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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following differential equation:
$\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big).(\text{y dx + x dy})=\text{y}\sin\Big(\frac{\text{y}}{\text{x}}\Big).(\text{x dy}-\text{y dx})$
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Answer

We have,
$\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big).(\text{y dx + x dy})=\text{y}\sin\Big(\frac{\text{y}}{\text{x}}\Big).(\text{x dy}-\text{y dx})$
$\Rightarrow\ \text{xy}\cos\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}+\text{x}^2\cos\Big(\frac{\text{y}}{\text{x}}\Big)\text{dy}=\text{xy}\sin\Big(\frac{\text{y}}{\text{x}}\Big)\text{dy}-\text{y}^2\sin\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}$
$\Rightarrow\ \Big[\text{xy}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{y}^2\sin\Big(\frac{\text{y}}{\text{x}}\Big)\Big]\text{dx}=\Big[\text{xy}\sin\Big(\frac{\text{y}}{\text{x}}\Big)-\text{x}^2\cos\Big(\frac{\text{y}}{\text{x}}\Big)\Big]\text{dy}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{xy}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{y}^2\sin\Big(\frac{\text{y}}{\text{x}}\Big)}{\text{xy}\sin\Big(\frac{\text{y}}{\text{x}}\Big)-\text{x}^2\cos\Big(\frac{\text{y}}{\text{x}}\Big)}$
This is a homogeneous differential equation.
Putting x = vy and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}^2\cos\text{v}+\text{v}^2\text{x}^2\sin\text{v}}{\text{vx}^2\sin\text{v}-\text{x}^2\cos\text{v}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}\cos\text{v}+\text{v}^2\sin\text{v}}{\text{v}\sin\text{v}-\cos\text{v}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}\cos\text{v}+\text{v}^2\sin\text{v}}{\text{v}\sin\text{v}-\cos\text{v}}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}\cos\text{v}+\text{v}^2\sin\text{v}-\text{v}^2\sin\text{v}+\text{v}\cos\text{v}}{\text{v}\sin\text{v}-\cos\text{v}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{v}\cos\text{v}}{\text{v}\sin\text{v}-\cos\text{v}}$
$\Rightarrow\ \frac{\text{v}\sin\text{v}-\cos\text{v}}{2\text{v}\cos\text{v}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{\text{v}\sin\text{v}-\cos\text{v}}{2\text{v}\cos\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{\text{v}\sin\text{v}-\cos\text{v}}{\text{v}\cos\text{v}}\text{dv}=2\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{\text{v}\sin\text{v}}{\text{v}\cos\text{v}}\text{dv}-\int\frac{\cos\text{v}}{\text{v}\cos\text{v}}\text{dv}=2\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\tan\text{v dv}-\int\frac{1}{\text{v}}\text{dv}=2\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \log|\sec\text{v}|-\log|\text{v}|=2\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \log\Big|\frac{\sec\text{v}}{\text{v}}\Big|=\log\big|\text{Cx}^2\big|$
$\Rightarrow\ \frac{\sec{\text{v}}}{\text{v}}=\text{Cx}^2$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\sec\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{\text{y}}{\text{x}}\times\text{C}\times\text{x}^2$
$\Rightarrow\ \sec\Big(\frac{\text{y}}{\text{x}}\Big)=\text{Cxy}$
Hence, $\sec\Big(\frac{\text{y}}{\text{x}}\Big)=\text{Cxy}$ is the required solution.

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