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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following differential equation:
$\Big(1+\text{e}^{\frac{\text{x}}{\text{y}}}\Big)\text{dx}+\text{e}^{\frac{\text{x}}{\text{y}}}\Big(1-\frac{\text{x}}{\text{y}}\Big)\text{dy}=0$
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Answer

We have,
$\Big(1+\text{e}^{\frac{\text{x}}{\text{y}}}\Big)\text{dx}+\text{e}^{\frac{\text{x}}{\text{y}}}\Big(1-\frac{\text{x}}{\text{y}}\Big)\text{dy}=0$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}=-\frac{\text{e}^{\frac{\text{x}}{\text{y}}}\Big(1-\frac{\text{x}}{\text{y}}\Big)}{1+\text{e}^{\frac{\text{x}}{\text{y}}}}$
This is a homogeneous differential equation.
Putting x = vy and $\frac{\text{dx}}{\text{dy}}=\text{v + y}\frac{\text{dv}}{\text{dy}}$, we get
$\text{v + y}\frac{\text{dv}}{\text{dy}}=-\frac{\text{e}^{\text{v}}(1-\text{v})}{1+\text{e}^{\text{v}}}$
$\Rightarrow\ \text{y}\frac{\text{dv}}{\text{dy}}=-\frac{\text{e}^{\text{v}}(1-\text{v})}{1+\text{e}^{\text{v}}}-\text{v}$
$\Rightarrow\ \text{y}\frac{\text{dv}}{\text{dy}}=\frac{-\text{e}^{\text{v}}+\text{e}^{\text{v}}\text{v}-\text{v}-\text{v}\text{e}^{\text{v}}}{1+\text{e}^{\text{v}}}$
$\Rightarrow\ \text{y}\frac{\text{dv}}{\text{dy}}=-\frac{\text{v}+\text{e}^{\text{v}}}{1+\text{e}^{\text{v}}}$
$\Rightarrow\ \frac{1+\text{e}^{\text{v}}}{\text{v}+\text{e}^{\text{v}}}\text{dv}=-\frac{1}{\text{y}}\text{dy}$
Integrating both sides, we get
$\int\frac{1+\text{e}^{\text{v}}}{\text{v}+\text{e}^{\text{v}}}\text{dv}=-\int\frac{1}{\text{y}}\text{dy}$
$\Rightarrow\ \log|\text{v}+\text{e}^{\text{v}}|=-\log|\text{y}|+\log\text{C}$
$\Rightarrow\ |\text{v}+\text{e}^{\text{v}}|=\Big|\frac{\text{C}}{\text{y}}\Big|$
$\Rightarrow\ \text{v}+\text{e}^{\text{v}}=\frac{\text{C}}{\text{y}}$
Putting $\text{v}=\frac{\text{x}}{\text{y}}$, we get
$\frac{\text{x}}{\text{y}}+\text{e}^{\frac{\text{x}}{\text{y}}}=\frac{\text{C}}{\text{y}}$
$\Rightarrow\ \text{x}+\text{ye}^{\frac{\text{x}}{\text{y}}}=\text{C}$
Hence, $\text{x}+\text{ye}^{\frac{\text{x}}{\text{y}}}=\text{C}$ is the required solution.

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