What is the solution for y^'=2.25sqrt(y) ?

The solution for y^'=2.25sqrt(y) is y=(81t^2}{64}+\frac{9c_{1}t)/8+(c_{1}^2)/4

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y^'=2.25sqrt(y)

y^{^{'}} = 2.25 y

y = \frac{81 t ^{2}}{64} + \frac{9 c _{1} t}{8} + \frac{c _{1}^{2}}{4}

Solution steps

y^{'} (t) = 2.25 y

Solve separable ODE:

y = \frac{81 t ^{2}}{64} + \frac{9 c _{1} t}{8} + \frac{c _{1}^{2}}{4}

y = \frac{81 t ^{2}}{64} + \frac{9 c _{1} t}{8} + \frac{c _{1}^{2}}{4}

\frac{d}{d y} (x (y)^{\frac{2}{3}}) + y^{\frac{2}{3}} = a^{\frac{2}{3}}

(1 + t^{2}) \frac{d y}{d t} = (1 + y^{2})

y^{^{'}} = x^{2} y^{- 3}

(x^{2} + 1) y^{^{'}} + 4 x y = 3 x

y^{^{'}} = - \frac{x - 7}{y - 3}, y (8) = 0

What is the solution for y^'=2.25sqrt(y) ?
The solution for y^'=2.25sqrt(y) is y=(81t^2}{64}+\frac{9c_{1}t)/8+(c_{1}^2)/4