Question

Solve: ydx – xdy + lnxdx = 0.

Answer 1

The given differential equation is 

ydx – xdy + lnxdx = 0

y - x(dy/dx) + lnx = 0

x(dy/dx) - y = lnx

dy/dx - (y/x) = lnx/x ...(i)

which is a linear differential equation

IF = e^-∫dx/x = e^{-log x} = 1/x

Multiplying both sides of Eq. (i) by IF and integrating, we get 

y.(IF) = ∫Q.(IF) dx + c

y.(1/x) = ∫(lnx/x²) dx + c

y/x = - ((log x/x) + (1/x)) + c

y = – (log x + 1) + c

which is the required solution of the given differential equation.