What is the solution for y^'=-3y^{4/3}sin(t) ?

The solution for y^'=-3y^{4/3}sin(t) is y=-(27)/(27cos^3(t)+27c_{1)cos^2(t)+9c_{1}^2cos(t)+c_{1}^3}

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y^'=-3y^{4/3}sin(t)

y^{^{'}} = - 3 y^{\frac{4}{3}} sin (t)

y = - \frac{27}{27 cos ^{3} ( t ) + 27 c _{1} cos ^{2} ( t ) + 9 c _{1}^{2} cos ( t ) + c _{1}^{3}}

Solution steps

y^{'} (t) = - 3 y^{\frac{4}{3}} sin (t)

Solve separable ODE:

y = - \frac{27}{27 cos ^{3} ( t ) + 27 c _{1} cos ^{2} ( t ) + 9 c _{1}^{2} cos ( t ) + c _{1}^{3}}

y = - \frac{27}{27 cos ^{3} ( t ) + 27 c _{1} cos ^{2} ( t ) + 9 c _{1}^{2} cos ( t ) + c _{1}^{3}}

\frac{d y}{d x} = \frac{x + 1}{e ^{y}}

\frac{d y}{d x} = - \frac{8}{9} x^{2} y^{6}

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

x^{5} \frac{d y}{d x} = 3 x^{2} 1 - y^{2}

x^{3} + 8 y \frac{d y}{d x} = 0

What is the solution for y^'=-3y^{4/3}sin(t) ?
The solution for y^'=-3y^{4/3}sin(t) is y=-(27)/(27cos^3(t)+27c_{1)cos^2(t)+9c_{1}^2cos(t)+c_{1}^3}