What is the solution for y^'=(2+3t)sqrt(y),y(0)=3 ?

The solution for y^'=(2+3t)sqrt(y),y(0)=3 is y=t^2+(3t^3)/2+(3sqrt(3)t^2)/2+2sqrt(3)t+(9t^4)/(16)+3

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y^'=(2+3t)sqrt(y),y(0)=3

y^{^{'}} = (2 + 3 t) y, y (0) = 3

y = t^{2} + \frac{3 t ^{3}}{2} + \frac{3 3 t ^{2}}{2} + 23 t + \frac{9 t ^{4}}{16} + 3

Solution steps

y^{'} (t) = (2 + 3 t) y

Solve separable ODE:

y = t^{2} + \frac{3 t ^{3}}{2} + \frac{3 3 t ^{2}}{2} + 23 t + \frac{9 t ^{4}}{16} + 3

y = t^{2} + \frac{3 t ^{3}}{2} + \frac{3 3 t ^{2}}{2} + 23 t + \frac{9 t ^{4}}{16} + 3

x^{^{'}} = - 4 x

cos^{3} (y) \frac{d y}{d x} = x csc (y)

y^{^{'}} = \frac{3 x}{4 y}

\frac{d y}{d x} = - \frac{5}{3} x^{5} y^{7}

y^{^{'}} = y \frac{1 - y}{1 + y}

What is the solution for y^'=(2+3t)sqrt(y),y(0)=3 ?
The solution for y^'=(2+3t)sqrt(y),y(0)=3 is y=t^2+(3t^3)/2+(3sqrt(3)t^2)/2+2sqrt(3)t+(9t^4)/(16)+3