What is the solution for y^{''}+6y^'+13y=10e^{-2t},y(0)=3 ?

The solution for y^{''}+6y^'+13y=10e^{-2t},y(0)=3 is y=e^{-3t}(cos(2t)+c_{2}sin(2t))+2e^{-2t}

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y^{''}+6y^'+13y=10e^{-2t},y(0)=3

y^{^{''}} + 6 y^{^{'}} + 13 y = 10 e^{- 2 t}, y (0) = 3

y = e^{- 3 t} (cos (2 t) + c_{2} sin (2 t)) + 2 e^{- 2 t}

Solution steps

y^{''} (t) + 6 y^{'} (t) + 13 y = 10 e^{- 2 t}

Solve linear ODE:

y = e^{- 3 t} (cos (2 t) + c_{2} sin (2 t)) + 2 e^{- 2 t}

y = e^{- 3 t} (cos (2 t) + c_{2} sin (2 t)) + 2 e^{- 2 t}

4 y^{^{''}} + y = sin (t), y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 16 y = cos (4 x) + sin (4 x), y^{0} = 3, y^{^{'}} 0 = 5

y^{^{'''}} + 289 y^{^{'}} = tan (17 x)

y^{^{''}} - 3 y^{^{'}} - 4 y = t^{2}

y^{^{''}} + y = 5 cos (x), y (0) = 0, y (0) = 1

What is the solution for y^{''}+6y^'+13y=10e^{-2t},y(0)=3 ?
The solution for y^{''}+6y^'+13y=10e^{-2t},y(0)=3 is y=e^{-3t}(cos(2t)+c_{2}sin(2t))+2e^{-2t}