Question

Solve xlogx dy/dx +y=logx^2

Answer 1

Answer:

This question is very simple. Divide the equation by x. logx, on both sides.

So we get :

dy/dx + y/ x.logx = 2. logx / x .logx

i.e. dy/dx + y/ x logx = 2/x

Clearly this is a linear differential equation. So firstly we find the integrating factor.

I.F. = e ^ ( intg. { 1/ xlogx dx } ) ,

I.F. = logx

Now the solution of this differential equation is given by

y. I.F = intg. ( Q. I.F dx ) where Q = 2/x

Hence solution is : y. logx = (logx)^2 + c

Hope it helps.

Answer 2

Answer:

Step-by-step explanation:

This is linear in y and writing the DE in standard form : y ' + P(x) y = Q(x) we get P(x) = 1 / (x log x) and Q(x) = log (x^2) / (x log x) = 2 / x.