What is the solution for xy^'=3x^3y^{1/2}-2y ?

The solution for xy^'=3x^3y^{1/2}-2y is y=(9x^6}{64}+(3c_{1}x^2)/4+\frac{c_{1}^2)/(x^2)

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xy^'=3x^3y^{1/2}-2y

x y^{^{'}} = 3 x^{3} y^{\frac{1}{2}} - 2 y

y = \frac{9 x ^{6}}{64} + \frac{3 c _{1} x ^{2}}{4} + \frac{c _{1}^{2}}{x ^{2}}

Solution steps

x y^{'} (x) = 3 x^{3} y^{\frac{1}{2}} - 2 y

Solve Bernoulli ODE:

y = \frac{9 x ^{6}}{64} + \frac{3 c _{1} x ^{2}}{4} + \frac{c _{1}^{2}}{x ^{2}}

y = \frac{9 x ^{6}}{64} + \frac{3 c _{1} x ^{2}}{4} + \frac{c _{1}^{2}}{x ^{2}}

\frac{s}{( s + 1 ) ^{2} + 1}

\frac{d}{d x} ((x - x + 6 x^{4})^{3})

\int (x^{3} (1 + x^{2}))^{- 1} d x

(t + 6)^{3}

h \to 0 lim (- \frac{5 h}{x ( x + h )})

What is the solution for xy^'=3x^3y^{1/2}-2y ?
The solution for xy^'=3x^3y^{1/2}-2y is y=(9x^6}{64}+(3c_{1}x^2)/4+\frac{c_{1}^2)/(x^2)