Question

Solve the following differential equations:
\[\frac{dy}{dx} = \frac{y}{x}\left\{ \log y - \log x + 1 \right\}\]

Answer 1

We have, 
\[\frac{dy}{dx} = \frac{y}{x}\left\{ \log \left( \frac{y}{x} \right) + 1 \right\}\]
This is a homogeneous differential equation . 
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx}, \text{ we get }\]
\[v + x\frac{dv}{dx} = v\left\{ \log v + 1 \right\}\]
\[ \Rightarrow x\frac{dv}{dx} = v\log v + v - v\]
\[ \Rightarrow \frac{1}{v\log v}dv = \frac{1}{x}dx\]
Integrating both sides, we get 
\[\int\frac{1}{v\log v}dv = \int\frac{1}{x}dx\]
\[\text{ Putting }\log v = t,\text{ we get }\]
\[dv = vdt\]
\[ \therefore \int\frac{1}{v \times t} \times v dt = \int\frac{1}{x}dx\]
\[ \Rightarrow \int\frac{dt}{t} = \int\frac{1}{x}dx\]
\[ \Rightarrow \log \left| t \right| = \log \left| x \right| + \log C\]
\[ \Rightarrow t = Cx . . . . . (1) \]
Substituting the value of t in (1), we get 
\[\log v = Cx\]
\[\text{ Putting }v = \frac{y}{x},\text{ we get }\]
\[ \Rightarrow \log \left( \frac{y}{x} \right) = Cx\]
\[\text{ Hence, }\log \left( \frac{y}{x} \right) = Cx\text{ is the required solution .}\]