Solve the following differential equation:-

\[\left( x + 3 y^2 \right)\frac{dy}{dx} = y\]

#### Solution

We have,

\[\left( x + 3 y^2 \right)\frac{dy}{dx} = y\]

\[ \Rightarrow \frac{dx}{dy} = \frac{1}{y}\left( x + 3 y^2 \right) \]

\[ \Rightarrow \frac{dx}{dy} - \frac{1}{y}x = 3y . . . . . \left( 1 \right)\]

Clearly, it is a linear differential equation of the form

\[\frac{dx}{dy} + Px = Q\]

\[\text{where }P = - \frac{1}{y}\text{ and }Q = 3y\]

\[ \therefore I . F . = e^{\int P\ dy} \]

\[ = e^{- \int\frac{1}{y}dy} \]

\[ = e^{- \log \left| y \right|} = \frac{1}{y}\]

Multiplying both sides of (1) by I . F . = `1/y`, we get

\[\frac{1}{y}\left( \frac{dx}{dy} - \frac{1}{y}x \right) = \frac{1}{y} \times 3y\]

\[ \Rightarrow \frac{1}{y}\left( \frac{dx}{dy} - \frac{1}{y}x \right) = 3\]

Integrating both sides with respect to y, we get

\[x\frac{1}{y} = \int 3dy + C\]

\[ \Rightarrow x\frac{1}{y} = 3y + C\]

\[ \Rightarrow x = 3 y^2 + Cy\]

\[\text{Hence, }x = 3 y^2 + Cy\text{ is the required solution.}\]