Solve the following differential equation: 3x^2dy=(3xy+y^2)dx - Vidyadip

Question

Solve the following differential equation:
$3\text{x}^2\text{dy}=(3\text{xy}+\text{y}^2)\text{dx}$

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Answer

We have,
$3\text{x}^2\text{dy}=(3\text{xy}+\text{y}^2)\text{dx}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{3\text{xy}+\text{y}^2}{3\text{x}^2}$
This is a homogeneous differential equation.
Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{3\text{vx}^2+\text{v}^2\text{x}^2}{3\text{x}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{3\text{v}+\text{v}^2}{3}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2}3$
$\Rightarrow\ \frac{3}{\text{v}^2}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$3\int\frac{1}{\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ -3\times\frac{1}{\text{v}}=\log|\text{x}|+\text{C}$
$\Rightarrow\ -\frac{3}{\text{v}}=\log|\text{x}|+\text{C}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \frac{-3\text{x}}{\text{y}}=\log|\text{x}|+\text{C}$
Hence, $\frac{-3\text{x}}{\text{y}}=\log|\text{x}|+\text{C}$ is the required solution.

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