Question
Solve the differential equation $\text{y e}^{\frac{\text{x}}{\text{y}}}\text{dx}=\Big(\text{x}\ \text{e}^{\frac{\text{x}}{\text{y}}}+\text{y}^2\Big)\text{dy}(\text{y}\neq0).$

Answer

Given: Differential equation $\text{y.e}^{\frac{\text{x}}{\text{y}}}\text{dx}=\Big(\text{x}.\text{e}^{\frac{\text{x}}{\text{y}}}+\text{y}^2\Big)\text{dy},\ \text{y}\neq0$
$\Rightarrow\ \ \frac{​​\text{dx}}{\text{dy}}=\frac{\text{x.e}^{\frac{\text{x}}{\text{y}}}+\text{y}^2}{\text{y.e}^{\frac{\text{x}}{\text{y}}}}=\frac{\text{x.e}^{\frac{\text{x}}{\text{y}}}}{\text{y.e}^\frac{\text{x}}{\text{y}}}+\frac{\text{y}^2}{\text{y.e}^{\frac{\text{x}}{\text{y}}}}$ $\Rightarrow\ \ \frac{\text{dx}}{\text{dy}}=\frac{\text{x}}{\text{y}}+\text{y.e}^{\frac{-\text{x}}{\text{y}}}\ \ ...\text{(i)}$
It is not a homogeneous differential equation because of presence of only y as a factor, yet it can be solved by putting $\frac{\text{x}}{\text{y}}=\text{v},\ \text{i.e}\ \text{x}=\text{vy}.\ \ \Rightarrow\ \ \frac{\text{dx}}{\text{dy}}=\text{v}+\text{y}\frac{\text{dv}}{\text{dy}}$
Putting these values in eq. (i), we get
$\text{v}+\text{y}\frac{\text{dv}}{\text{dy}}=\text{v}+\text{y e}^{-\text{v}}\ \ \Rightarrow\ \ \text{y}\frac{\text{dv}}{\text{dy}}=\text{y.e}^{-\text{x}}$ $\Rightarrow\ \ \text{y}\frac{\text{dv}}{\text{dy}}=\frac{\text{y}}{\text{e}^{\text{v}}}$
$\Rightarrow\ \ \text{e}^\text{v}=\text{y}+\text{c}\ \ \Rightarrow\ \ \text{e}^{\frac{\text{x}}{\text{y}}}=\text{y}+\text{c}$

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