What is the solution for y^{''}+4y^'+5y=10e^{-3x} ?

The solution for y^{''}+4y^'+5y=10e^{-3x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+5e^{-3x}

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y^{''}+4y^'+5y=10e^{-3x}

y^{^{''}} + 4 y^{^{'}} + 5 y = 10 e^{- 3 x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 5 e^{- 3 x}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = 10 e^{- 3 x}

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 5 e^{- 3 x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 5 e^{- 3 x}

y^{^{''}} + 7 y^{^{'}} + 12 y = - 432 t^{3}

y^{^{''}} + v y^{^{'}} = c e^{(- r^{2})}

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- 4 t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} + 3 y = 6 e^{- 5 t}

y^{^{''}} + 4 y^{^{'}} + 5 y = sin (5 t)

What is the solution for y^{''}+4y^'+5y=10e^{-3x} ?
The solution for y^{''}+4y^'+5y=10e^{-3x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+5e^{-3x}