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# Differential equation is ylogy dx + (x-log y) dy = 0. Can anyone tell me the approach to find general solution?

Sai Bhargavi
18 Points
4 years ago
By rearranging the give differential equation we get , dx/dy =1/y -x/(ylogy)
dx/dy + x/(ylogy) =1/y
Now, calculate the integrating factor (IF),
e∫1/(ylogy)=I.F
Taking, logy=t (say)
differentiating on both sides we get, dy=ydt
IF=e∫dt/t=elogt=t, where t=logy
now, solving in linear differential equation form
x(t)= ∫t/y dy, since dy=ydt
xt = ∫t
xt=t2/2 +c
ie x(logy)=(logy)2/2 =c
ankit singh
10 months ago

Given,

$y\log y dx - x dy = 0$

$\\ \implies \frac{1}{ylog y}dy = \frac{1}{x}dx$

let logy = t

=> 1/ydy = dt

$\\ \implies \int \frac{dt}{t} = \int \frac{1}{x}dx \\ \implies \log t = \log x + \log k \\ \implies t = kx \\ \implies \log y = kx$thank u regards ankit singh manit bhopal 3 rd year student forum answer expert at askiitians

This is the general solution