What is the solution for y^'=(2y^2-9)/((3xy)) ?

The solution for y^'=(2y^2-9)/((3xy)) is y=sqrt((c_{1)x^{4/3}+9)/2},y=-sqrt((c_{1)x^{4/3}+9)/2}

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y^'=(2y^2-9)/((3xy))

y^{^{'}} = \frac{2 y ^{2} - 9}{( 3 x y )}

y = \frac{c _{1} x ^{\frac{4}{3}} + 9}{2}, y = - \frac{c _{1} x ^{\frac{4}{3}} + 9}{2}

Solution steps

y^{'} (x) = \frac{2 y ^{2} - 9}{( 3 x y )}

Solve separable ODE:

y = \frac{c _{1} x ^{\frac{4}{3}} + 9}{2}, y = - \frac{c _{1} x ^{\frac{4}{3}} + 9}{2}

y = \frac{c _{1} x ^{\frac{4}{3}} + 9}{2}, y = - \frac{c _{1} x ^{\frac{4}{3}} + 9}{2}

What is the solution for y^'=(2y^2-9)/((3xy)) ?
The solution for y^'=(2y^2-9)/((3xy)) is y=sqrt((c_{1)x^{4/3}+9)/2},y=-sqrt((c_{1)x^{4/3}+9)/2}