Question

Find general solution: \((x+3y^2) \frac {dy}{dx} = y,\  (y>0)\) 

Answer 1

(x+3y²)\(\frac {dy}{dx}\) = y

⇒\(\frac {dy}{dx}\)= \(\frac yx\)+3y²

⇒\(\frac {dx}{dy}\) = \(\frac {x+3y^2}{y}\) = \(\frac xy\)+3y

⇒\(\frac {dx}{dy}\)-\(\frac xy\) = 3y

This is a linear differential equation of the form:

\(\frac {dx}{dy}\)+px = Q (where p=-\(\frac 1y\) and Q=3y)

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

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

x(I.F.) = ∫(Q×I.F.)dy+C

⇒x.\(\frac 1y\) = ∫(3y.\(\frac 1y\))dy+C

⇒\(\frac xy\) = 3y+C

⇒x = 3y²+Cy