Question

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

Answer 1

We have,
\[\left( y + 3 x^2 \right)\frac{dx}{dy} = x\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y + 3 x^2}{x}\]
\[ \Rightarrow \frac{dy}{dx} - \frac{1}{x}y = 3x . . . . . . . . . . \left( 1 \right)\]
Clearly, it is a linear differential equation of the form 
\[\frac{dy}{dx} + Py = Q\]
\[\text{ where }P = - \frac{1}{x}\text{ and }Q = 3x\]
\[ \therefore I . F . = e^{\int P\ dx} \]
\[ = e^{- \int\frac{1}{x} dx} \]
\[ = e^{- \log x} \]
\[ = \frac{1}{x}\]
\[\text{ Multiplying both sides of }(1)\text{ by }I . F . = \frac{1}{x}, \text{ we get }\]
\[\frac{1}{x} \left( \frac{dy}{dx} - \frac{1}{x}y \right) = \frac{1}{x}3x \]
\[ \Rightarrow \frac{1}{x}\frac{dy}{dx} - \frac{1}{x^2}y = 3\]
Integrating both sides with respect to x, we get
\[\frac{1}{x}y = 3\int dx + C\]
\[ \Rightarrow \frac{y}{x} = 3x + C\]
\[\text{ Hence, }\frac{y}{x} = 3x + C\text{ is the required solution.}\]