The curve satisfying the differential equation $y \mathrm{~d} x-\left(x+3 y^2\right) \mathrm{dy}=0$ and…
The curve satisfying the differential equation $y \mathrm{~d} x-\left(x+3 y^2\right) \mathrm{dy}=0$ and passing through the point $(1,1)$ also passes through the point
$\left(\frac{1}{4}, \frac{1}{2}\right)$
$\left(\frac{1}{4},-\frac{1}{2}\right)$
$\left(\frac{1}{3},-\frac{1}{3}\right)$
$\left(-\frac{1}{3}, \frac{1}{3}\right)$
Solution
$\begin{aligned} & y \mathrm{~d} x-\left(x+3 y^2\right) \mathrm{d} y=0 \\ & \Rightarrow y \mathrm{~d} x=\left(x+3 y^2\right) \mathrm{d} y \\ & \Rightarrow \frac{\mathrm{~d} x}{\mathrm{~d} y}=\frac{x+3 y^2}{y} \\ & \Rightarrow \frac{\mathrm{~d} x}{\mathrm{~d} y}=\frac{x}{y}+3 y \\ & \Rightarrow \frac{\mathrm{~d} x}{\mathrm{~d} y}-\left(\frac{1}{y}\right) x=3 y\end{aligned}$
Which is a linear equation
$\therefore \quad \mathrm{IF}=\mathrm{e}^{\int \frac{-1}{y} \mathrm{~d} y}=\mathrm{e}^{-\log y}=\frac{1}{y}$
$\therefore \quad$ The required solution is
$x \frac{1}{y}=\int 3 y \times \frac{1}{y} \mathrm{~d} y+\mathrm{c}$
$\begin{aligned}
\therefore \quad & \frac{x}{y}=3 y+c \\
& \Rightarrow x=3 y^2+c y
\end{aligned}$ Curve passes through $(1,1)$
$\begin{aligned}
& \Rightarrow 1=3+c \\
& \Rightarrow c=-2
\end{aligned}$ Equation (i) becomes,
$x=3 y^2-2 y$ Option (D) i.e., $\left(\frac{-1}{3}, \frac{1}{3}\right)$
Satisfies above equation.