What is the solution for y^'+cos(x)y=2cos(x) ?

The solution for y^'+cos(x)y=2cos(x) is y=e^{-sin(x)+c_{1}}+2

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y^'+cos(x)y=2cos(x)

y^{^{'}} + cos (x) y = 2 cos (x)

y = e^{- s i n (x) + c_{1}} + 2

Solution steps

y^{'} (x) + cos (x) y = 2 cos (x)

Solve separable ODE:

y = e^{- s i n (x) + c_{1}} + 2

y = e^{- s i n (x) + c_{1}} + 2

x \to - 1 - lim (x)

f (x) = 5 arcsin (x^{4})

\frac{d}{d x} (1 - 25 x^{2} arccos (5 x))

x^{^{'}} = 2 (x + 1), x (0) = 0

\int cos (\frac{x y}{2}) d y

What is the solution for y^'+cos(x)y=2cos(x) ?
The solution for y^'+cos(x)y=2cos(x) is y=e^{-sin(x)+c_{1}}+2