What is the solution for y^'=(2+3x)sqrt(y),y(0)=3 ?

The solution for y^'=(2+3x)sqrt(y),y(0)=3 is y=x^2+(3x^3)/2+(3sqrt(3)x^2)/2+2sqrt(3)x+(9x^4)/(16)+3

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y^'=(2+3x)sqrt(y),y(0)=3

y^{^{'}} = (2 + 3 x) y, y (0) = 3

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

Solution steps

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

Solve separable ODE:

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

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

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

y^{^{'}} + (cos (x)) y = 5 cos (x)

y^{^{'}} = \frac{3}{2} y^{\frac{1}{3}}, y (0) = 0

x^{^{'}} = \frac{( x ^{2} + x )}{( 2 x + 1 )}

3 = \frac{π h ^{2}}{4} \frac{d h}{d t}

What is the solution for y^'=(2+3x)sqrt(y),y(0)=3 ?
The solution for y^'=(2+3x)sqrt(y),y(0)=3 is y=x^2+(3x^3)/2+(3sqrt(3)x^2)/2+2sqrt(3)x+(9x^4)/(16)+3