Question
The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{y}}{\text{x}}+\frac{\phi(\frac{\text{y}}{\text{x}})}{\phi'(\frac{\text{y}}{\text{x}})}$ is:
  1. $\phi(\frac{\text{y}}{\text{x}})=\text{Kx}$
  2. $\text{x}\phi(\frac{\text{y}}{\text{x}})=\text{K}$
  3. $\phi(\frac{\text{y}}{\text{x}})=\text{Ky}$
  4. $\text{y}\phi(\frac{\text{y}}{\text{x}})=\text{K}$ 

Answer

  1. $\phi(\frac{\text{y}}{\text{x}})=\text{Kx}$

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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