68. ydx=xlog(yx)dy−2xdy=0

The given D.E is ydx+xlog(yx).dy−2xdy=0⇒ydx=[2xdy−xlog(yx)dy]⇒ydx=[2x−xlog(yx)]dy⇒dydx=y2x−xlog(yx)=yx(2−logyx)=yx2−logyx=f(yx) Hence, the given D.E is homogenous. Let, y=vx=yx=v so that dydx=v+xdvdx in the D.E. Then, v+xdvdx=v2−logv ⇒xdvdx=v2−logv−v=v−2v+vlogv2−logv=vlogv−v2−logv⇒2logvv[logv−1]dv=dxx⇒1+1−logvv[logv−1]dv=dxx⇒1−[logv−1]v[logv−1]dv=dxx ⇒[1v(logv−1)−1v]dv=dxx Integrating both sides we get, ∫[1v(logv−1)−1v]dv=∫dxx∫dvv(logv−1)−logv=logx+logc Let, logv−1=t,so,ddv(logv−1)=dtdv ⇒1v=dtdv⇒dvv=dt⇒∫dtt−logv=logx+logc⇒logt−logv=logx+logc⇒log|logv−1|−logv=logcx Putting back v=yx we get, log|log(yx)−1|−logyx=logcx=log[log(yx)−1yx]=logcx=yx[log(yx)−1]=cx =log(yx)−1=cy is the required solution.

68. $y d x = x l o g (\frac{y}{x}) d y - 2 x d y = 0$

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Vishal Baghel | Contributor-Level 10

8 months ago

The given D.E is

$\begin{array}{l} y d x + x l o g (\frac{y}{x}) . d y - 2 x d y = 0 \\ \Rightarrow y d x = [2 x d y - x l o g (\frac{y}{x}) d y] \\ \Rightarrow y d x = [2 x - x l o g (\frac{y}{x})] d y \\ \Rightarrow \frac{d y}{d x} = \frac{y}{2 x - x l o g (\frac{y}{x})} = \frac{y}{x (2 - l o g \frac{y}{x})} \\ = \frac{\frac{y}{x}}{2 - l o g \frac{y}{x}} = f (\frac{y}{x}) \end{array}$

Hence, the given D.E is homogenous.

Let, $y = v x = \frac{y}{x} = v$ so that $\frac{d y}{d x} = v + x \frac{d v}{d x}$ in the D.E.

Then, $v + x \frac{d v}{d x} = \frac{v}{2 - l o g v}$

$\begin{array}{l} \Rightarrow x \frac{d v}{d x} = \frac{v}{2 - l o g v} - v = \frac{v - 2 v + v l o g v}{2 - l o g v} = \frac{v l o g v - v}{2 - l o g v} \\ \Rightarrow \frac{2 l o g v}{v [l o g v - 1]} d v = \frac{d x}{x} \\ \Rightarrow \frac{1 + 1 - l o g v}{v [l o g v - 1]} d v = \frac{d x}{x} \\ \Rightarrow \frac{1 - [l o g v - 1]}{v [l o g v - 1]} d v = \frac{d x}{x} \end{array}$

$\Rightarrow [\frac{1}{v (l o g v - 1)} - \frac{1}{v}] d v = \frac{d x}{x}$

Integrating both sides we get,

$\begin{array}{l} \int [\frac{1}{v (l o g v - 1)} - \frac{1}{v}] d v = \int \frac{d x}{x} \\ \int \frac{d v}{v (l o g v - 1)} - l o g v = l o g x + l o g c \end{array}$

Let, $l o g v - 1 = t, s o, \frac{d}{d v} (l o g v - 1) = \frac{d t}{d v}$

$\begin{array}{l} \Rightarrow \frac{1}{v} = \frac{d t}{d v} \\ \Rightarrow \frac{d v}{v} = d t \\ \Rightarrow \int \frac{d t}{t} - l o g v = l o g x + l o g c \\ \Rightarrow l o g t - l o g v = l o g x + l o g c \\ \Rightarrow l o g | l o g v - 1 | - l o g v = l o g c x \end{array}$

Putting back $v = \frac{y}{x}$ we get,

$\begin{array}{l} l o g | l o g (\frac{y}{x}) - 1 | - l o g \frac{y}{x} = l o g \frac{c}{x} \\ = l o g [\frac{l o g (\frac{y}{x}) - 1}{\frac{y}{x}}] = l o g \frac{c}{x} \\ = \frac{y}{x} [l o g (\frac{y}{x}) - 1] = c x \end{array}$