What is the solution for y^'+ycos(t)= 1/2 sin(2t) ?

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

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y^'+ycos(t)= 1/2 sin(2t)

y^{^{'}} + y cos (t) = \frac{1}{2} sin (2 t)

y = sin (t) - 1 + \frac{c _{1}}{e ^{s i n (t)}}

Solution steps

y^{'} (t) + y cos (t) = \frac{1}{2} sin (2 t)

Solve linear ODE:

y = sin (t) - 1 + \frac{c _{1}}{e ^{s i n (t)}}

y = sin (t) - 1 + \frac{c _{1}}{e ^{s i n (t)}}

\frac{d ^{2}}{d x ^{2}} (x^{6} e^{x})

\int_{0}^{1} 7

\frac{\partial}{\partial x} ((x + v (x, t) t)^{3})

x \to - 3 lim (- x^{2} + 6 x - 1)

n = 1 \sum \infty \frac{8}{5 ^{n}}

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