What is the solution for y^'+ycos(x)=4cos(x) ?

The solution for y^'+ycos(x)=4cos(x) is y=e^{-sin(x)+c_{1}}+4

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y^'+ycos(x)=4cos(x)

y^{^{'}} + y cos (x) = 4 cos (x)

y = e^{- s i n (x) + c_{1}} + 4

Solution steps

y^{'} (x) + y cos (x) = 4 cos (x)

Solve separable ODE:

y = e^{- s i n (x) + c_{1}} + 4

y = e^{- s i n (x) + c_{1}} + 4

\frac{d}{d x} (\frac{1 - cos ( x )}{x ^{2}})

\int sin (2 x) 1 + 2 cos (2 x) d x

x \to 2 lim (2 (3))

\int cos^{4} (5 x + 7) d x

1 - 4 x^{2} \frac{d y}{d x} = x

What is the solution for y^'+ycos(x)=4cos(x) ?
The solution for y^'+ycos(x)=4cos(x) is y=e^{-sin(x)+c_{1}}+4