Question
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\sin\Big(\frac{\text{y}}{\text{x}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\sin\Big(\frac{\text{y}}{\text{x}}\Big)$
✓
Answer
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\sin\Big(\frac{\text{y}}{\text{x}}\Big)$
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
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\text{v}+\sin\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\text{v}+\sin\text{v}-\text{v}$
$\Rightarrow\ \frac{1}{\sin\text{v}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{\sin\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\text{cosec v dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \log\Big|\tan\frac{\text{v}}{2}\Big|=\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \log\Big|\tan\frac{\text{v}}{2}\Big|=\log|\text{Cx}|$
$\Rightarrow\ \tan\frac{\text{v}}{2}=\text{Cx}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \tan\Big(\frac{\text{y}}{2\text{x}}\Big)=\text{Cx}$
Hence, $\tan\Big(\frac{\text{y}}{2\text{x}}\Big)=\text{Cx}$ is the required solution.
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
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\text{v}+\sin\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\text{v}+\sin\text{v}-\text{v}$
$\Rightarrow\ \frac{1}{\sin\text{v}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{1}{\sin\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\text{cosec v dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \log\Big|\tan\frac{\text{v}}{2}\Big|=\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \log\Big|\tan\frac{\text{v}}{2}\Big|=\log|\text{Cx}|$
$\Rightarrow\ \tan\frac{\text{v}}{2}=\text{Cx}$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\Rightarrow\ \tan\Big(\frac{\text{y}}{2\text{x}}\Big)=\text{Cx}$
Hence, $\tan\Big(\frac{\text{y}}{2\text{x}}\Big)=\text{Cx}$ is the required solution.
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