What is the solution for y^'-8y^2cos(t)-7y^2sin(4t)=0 ?

The solution for y^'-8y^2cos(t)-7y^2sin(4t)=0 is y=-4/(32sin(t)-7cos(4t)+c_{1)}

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y^'-8y^2cos(t)-7y^2sin(4t)=0

y^{^{'}} - 8 y^{2} cos (t) - 7 y^{2} sin (4 t) = 0

y = - \frac{4}{32 sin ( t ) - 7 cos ( 4 t ) + c _{1}}

Solution steps

y^{'} (t) - 8 y^{2} cos (t) - 7 y^{2} sin (4 t) = 0

Solve separable ODE:

y = - \frac{4}{32 sin ( t ) - 7 cos ( 4 t ) + c _{1}}

y = - \frac{4}{32 sin ( t ) - 7 cos ( 4 t ) + c _{1}}

\int (\frac{e ^{x}}{1 + e ^{x}}) d x

\int 2 x (x^{2} + 14)^{9} d x

\frac{( d y )}{( d x )} = \frac{x}{y}

\int_{1}^{2} (ln (x))^{2} d x

y = - x, x = - 1, x = 2, y = 0

What is the solution for y^'-8y^2cos(t)-7y^2sin(4t)=0 ?
The solution for y^'-8y^2cos(t)-7y^2sin(4t)=0 is y=-4/(32sin(t)-7cos(4t)+c_{1)}