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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

3 y y^{^{'}} = x, y (3) = 7

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

3 x - 4 y x^{2} + 1 \frac{d y}{d x} = 0, y (0) = - 8

3 \frac{d y}{d x} - 7 \frac{d y}{d x} + 3 y = 0

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

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