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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

(x^{2} + 1) (tan (y)) \frac{d y}{d x} = x

x (2 y + 1) d x - (x^{2} + 1) d y = 0

(8 x - 1) y^{^{'}} + 8 y = 0

cos (y) d x + e^{- x} sin (y) d y = 0

\frac{d}{df} (H (f)) = \frac{1 + u ( f ) ^{2}}{1 + f ^{2}}

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