Q87 / 150 · MHT CET 2008 · Mathematics
The solution of the differential equation $\left(3 x y+y^{2}\right) d x+\left(x^{2}+x y\right) d y=0$ is
- Correct answerA$x^{2}\left(2 x y+y^{2}\right)=c^{2}$
- B$x^{2}\left(2 x y-y^{2}\right)=c^{2}$
- C$x^{2}\left(y^{2}-2 x y\right)=c^{2}$
- DNone of these
Solution
Homogeneous equation can be written in the form of
Put
$$
\begin{aligned}
\frac{d y}{d x} &=-\frac{3 x y+y^{2}}{x^{2}+x y} \\
y=& v x \text { and } \frac{d y}{d x}=v+x \frac{d v}{d x}, \text { we get }
\end{aligned}
$$
$$
v+x \frac{d v}{d x}=-\frac{3 x^{2} v+x^{2} v^{2}}{x^{2}+x^{2} v}
$$
$$
\Rightarrow \quad x \frac{d v}{d x}=\frac{-2 v(v+2)}{v+1}
$$
$$
\begin{array}{ll}
\Rightarrow \quad & \frac{1}{x} d x=-\frac{(v+1)}{2 v(v+2)} d v \\
\Rightarrow \quad & -\frac{2}{x}=-\left[\frac{1}{2(v+2)}+\frac{1}{2 v}\right] d v
\end{array}
$$
On integrating, we get
$$
\begin{aligned}
&-2 \log _{e} x=\frac{1}{2} \log (v+2)+\frac{1}{2} \log v-\log c \\
\Rightarrow & v(v+2) x^{4}=c^{2} \\
\Rightarrow & \frac{y}{x}\left(\frac{y}{x}+2\right) x^{4}=c^{2} &\left(\because v=\frac{y}{x}\right)
\end{aligned}
$$
Hence, required solution is $\left(y^{2}+2 x y\right) x^{2}=c^{2}$
Hence, the correct answer is option: 'A'.