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

Question

Solve the following differential equation:
$\text{xy}\frac{\text{dy}}{\text{dx}}=\text{x}^2-\text{y}^2$

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Answer

We have,
$\text{xy}\frac{\text{dy}}{\text{dx}}=\text{x}^2-\text{y}^2$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2-\text{y}^2}{\text{xy}}$
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
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2-\text{v}^2\text{x}^2}{\text{vx}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{1-\text{v}^2}{\text{v}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-\text{v}^2}{\text{v}}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-2\text{v}^2}{\text{v}}$
$\Rightarrow\ \frac{\text{v}}{1-2\text{v}^2}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{\text{v}}{1-2\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \frac{-1}4\log\big|1-2\text{v}^2\big|=\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \log\big|1-2\text{v}^2\big|=-4\log|\text{x}|-4\log\text{C}$
$\Rightarrow\ \log\big|\big(1-2\text{v}^2\big)\big(\text{x}^4\big)\big|=\log\frac{1}{\text{C}^4}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \log\big|\text{x}^2\big(\text{x}^2-2\text{y}^2\big)\big|=\log\frac{1}{\text{C}^4}$
$\Rightarrow\ \text{x}^2\big(\text{x}^2-2\text{y}^2\big)=\text{C}_1$
where
$\text{C}_1=\frac{1}{\text{C}^4}$
Hence, $\text{x}^2\big(\text{x}^2-2\text{y}^2\big)=\text{C}_1$ is the required solution.

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