Question
Solve the following differential equation:
$(\text{x}^2-2\text{xy})\text{dy}+(\text{x}^2-3\text{xy}+2\text{y}^2)\text{dx}=0$
$(\text{x}^2-2\text{xy})\text{dy}+(\text{x}^2-3\text{xy}+2\text{y}^2)\text{dx}=0$
✓
Answer
Here, $(\text{x}^2-2\text{xy})\text{dy}+(\text{x}^2-3\text{xy}+2\text{y}^2)\text{dx}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2-3\text{xy}+2\text{y}^2}{2\text{xy}-\text{x}^2}$
It is a homogeneous equation.
Put x = vy
and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2-3\text{xvx}+2\text{v}^2\text{x}^2}{2\text{xvx}-\text{x}^2}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-3\text{v}+2\text{v}^2}{2\text{v}-1}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-3\text{v}+2\text{v}^2-2\text{v}^2+\text{v}}{2\text{v}-1}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-2\text{v}}{2\text{v}-1}$
$\frac{2\text{v}-1}{1-2\text{v}}\text{dv}=\frac{\text{dx}}{\text{x}}$
$\frac{1-2\text{v}}{1-2\text{v}}\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\int\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\text{v}=-\log|\text{x}|+\text{C}$
$\frac{\text{y}}{\text{x}}+\log\text{x}=\text{C}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2-3\text{xy}+2\text{y}^2}{2\text{xy}-\text{x}^2}$
It is a homogeneous equation.
Put x = vy
and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2-3\text{xvx}+2\text{v}^2\text{x}^2}{2\text{xvx}-\text{x}^2}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-3\text{v}+2\text{v}^2}{2\text{v}-1}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-3\text{v}+2\text{v}^2-2\text{v}^2+\text{v}}{2\text{v}-1}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-2\text{v}}{2\text{v}-1}$
$\frac{2\text{v}-1}{1-2\text{v}}\text{dv}=\frac{\text{dx}}{\text{x}}$
$\frac{1-2\text{v}}{1-2\text{v}}\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\int\text{dv}=-\int\frac{\text{dx}}{\text{x}}$
$\text{v}=-\log|\text{x}|+\text{C}$
$\frac{\text{y}}{\text{x}}+\log\text{x}=\text{C}$
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