What is the solution for 4(sin(t)y^'+cos(t)y)=cos(t)sin^5(t) ?

The solution for 4(sin(t)y^'+cos(t)y)=cos(t)sin^5(t) is y=(sin^5(t))/(24)+(c_{1})/(sin(t))

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4(sin(t)y^'+cos(t)y)=cos(t)sin^5(t)

4 (sin (t) y^{^{'}} + cos (t) y) = cos (t) sin^{5} (t)

y = \frac{sin ^{5} ( t )}{24} + \frac{c _{1}}{sin ( t )}

Solution steps

4 (sin (t) y^{'} (t) + cos (t) y) = cos (t) sin^{5} (t)

Solve linear ODE:

y = \frac{sin ^{5} ( t )}{24} + \frac{c _{1}}{sin ( t )}

y = \frac{sin ^{5} ( t )}{24} + \frac{c _{1}}{sin ( t )}

\frac{2 x - 1}{x + 3}, at x = 4

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\int sin^{3} (θ) cos^{6} (θ) d θ

f (2.1) = 3 x^{2} - 4 x + 1, (2)

What is the solution for 4(sin(t)y^'+cos(t)y)=cos(t)sin^5(t) ?
The solution for 4(sin(t)y^'+cos(t)y)=cos(t)sin^5(t) is y=(sin^5(t))/(24)+(c_{1})/(sin(t))