Question

Find general solution: \(y dx+(x-y^2)dy=0\)

Answer 1

y dx+(x-y²)dy = 0

⇒ydx = (y²-x)dy

\(\implies\)\(\frac {dx}{dy}\) = y²-\(\frac xy\) = y-\(\frac xy\)

\(\implies\)\(\frac {dx}{dy}\)+\(\frac xy\) = y

This is a linear differential equation of the form:

\(\frac {dy}{dx}\)+px = Q (where p=\(\frac 1y\) and Q=y)

Now, I.F = e^∫pdy = \(e^{∫\frac 1y dy}\) = e^{log y} = y

The general solution of the given differential equation is given by the relation,

x(I.F.) = ∫(Q×I.F.)dy + C

\(\implies\)xy = ∫(y.y)dy + C

\(\implies\)xy = ∫y²dy + C

\(\implies\)xy = \(\frac {y^3}{3}\) + C

\(\implies\)x = \(\frac {y^2}{3}\) + \(\frac Cy\)