What is the solution for y^{''}+6y^'+10y=e^{-3t},y(0)=4,y^'(0)=5 ?

The solution for y^{''}+6y^'+10y=e^{-3t},y(0)=4,y^'(0)=5 is y=e^{-3t}(3cos(t)+17sin(t))+e^{-3t}

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y^{''}+6y^'+10y=e^{-3t},y(0)=4,y^'(0)=5

y^{^{''}} + 6 y^{^{'}} + 10 y = e^{- 3 t}, y (0) = 4, y^{^{'}} (0) = 5

y = e^{- 3 t} (3 cos (t) + 17 sin (t)) + e^{- 3 t}

Solution steps

y^{''} (t) + 6 y^{'} (t) + 10 y = e^{- 3 t}

Solve linear ODE:

y = e^{- 3 t} (3 cos (t) + 17 sin (t)) + e^{- 3 t}

y = e^{- 3 t} (3 cos (t) + 17 sin (t)) + e^{- 3 t}

y^{^{'''}} = 3 y^{^{''}} + 9 (e^{(5 x)}) + 2 (cos (x))

y^{^{''}} - y^{^{'}} - y = x^{2} - 2

y^{^{''}} - 16 y^{^{'}} = 32 + 45 e^{x}

y^{^{'''}} - 4 y = e^{t} + t, y (0) = 0, y^{^{'}} (0) = 0, y^{^{''}} (0) = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 1, y (0) = 1, y^{^{'}} (0) = - 1

What is the solution for y^{''}+6y^'+10y=e^{-3t},y(0)=4,y^'(0)=5 ?
The solution for y^{''}+6y^'+10y=e^{-3t},y(0)=4,y^'(0)=5 is y=e^{-3t}(3cos(t)+17sin(t))+e^{-3t}