What is the solution for 4sin(t)(dy)/(dt)+4ycos(t)=cos(t)sin^4(t) ?

The solution for 4sin(t)(dy)/(dt)+4ycos(t)=cos(t)sin^4(t) is y=(sin^4(t))/(20)+(c_{1})/(sin(t))

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4sin(t)(dy)/(dt)+4ycos(t)=cos(t)sin^4(t)

4 sin (t) \frac{d y}{d t} + 4 y cos (t) = cos (t) sin^{4} (t)

y = \frac{sin ^{4} ( t )}{20} + \frac{c _{1}}{sin ( t )}

Solution steps

4 sin (t) \frac{d y}{d t} + 4 y cos (t) = cos (t) sin^{4} (t)

Solve linear ODE:

y = \frac{sin ^{4} ( t )}{20} + \frac{c _{1}}{sin ( t )}

y = \frac{sin ^{4} ( t )}{20} + \frac{c _{1}}{sin ( t )}

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 x} + x^{2}

x \to 0 lim (x^{2} - 2)

\frac{3}{x ^{2} - x - 2}, 1

\frac{\partial}{\partial y} (4 - x^{2} - 6 y^{2})

(8 t + y + 2) d t - d y = 0

What is the solution for 4sin(t)(dy)/(dt)+4ycos(t)=cos(t)sin^4(t) ?
The solution for 4sin(t)(dy)/(dt)+4ycos(t)=cos(t)sin^4(t) is y=(sin^4(t))/(20)+(c_{1})/(sin(t))