What is the solution for y^'=t^3(y^2+3y+16) ?

The solution for y^'=t^3(y^2+3y+16) is y=(tan((sqrt(55)(t^4+c_{1}))/8)sqrt(55))/(2)-3/2

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y^'=t^3(y^2+3y+16)

y^{^{'}} = t^{3} (y^{2} + 3 y + 16)

y = \frac{tan ( \frac{55 ( t ^{4} + c _{1} )}{8} ) 55}{2} - \frac{3}{2}

Solution steps

y^{'} (t) = t^{3} (y^{2} + 3 y + 16)

Solve separable ODE:

y = \frac{tan ( \frac{55 ( t ^{4} + c _{1} )}{8} ) 55}{2} - \frac{3}{2}

y = \frac{tan ( \frac{55 ( t ^{4} + c _{1} )}{8} ) 55}{2} - \frac{3}{2}

\frac{d y}{d t} + t^{2} y = t^{2}

d y = - y d x

x^{^{'}} = (\frac{2}{t} - t) x

\frac{2 x - 6}{25} + \frac{2 ( y - 5 ) ( \frac{d y}{d x} )}{9} = 1

(x^{2} + 6 x^{2} y^{2}) d x + e^{x^{3}} y d y = 0

What is the solution for y^'=t^3(y^2+3y+16) ?
The solution for y^'=t^3(y^2+3y+16) is y=(tan((sqrt(55)(t^4+c_{1}))/8)sqrt(55))/(2)-3/2