Solve the following differential equations: dy dx = y-x y + x - Vidyadip

Question

Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}-\text{x}}{\text{y + x}}$

✓

Answer

$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}-\text{x}}{\text{y + x}}$
It is homogeneous equation
Put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}-\text{x}}{\text{vx + x}}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}-1}{\text{v}+1}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}-1}{\text{v}+1}-\text{v}$
$=\frac{\text{v}-1-\text{v}^2-\text{v}}{\text{v}+1}$
$\text{x}\frac{\text{dv}}{\text{dx}}=-\frac{(1+\text{v}^2)}{\text{v}+1}$
$\int\frac{\text{v}+1}{\text{v}^2+1}\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\int\frac{\text{v}}{\text{v}^2+1}\text{dv}+\int\frac{1}{\text{v}^2+1}\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\frac{1}2\int\frac{2\text{v}}{\text{v}^2+1}\text{dv}+\int\frac{1}{\text{v}^2+1}\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\frac{1}2\log\big|\text{v}^2+1\big|+\tan^{-1}\text{v}=-\log|\text{x}|+\log|\text{C}|$
$\log\Big|\frac{\text{y}^2+\text{x}^2}{\text{x}^2}\Big|+2\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=2\log\Big|\frac{\text{C}}{\text{x}}\Big|$
$\log\big|\text{x}^2+\text{y}^2\big|-2\log|\text{x}|+2\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=2\log\Big|\frac{\text{C}}{\text{x}}\Big|$
$\log\big|\text{x}^2+\text{y}^2\big|+2\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=2\log|\text{C}|$
$\log\big|\text{x}^2+\text{y}^2\big|+2\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\text{K}$

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