Solve the following differential equation: x^2 dy dx =x^2-2y^2+xy

Question

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

✓

Answer

Consider the given differential equation
$\text{x}^2\frac{\text{dy}}{\text{dx}}=\text{x}^2-2\text{y}^2+\text{xy}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}^2-2\text{y}^2+\text{xy}}{\text{x}^2}$
This is a homogeneous differential equation.
Substituting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we have
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2-2\text{v}^2\times\text{x}^2+\text{x}\times\text{v}\times\text{x}}{\text{x}^2}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=1-2\text{v}^2+\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=1-2\text{v}^2$
$\Rightarrow\ \frac{\text{dv}}{1-2\text{v}^2}=\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{\text{dv}}{\text{v}^2-\frac{1}2}=-2\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \int\frac{\text{dv}}{\big(\frac{1}{\sqrt2}\big)^2-\text{v}^2}=2\int\frac{\text{dx}}{\text{x}}$
$\Rightarrow\ \frac{\sqrt2}2\log\bigg(\frac{\frac{1}{\sqrt2}+\text{v}}{\frac{1}{\sqrt2}-\text{v}}\bigg)=2\log\text{x}+\log\text{C}$
$\Rightarrow\ \frac{1}{\sqrt2}\log\Bigg(\frac{\frac{1}{\sqrt2}+\frac{\text{y}}{\text{x}}}{\frac{1}{\sqrt2}-\frac{\text{y}}{\text{x}}}\Bigg)2\log\text{x}+\log\text{C}$
$\Rightarrow\ \frac{1}{\sqrt2}\log\Big(\frac{\text{x + y}\sqrt2}{\text{x}-\text{y}\sqrt2}\Big)2\log\text{x}+\log\text{C}$
$\Rightarrow\ \frac{1}{\sqrt2}\log\Big(\frac{\text{x + y}\sqrt2}{\text{x}-\text{y}\sqrt2}\Big)\log\text{x}^2+\log\text{C}$
$\Rightarrow\ \log\Big(\frac{\text{x + y}\sqrt2}{\text{x}-\text{y}\sqrt2}\Big)^{\frac{1}{\sqrt2}}=\log\text{Cx}^2$
$\Rightarrow\ \Big(\frac{\text{x + y}\sqrt2}{\text{x}-\text{y}\sqrt2}\Big)^{\frac{1}{\sqrt2}}=\text{Cx}^2$
$\Rightarrow\ \Big(\frac{\text{x + y}\sqrt2}{\text{x}-\text{y}\sqrt2}\Big)=\big(\text{Cx}^2\big)^{\sqrt2}$