Grade 12th passDifferential Calculus

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

Nitesh

9 Years agoGrade 12th pass

2 Answers

Sai Bhargavi

9 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=e^logt=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

5 Years 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

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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

2 Answers

Sai Bhargavi

ankit singh

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