Question
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2-\text{x}^2}{2\text{xy}}$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2-\text{x}^2}{2\text{xy}}$
✓
Answer
Here, $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}^2-\text{x}^2}{2\text{xy}}$It is homogeneous equation
Put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2\text{x}^2-\text{x}^2}{2\text{xvx}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2-1}{2\text{v}}-\frac{\text{v}}1$
$=\frac{\text{v}^2-1-2\text{v}^2}{2\text{v}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{-1-\text{v}^2}{2\text{v}}$
$\int\frac{2\text{v}}{1+\text{v}^2}\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\log\big|1+\text{v}^2\big|=-\log|\text{x}|+\log|\text{C}|$
$1+\text{v}^2=\frac{\text{C}}{\text{x}}$
$1+\frac{\text{y}^2}{\text{x}^2}=\frac{\text{C}}{\text{x}}$
$\text{x}^2+\text{y}^2=\text{Cx}$
Put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2\text{x}^2-\text{x}^2}{2\text{xvx}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}^2-1}{2\text{v}}-\frac{\text{v}}1$
$=\frac{\text{v}^2-1-2\text{v}^2}{2\text{v}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{-1-\text{v}^2}{2\text{v}}$
$\int\frac{2\text{v}}{1+\text{v}^2}\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\log\big|1+\text{v}^2\big|=-\log|\text{x}|+\log|\text{C}|$
$1+\text{v}^2=\frac{\text{C}}{\text{x}}$
$1+\frac{\text{y}^2}{\text{x}^2}=\frac{\text{C}}{\text{x}}$
$\text{x}^2+\text{y}^2=\text{Cx}$
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