x*z*dz/dx = x^2 + 3*z^2?

Question

Use the fact that if a new dependant variable y = z^(1-n) is defined, the DE dz/dx + p(x)*z = q(x)*z^n will become linear in y(x).

Answer 1

Well, they've given us a little hint there, so let's apply it:
First write the ODE as
dz/dx - (3/x) z = x/z = xz^(-1), so use n = -1 and let y = z^2,
dy/dx = 2z dz/dx = 2(x + 3z^2 / x) = 2(x + 3y/x)
So dy/dx + (-6/x) y = 2x.
This is a linear first-order ODE in y, so take an integrating factor of e^∫(-6/x) dx = x^(-6).
x^(-6) dy/dx + (-6x^(-7)) y = 2x^(-5)
=> d/dx (x^(-6).y) = 2x^(-5)
=> x^(-6) . y = 2x^(-4)/(-4) + c
=> y = -x^2 / 2 + cx^6
Now y = z^2, so the solution is
z^2 = cx^6 - x^2 / 2
or z = ±x √(cx^4 - 1/2)