Q120 / 160 · AP EAMCET 2024 (22 May Shift 2) · Mathematics
The general solution of the differential equation $\left(x y+y^2\right) d x-\left(x^2-2 x y\right) d y=0$ is
- A$c x y^2=e^{\frac{x}{y}}$
- Correct answerB$c x y^2 e^{\frac{x}{y}}=1$
- C$c x y e^{\frac{x}{y}}=1$
- D$c x y=e^{\frac{x}{y}}$
Solution
$\begin{array}{r}
\left(x y+y^2\right) d x-\left(x^2-2 x y\right) d y=0 \\
\frac{d y}{d x}=\frac{\left(x y+y^2\right)}{\left(x^2-2 x y\right)} \Rightarrow \frac{d y}{d x}=\frac{\frac{y}{x}+\left(\frac{y}{x}\right)^2}{1-2\left(\frac{y}{x}\right)}
\end{array}$
Let $\mathrm{y}=v x \Rightarrow \frac{d y}{d x}=v+\frac{x d v}{d x}$ $\begin{aligned} & \therefore+\frac{x d v}{d x}=\frac{v v}{12 v} \Rightarrow \frac{x d v}{d x}=\frac{v+v^2-v+2 v^2}{1-2 v} \\ & \Rightarrow \frac{x d v}{d x}=\frac{3 v^2}{1-2 v} \Rightarrow\left(\frac{1}{3 v^2}-\frac{2}{3 v}\right) d v=\frac{d x}{x} \end{aligned}$
Integrating we get, $\frac{-1}{3 v}-\frac{2}{3} \log v=\log x+\log c_1$ $\begin{aligned} & \Rightarrow \frac{-x}{y}-2 \log \frac{y}{x}=3 \log x+\log c_2 \\ & \Rightarrow \frac{-x}{y}=\log c_2 x+2 \log y \Rightarrow \frac{-x}{y}=\log c_2 x y^2 \\ & \Rightarrow e^{-\frac{x}{y}}=c x y^2 \Rightarrow c x y^2 e^{\frac{x}{y}}=1 \end{aligned}$
Let $\mathrm{y}=v x \Rightarrow \frac{d y}{d x}=v+\frac{x d v}{d x}$ $\begin{aligned} & \therefore+\frac{x d v}{d x}=\frac{v v}{12 v} \Rightarrow \frac{x d v}{d x}=\frac{v+v^2-v+2 v^2}{1-2 v} \\ & \Rightarrow \frac{x d v}{d x}=\frac{3 v^2}{1-2 v} \Rightarrow\left(\frac{1}{3 v^2}-\frac{2}{3 v}\right) d v=\frac{d x}{x} \end{aligned}$
Integrating we get, $\frac{-1}{3 v}-\frac{2}{3} \log v=\log x+\log c_1$ $\begin{aligned} & \Rightarrow \frac{-x}{y}-2 \log \frac{y}{x}=3 \log x+\log c_2 \\ & \Rightarrow \frac{-x}{y}=\log c_2 x+2 \log y \Rightarrow \frac{-x}{y}=\log c_2 x y^2 \\ & \Rightarrow e^{-\frac{x}{y}}=c x y^2 \Rightarrow c x y^2 e^{\frac{x}{y}}=1 \end{aligned}$
Hence, the correct answer is option: 'B'.