- $\phi(\frac{\text{y}}{\text{x}})=\text{Kx}$
- $\text{x}\phi(\frac{\text{y}}{\text{x}})=\text{K}$
- $\phi(\frac{\text{y}}{\text{x}})=\text{Ky}$
- $\text{y}\phi(\frac{\text{y}}{\text{x}})=\text{K}$
$\phi(\frac{\text{y}}{\text{x}})=\text{Kx}$
Solution:
We have,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\frac{\phi(\frac{\text{y}}{\text{x}})}{\phi'(\frac{\text{y}}{\text{x}})}$
Let $\text{y}=\text{ux}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{u}+\text{x}\frac{\text{du}}{\text{dx}}$
$\therefore \text{u}+\text{x}\frac{\text{du}}{\text{dx}}=\text{u}+\frac{\phi(\text{u})}{\phi'(\text{u})}$
$\Rightarrow \text{x}\frac{\text{du}}{\text{dx}}=\frac{\phi(\text{u})}{\phi'(\text{u})}$
$\Rightarrow \frac{\phi(\text{u})}{\phi'(\text{u})}\text{du}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$ \int\frac{\phi(\text{u})}{\phi'(\text{u})}\text{du}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow \log|\phi(\text{v})|=\log|\text{x}|+\log|\text{K}|$
$\Rightarrow \log|\phi(\frac{\text{y}}{2})|-\log|\text{x}|=\log\text{K}$
$\Rightarrow \log|\phi(\frac{\text{y}}{2})|=\log\text{K}$
$\Rightarrow\phi(\frac{\text{y}}{2})|=\text{Kx}$
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