What is the solution for y^{''}+4y+3y=9e^{-3x} ?

The solution for y^{''}+4y+3y=9e^{-3x} is y=c_{1}cos(sqrt(7)x)+c_{2}sin(sqrt(7)x)+9/16 e^{-3x}

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y^{''}+4y+3y=9e^{-3x}

y^{^{''}} + 4 y + 3 y = 9 e^{- 3 x}

y = c_{1} cos (7 x) + c_{2} sin (7 x) + \frac{9}{16} e^{- 3 x}

Solution steps

y^{''} (x) + 4 y + 3 y = 9 e^{- 3 x}

Solve linear ODE:

y = c_{1} cos (7 x) + c_{2} sin (7 x) + \frac{9}{16} e^{- 3 x}

y = c_{1} cos (7 x) + c_{2} sin (7 x) + \frac{9}{16} e^{- 3 x}

y^{^{'''}} + 4 y^{^{'}} = sec (x) cos (x)

y^{^{''}} + 3 y^{^{'}} + 2 y = 6 e^{t}, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} + 6 y^{^{'}} + 9 y = 5 + 5 x

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 15 t

y^{^{''}} + 9 y = e^{x}

What is the solution for y^{''}+4y+3y=9e^{-3x} ?
The solution for y^{''}+4y+3y=9e^{-3x} is y=c_{1}cos(sqrt(7)x)+c_{2}sin(sqrt(7)x)+9/16 e^{-3x}