What is the solution for y^{''}+2y^'+2y=3cos(t)-sin(t) ?

The solution for y^{''}+2y^'+2y=3cos(t)-sin(t) is y=e^{-t}(c_{1}cos(t)+c_{2}sin(t))+sin(t)+cos(t)

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y^{''}+2y^'+2y=3cos(t)-sin(t)

y^{^{''}} + 2 y^{^{'}} + 2 y = 3 cos (t) - sin (t)

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + sin (t) + cos (t)

Solution steps

y^{''} (t) + 2 y^{'} (t) + 2 y = 3 cos (t) - sin (t)

Solve linear ODE:

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + sin (t) + cos (t)

y = e^{- t} (c_{1} cos (t) + c_{2} sin (t)) + sin (t) + cos (t)

\int_{1}^{3} x e^{- x^{2}} d x

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

(0.2) (6.6)

x \to 3 lim (\frac{x ^{2}}{9 - x ^{2}})

\frac{d}{d x} (\frac{( e ^{x} )}{x ^{2}})

What is the solution for y^{''}+2y^'+2y=3cos(t)-sin(t) ?
The solution for y^{''}+2y^'+2y=3cos(t)-sin(t) is y=e^{-t}(c_{1}cos(t)+c_{2}sin(t))+sin(t)+cos(t)