Question
Solve the following differential equations:
$(\text{x}^2-\text{y}^2)\text{dx}-2\text{xy dy}=0$
$(\text{x}^2-\text{y}^2)\text{dx}-2\text{xy dy}=0$
$(\text{x}^2-\text{y}^2)\text{dx}-2\text{xy dy}=0$ 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 $\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2-(\text{vx})^2}{2\text{x(vx)}}$ $\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2-\text{v}^2\text{x}^2}{2\text{vx}^2}$ $\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{1-\text{v}^2}{2\text{v}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-\text{v}^2}{2\text{v}}-\text{v}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-3\text{v}^2}{2\text{v}}$ $\Rightarrow\ \frac{2\text{v}}{1-3\text{v}^2}\text{dv}=\frac{1}{\text{x}}\text{dx}$ Integrating both sides, we get $\int\frac{2\text{v}}{1-3\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ -\frac{1}3\int\frac{-6\text{v}}{1-3\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow-\frac{1}3\log\big|1-3\text{v}^2\big|=\log|\text{x}|+\log|\text{C}|$ $\Rightarrow\ \log\big|1-3\text{v}^2\big|=-3\log|\text{Cx}|$ $\Rightarrow\ \log\big|1-3\text{v}^2\big|=\log\bigg|\frac{1}{(\text{Cx})^3}\bigg|$ $\Rightarrow\ 1-3\text{v}^2=\frac{1}{(\text{Cx})^3}$ Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get $1-3\Big(\frac{\text{y}}{\text{x}}\Big)^2=\frac{1}{(\text{Cx})^3}$ $\Rightarrow\ \frac{\text{x}^2-3\text{y}^2}{\text{x}^2}=\frac{1}{\text{C}^3\text{x}^3}$ $\Rightarrow\ \text{x}(\text{x}^2-3\text{y}^2)=\frac{1}{\text{C}^3}$ $\Rightarrow\ \text{x}(\text{x}^2-3\text{y}^2)=\text{K}$ $\Big($where, $\text{K}=\frac{1}{\text{C}^3}\Big)$
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