What is the solution for y^'=k(y-9-10cos(2*pi*t)) ?

The solution for y^'=k(y-9-10cos(2*pi*t)) is y=e^{kt}*integral of-9ke^{-kt}-10ke^{-kt}cos(2pit)

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y^'=k(y-9-10cos(2pit))

y^{^{'}} = k (y - 9 - 10 cos (2 \cdot π \cdot t))

y = e^{k t} \cdot \int - 9 k e^{- k t} - 10 k e^{- k t} cos (2 π t) d t

Solution steps

y^{'} (t) = k (y - 9 - 10 cos (2 \cdot π \cdot t))

Solve linear ODE:

y = e^{k t} \cdot \int - 9 k e^{- k t} - 10 k e^{- k t} cos (2 π t) d t

y = e^{k t} \cdot \int - 9 k e^{- k t} - 10 k e^{- k t} cos (2 π t) d t

\frac{d ^{2}}{d x ^{2}} (2 sin (- 4 x - 3))

n = 2 \sum \infty \frac{1}{5 ^{n}}

x \to 0 lim (\frac{sin ( x )}{8 x})

f (x) = \frac{sin ( x ) + cos ( x )}{e ^{x}}

\int 3 x^{2} - 8 x + C d x

What is the solution for y^'=k(y-9-10cos(2*pi*t)) ?
The solution for y^'=k(y-9-10cos(2*pi*t)) is y=e^{kt}*integral of-9ke^{-kt}-10ke^{-kt}cos(2pit)