mathrm{y}^2+2 \mathrm{xy}}{\mathrm{x}^2}\right)+\log \mathrm{x}=\log \mathrm{c}_1 \Rightarrow \log \left(\frac{\mathrm{y}^2+2 \mathrm{xy}}{\mathrm{x}^2}\right)+\log \mathrm{x}^4=4 \log \mathrm{c}_1 \\
& \therefore \log \left[\left(\frac{\mathrm{y}^2+2 \mathrm{xy}}{\mathrm{x}^2}\right)\left(\mathrm{x}^4\right)\right]=\log \mathrm{c}_1^4 \Rightarrow \log \left[\mathrm{x}^2\left(\mathrm{y}^2+2 \mathrm{xy}\right)\right]=\log \mathrm{c}_1^4 \\
& \therefore \mathrm{x}^2\left(\mathrm{y}^2+2 \mathrm{xy}\right)=\mathrm{c}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.