Solve-ydx-xdy-log-xdx-0-

Question Number 40422 by rahul 19 last updated on 21/Jul/18

Solve :

yd x - x d y + \log x d x = 0

Commented by prof Abdo imad last updated on 22/Jul/18

\Rightarrow y - x \frac{d y}{d x} + l n (x) = 0 \Rightarrow {x y}^{^{'}} - y - l n (x) = 0 \Rightarrow

{x y}^{^{'}} - y = l n (x)

h e \Rightarrow {x y}^{^{'}} - y = 0 \Rightarrow \frac{y^{^{'}}}{y} = \frac{1}{x} \Rightarrow l n ∣ y ∣= l n ∣ x ∣ + c \Rightarrow

\Rightarrow y = k ∣ x ∣

c a s e 1 x > 0 m v c m e t h o d g i v e

y^{^{'}} = k^{^{'}} x + k

e) \Rightarrow x (k^{^{'}} x + k) - k x = l n (x) \Rightarrow

k^{^{'}} x^{2} = l n (x) \Rightarrow k^{^{'}} = \frac{l n (x)}{x^{2}} \Rightarrow k (x) = \int_{.}^{x} \frac{l n (t)}{t^{2}} d t + c

b y p a r t s \int \frac{l n (t)}{t^{2}} d t = - \frac{1}{t} l n (t) + \int \frac{1}{t} \frac{d t}{t}

= - \frac{1}{t} l n (t) - \frac{1}{t} \Rightarrow k (x) = - \frac{l n (x)}{x} - \frac{1}{x} + c \Rightarrow

y (x) = x (- \frac{l n (x)}{x} - \frac{1}{x} + c)

= c x - 1 - l n (x)

c a s e 2 x < 0 w e f o l l o w t h e s a m e m e t h o d .

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Jul/18

\frac{x d y - y d x}{x^{2}} = \frac{l n x d x}{x^{2}}

d (\frac{y}{x}) = \frac{l n x d x}{x^{2}}

\int d (\frac{y}{x}) = l n x \int x^{- 2} d x - \int \frac{1}{x} . (\frac{x^{- 1}}{- 1}) d x

\frac{y}{x} = - \frac{l n x}{x} + \int \frac{d x}{x^{2}} + c

\frac{y}{x} = - \frac{l n x}{x} - \frac{1}{x} + c

\frac{y + 1 + l n x}{x} = c

Commented by rahul 19 last updated on 21/Jul/18

thank you sir ��