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

Question

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

✓

Answer

Here, $(\text{x}-\text{y})\frac{\text{dy}}{\text{dx}}=\text{x + 2y}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2\text{y}}{\text{x}-\text{y}}$
It is a homogeneous equation.
Put x = vy
and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}+2\text{vx}}{\text{x}-\text{vx}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}}{1+\text{v}}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}-\text{v + v}^2}{1-\text{v}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v + v}^2}{1-\text{v}}$
$\frac{1-\text{v}}{\text{v}^2+\text{v}+1}\text{dv}=\frac{\text{dx}}{\text{x}}$
$-\frac{\text{v}-1}{\text{v}^2+\text{v}+1}\text{dv}=\frac{\text{dx}}{\text{x}}$
$\frac{1}2\times\frac{2\text{v}-2}{\text{v}^2+\text{v}+1}\text{dv}=\frac{-\text{dx}}{\text{x}}$
$\int\frac{(2\text{v}+1)-3}{\text{v}^2+\text{v}+1}\text{dv}=-\int\frac{2\text{dx}}{\text{x}}$
$\int\frac{2\text{v}+1}{\text{v}^2+\text{v}+1}\text{dv}-\int\frac{3}{\text{v}^2+2\text{v}\big(\frac{1}2\big)+\big(\frac{1}2\big)^2-\big(\frac{1}2\big)^2+1}=-2\int\frac{\text{dx}}{\text{x}}$
$\int\frac{2\text{v}+1}{\text{v}^2+\text{v}+1}\text{dv}-\int\frac{3}{\big(\text{v}+\frac{1}2\big)^2+\big(\frac{\sqrt3}2\big)^2}\text{dv}=-2\int\frac{\text{dx}}{\text{x}}$
$\log|\text{v}^2+\text{v}+1|-3\Big(\frac{2}{\sqrt3}\Big)\tan^{-1}\Bigg(\frac{\text{v}+\frac{1}2}{\frac{\sqrt3}{2}}\Bigg)=-2\log|\text{x}|+\text{C}$
$\log|\text{y}^2+\text{xy}+\text{x}^2|=2\sqrt3\tan^{-1}\Big(\frac{2\text{y + x}}{\text{x}\sqrt3}\Big)+\text{C}$

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