Q122 / 200 · WBJEE 2013 · Mathematics
The solution of the differential equation $\left(y^{2}+2 x\right) \frac{d y}{d x}=y$ satisfies $x=1, y=1$. Then the solution is
- Correct answerA$x=y^{2}\left(1+\log _{e} y\right)$
- B$y=x^{2}\left(1+\log _{e} x\right)$
- C$x=y^{2}\left(1-\log _{e} y\right)$
- D$y=x^{2}\left(1-\log _{e} x\right)$
Solution
Given differential equation is
$\left(y^{2}+2 x\right) \frac{d y}{d x}=y \Rightarrow \frac{d x}{d y}=\left(\frac{y^{2}+2 x}{y}\right)$
$\Rightarrow \quad \frac{d x}{d y}=y+\frac{2 x}{y}$
$\Rightarrow \quad \frac{d x}{d y}-\frac{2}{y} \cdot x=y$
$\mathrm{IF}=e^{\int \frac{2}{y} d y}=e^{-2 \mathrm{bg} \gamma}=\mathrm{y}^{-2}=\frac{1}{\mathrm{y}^{2}}$
$\therefore$ Complete solution is $x \cdot \frac{1}{y^{2}}=\int y \cdot \frac{1}{y^{2}} d y+C$
$\Rightarrow \quad \frac{x}{y^{2}}=\int \frac{d y}{y}+C=\log _{e} y+c$
$\Rightarrow \quad x=y^{2} \log _{e} y+C y^{2}$
$A t+x=1 y=1$ then from Eq $(i),$ we get
$$
1=0+C \Rightarrow C=1
$$
$\therefore$ From Eq. (i), we get
$$
x=y^{2}\left(\log _{e} y+1\right)
$$
Which is the required solution.
Hence, the correct answer is option: 'A'.