# Solve the Following Differential Equation:- ( X + 3 Y 2 ) D Y D X = Y - Mathematics

Sum

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

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

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

$\Rightarrow x = 3 y^2 + Cy$

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

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 66.15 | Page 147