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Differential Equations — Maths STD 12 Science — Question

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

We have, $\text{x}\frac{\text{dy}}{\text{dx}}-\text{y + x}\sin\Big(\frac{\text{y}}{\text{x}}\Big)=0$ $\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y + x}\sin\Big(\frac{\text{y}}{\text{x}}\Big)}{\text{x}}$ This is a homogeneous differential equation. Put y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v +x}\frac{\text{dv}}{\text{dx}},$ we get$\text{v +x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}-\text{x}\sin\text{v}}{\text{x}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\text{v}-\sin\text{v}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=-\sin\text{v}$
$\Rightarrow\ \text{cosec v dv}=-\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get$\int\text{cosec v dv}=-\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ -\int\text{cosec v dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ -\log|\text{cosec v}-\cot\text{v}|=\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \log\Big|\frac{1}{\text{cosec v}-\cot\text{v}}\Big|=\log|\text{Cx}|$
$\Rightarrow\ \log|\text{cosec v}+\cot\text{v}|=\log|\text{Cx}|$
$\Rightarrow\ \log\Big|\frac{1+\cos\text{v}}{\sin\text{v}}\Big|=\log|\text{Cx}|$
$\Rightarrow\ \frac{1+\cos\text{v}}{\sin\text{v}}=\text{Cx}$
$\Rightarrow\ \text{x}\sin\text{v}=\frac{1}{\text{C}}(1+\cos\text{v})$
$\Rightarrow\ \text{x}\sin\text{v}=\text{K}(1+\cos\text{v})$ $\Big($where, $\text{K}=\frac{1}{\text{C}}\Big)$
Putting $\text{v}=\frac{\text{y}}{\text{x}},$ we get
$\Rightarrow\ \text{x}\sin\Big(\frac{\text{y}}{\text{x}}\Big)=\text{K}\Big[1+\cos\Big(\frac{\text{y}}{\text{x}}\Big)\Big]$
Hence, $\text{x}\sin\Big(\frac{\text{y}}{\text{x}}\Big)=\text{K}\Big[1+\cos\Big(\frac{\text{y}}{\text{x}}\Big)\Big]$ is the required solution.

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