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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

y = \frac{10}{x ^{2} - 2 x - 24}, x = - 2, x = 2

h^{^{'}} = 6 - 5 cos (\frac{π}{15} t)

\int \frac{1 - sin ( x )}{x + cos ( x )} d x

x \to + (- 1) lim (x)

\frac{d}{d x} (- \frac{16}{x ^{3}})

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