What is the solution for y^'+ycos(x)=4cos(x),y(0)=6 ?

The solution for y^'+ycos(x)=4cos(x),y(0)=6 is y=2e^{-sin(x)}+4

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y^'+ycos(x)=4cos(x),y(0)=6

y^{^{'}} + y cos (x) = 4 cos (x), y (0) = 6

y = 2 e^{- s i n (x)} + 4

Solution steps

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

Solve separable ODE:

y = 2 e^{- s i n (x)} + 4

y = 2 e^{- s i n (x)} + 4

(- 2.2) (1.2)

\frac{d}{d x} (ln (\frac{x + 8}{x - 8}))

\int \frac{2 x ^{2} + 4 x - 3}{x ^{2}} d x

y^{^{''}} + 121 y = cos (10 t)

\int 2^{\frac{1}{2}}

What is the solution for y^'+ycos(x)=4cos(x),y(0)=6 ?
The solution for y^'+ycos(x)=4cos(x),y(0)=6 is y=2e^{-sin(x)}+4