What is the solution for y^'+ycos(x)=2cos(x),y(0)=4 ?

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

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y^'+ycos(x)=2cos(x),y(0)=4

y^{^{'}} + y cos (x) = 2 cos (x), y (0) = 4

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

Solution steps

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

Solve separable ODE:

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

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

\int x^{3} e^{2 x^{2}} d x

\frac{d}{d x} (e^{s i n (x) + c o s (x)})

\frac{d}{d x} ((sin (20 x))^{3})

x + 3

\int_{- 1}^{- 4} \frac{1 - x ^{4}}{x ^{2}} d x

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