Question

Solve the following differential equation:- `y dx + x log  (y)/(x)dy-2x dy=0`

Answer 1

We have,

\[y dx + x \log \left( \frac{y}{x} \right)dy - 2x dy = 0\]

\[ \Rightarrow x \log \left( \frac{y}{x} \right)dy - 2x dy = - y dx\]

\[ \Rightarrow \left[ \log \left( \frac{y}{x} \right) - 2 \right]x dy = - y dx\]

\[ \Rightarrow \frac{dy}{dx} = \frac{- y}{\left[ \log \left( \frac{y}{x} \right) - 2 \right]x}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{\frac{y}{x}}{2 - \log \left( \frac{y}{x} \right)} . . . . . . . . \left( 1 \right)\]

Clearly this is a homogenous equation,

Putting y = vx

\[ \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\]

\[\text{Substituting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx}\text{ in (1) we get}\]

\[v + x\frac{dv}{dx} = \frac{v}{2 - \log \left( v \right)}\]

\[ \Rightarrow x\frac{dv}{dx} = \frac{v}{2 - \log \left( v \right)} - v\]

\[ \Rightarrow x\frac{dv}{dx} = \frac{v - 2v + v \log \left( v \right)}{2 - \log \left( v \right)}\]

\[ \Rightarrow x\frac{dv}{dx} = \frac{- v + v \log \left( v \right)}{2 - \log \left( v \right)}\]

\[ \Rightarrow \frac{2 - \log \left( v \right)}{- v + v \log \left( v \right)}dv = \frac{1}{x}dx\]

\[ \Rightarrow \frac{\log \left( v \right) - 2}{v \log \left( v \right) - v}dv = - \frac{1}{x}dx\]

\[ \Rightarrow \frac{\log \left( v \right) - 1 - 1}{v \left[ \log \left( v \right) - 1 \right]}dv = - \frac{1}{x}dx\]

\[ \Rightarrow \frac{\log \left( v \right) - 1}{v \left[ \log \left( v \right) - 1 \right]}dv - \frac{1}{v \left[ \log \left( v \right) - 1 \right]}dv = - \frac{1}{x}dx\]

\[ \Rightarrow \frac{1}{v}dv - \frac{1}{v \left[ \log \left( v \right) - 1 \right]}dv = - \frac{1}{x}dx\]

Integrating both sides we get

\[\int\frac{1}{v}dv - \int\frac{1}{v \left[ \log \left( v \right) - 1 \right]}dv = - \int\frac{1}{x}dx\]

\[ \Rightarrow \log \left| v \right| - I = - \log \left| x \right| - \log C . . . . . . . . \left( 2 \right)\]

Where,

\[I = \int\frac{1}{v \left[ \log \left| \left( v \right) \right| - 1 \right]}dv\]

Puting log v = t

\[\frac{1}{v}dv = dt\]

\[ \therefore I = \int\frac{1}{t - 1}dt\]

\[ \Rightarrow I = \log \left| t - 1 \right|\]

\[ \Rightarrow I = \log \left| \log \left| v \right| - 1 \right| . . . . . \left( 3 \right)\]

From (2) and (3) we get

\[\log \left| v \right| - \log \left| \log \left| v \right| - 1 \right| = - \log \left| x \right| - \log C\]

\[ \Rightarrow \log \left| \frac{v}{\log \left| v \right| - 1} \right| = - \log \left| Cx \right|\]

\[ \Rightarrow \frac{v}{\log \left| v \right| - 1} = \frac{1}{Cx}\]

\[ \Rightarrow \log \left| v \right| - 1 = vCx\]

\[ \Rightarrow \log \left| \frac{y}{x} \right| - 1 = Cy\]