What is the solution for y^{''}+2y^'+5y=4cos(2t)e^{-t} ?

The solution for y^{''}+2y^'+5y=4cos(2t)e^{-t} is y=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))+e^{-t}tsin(2t)

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y^{''}+2y^'+5y=4cos(2t)e^{-t}

y^{^{''}} + 2 y^{^{'}} + 5 y = 4 cos (2 t) e^{- t}

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 4 cos (2 t) e^{- t}

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (sin (x^{2} - 9 x y))

\int \frac{1}{cos ^{5} ( x )} d x

2 y y^{^{'}} + 2 = y^{2} + 2 x, y (0) = 6

\frac{2}{x ^{\frac{1}{3}}} - 2

2 y^{^{'}} + t y = 7

What is the solution for y^{''}+2y^'+5y=4cos(2t)e^{-t} ?
The solution for y^{''}+2y^'+5y=4cos(2t)e^{-t} is y=e^{-t}(c_{1}cos(2t)+c_{2}sin(2t))+e^{-t}tsin(2t)