What is the solution for y^{''}+4y^'+5y=35e^{-4x},y(0)=-2 ?

The solution for y^{''}+4y^'+5y=35e^{-4x},y(0)=-2 is y=e^{-2x}(-9cos(x)+c_{2}sin(x))+7e^{-4x}

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y^{''}+4y^'+5y=35e^{-4x},y(0)=-2

y^{^{''}} + 4 y^{^{'}} + 5 y = 35 e^{- 4 x}, y (0) = - 2

y = e^{- 2 x} (- 9 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} (- 9 cos (x) + c_{2} sin (x)) + 7 e^{- 4 x}

y = e^{- 2 x} (- 9 cos (x) + c_{2} sin (x)) + 7 e^{- 4 x}

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

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

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

y^{^{''}} - y^{^{'}} + 14 y = 5 + e^{\frac{x}{2}}

y^{^{''}} - 16 y = 9.6 e^{4 x} + 30 e^{x}

What is the solution for y^{''}+4y^'+5y=35e^{-4x},y(0)=-2 ?
The solution for y^{''}+4y^'+5y=35e^{-4x},y(0)=-2 is y=e^{-2x}(-9cos(x)+c_{2}sin(x))+7e^{-4x}