What is the solution for (dy)/(dx)=5x^2sqrt(y),y>0 ?

The solution for (dy)/(dx)=5x^2sqrt(y),y>0 is y=(25x^6}{36}+\frac{5c_{1}x^3)/6+(c_{1}^2)/4

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(dy)/(dx)=5x^2sqrt(y),y>0

\frac{d y}{d x} = 5 x^{2} y, y > 0

y = \frac{25 x ^{6}}{36} + \frac{5 c _{1} x ^{3}}{6} + \frac{c _{1}^{2}}{4}

Solution steps

\frac{d y}{d x} = 5 x^{2} y

Solve separable ODE:

y = \frac{25 x ^{6}}{36} + \frac{5 c _{1} x ^{3}}{6} + \frac{c _{1}^{2}}{4}

y = \frac{25 x ^{6}}{36} + \frac{5 c _{1} x ^{3}}{6} + \frac{c _{1}^{2}}{4}

e^{- y} y^{^{'}} = e^{x} + 4 x^{2}

y^{^{'}} = \frac{x ^{2} - 8}{y ^{2}}, y (0) = 2

y^{^{'}} = - (sin (x)) (cos^{2} (y))

y^{^{'}} = - \frac{9 x}{y}

6 (y^{2} + 2 x y \frac{d y}{d x}) + 18 x = - 4 y^{3} \frac{d y}{d x}

What is the solution for (dy)/(dx)=5x^2sqrt(y),y>0 ?
The solution for (dy)/(dx)=5x^2sqrt(y),y>0 is y=(25x^6}{36}+\frac{5c_{1}x^3)/6+(c_{1}^2)/4