Question
Solve the following differential equation:
$(\text{x}+2\text{y})\text{dx}-(2\text{x}-\text{y})\text{dy}=0$
$(\text{x}+2\text{y})\text{dx}-(2\text{x}-\text{y})\text{dy}=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2\text{y}}{2\text{x}-\text{y}}$
This is a homogeneous differential equation. Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get $\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}+2\text{vx}}{2\text{x}-\text{vx}}$$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}}{2-\text{v}}-\text{v}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2}{2-\text{v}}$ $\Rightarrow\ \frac{2-\text{v}}{1+\text{v}^2}\text{dv}=\frac{1}{\text{x}}\text{dx}$ Integrating both sides, we get $\int\frac{2-\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}\ \dots(1)$ $\Rightarrow\ \int\frac{2}{1+\text{v}^2}\text{dv}-\int\frac{\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ \int\frac{2}{1+\text{v}^2}\text{dv}-\frac{1}2\int\frac{2\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ 2\tan^{-1}\text{v}-\frac{1}2\log|1+\text{v}^2|=\log|\text{x}|+\log\text{C}$ $\Rightarrow\ 2\tan^{-1}\text{v}=\log|\text{x}|+\log\text{C}+\log\Big|(1+\text{v}^2)^{\frac{1}2}\Big|$ $\Rightarrow\ 2\tan^{-1}\text{v}=\log\Big|\text{Cx}\sqrt{1+\text{v}^2}\Big|$ $\Rightarrow\ \Big|\text{Cx}\sqrt{1+\text{v}^2}\Big|=\text{e}^{2\tan^{-1}\text{v}}$ Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get $\Rightarrow\ \Bigg|\text{Cx}\sqrt{1+\Big(\frac{\text{y}}{\text{x}}\Big)^2}\Bigg|=\text{e}^{2\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)}$ $\Rightarrow\ \text{C}\sqrt{\text{x}^2+\text{y}^2}=\text{e}^{2\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)}$ Hence, $\sqrt{\text{x}^2+\text{y}^2}=\text{Ke}^{2\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)}$ is the required solution.
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$\tan\text{y}\frac{\text{dy}}{\text{dx}}=\sin(\text{x}+\text{y})+\sin(\text{x}-\text{y})$