What is the solution for y^{''}+4y^'+5y=35e^{(-4x)} ?

The solution for y^{''}+4y^'+5y=35e^{(-4x)} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+7e^{-4x}

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y^{''}+4y^'+5y=35e^{(-4x)}

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e^{(- 4 x)}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 7 e^{- 4 x}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = 35 e^{(- 4 x)}

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 7 e^{- 4 x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + 7 e^{- 4 x}

y^{^{''}} - 2 y^{^{'}} + y = \frac{( e ^{x} )}{( x ^{2} + 4 )}

y^{^{''}} - y = 12 t, y (0) = - 12 t

y^{^{''}} + 4 y^{^{'}} = e^{x} cos (2 x)

(D^{2} - 6 D + 13) y = 5^{x}

y^{^{''}} + 6 y^{^{'}} + 5 y = 10 cos (5 x)

What is the solution for y^{''}+4y^'+5y=35e^{(-4x)} ?
The solution for y^{''}+4y^'+5y=35e^{(-4x)} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+7e^{-4x}