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

Nitesh , 8 Years ago
Grade 12th pass
anser 2 Answers
Sai Bhargavi
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
Last Activity: 8 Years ago
ankit singh

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 = kxthank u regards ankit singh manit bhopal 3 rd year student forum answer expert at askiitians

This is the general solution

Last Activity: 5 Years ago
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