What is the solution for y^{''}+5y^'+4y=2e^{-5x},y(0)=4,y^'(0)=3 ?

The solution for y^{''}+5y^'+4y=2e^{-5x},y(0)=4,y^'(0)=3 is y=(3+7/2)e^{-x}-3e^{-4x}+1/2 e^{-5x}

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y^{''}+5y^'+4y=2e^{-5x},y(0)=4,y^'(0)=3

y^{^{''}} + 5 y^{^{'}} + 4 y = 2 e^{- 5 x}, y (0) = 4, y^{^{'}} (0) = 3

y = (3 + \frac{7}{2}) e^{- x} - 3 e^{- 4 x} + \frac{1}{2} e^{- 5 x}

Solution steps

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

Solve linear ODE:

y = (3 + \frac{7}{2}) e^{- x} - 3 e^{- 4 x} + \frac{1}{2} e^{- 5 x}

y = (3 + \frac{7}{2}) e^{- x} - 3 e^{- 4 x} + \frac{1}{2} e^{- 5 x}

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - 2 y = 9 x e^{3 x}

D^{2} (D + 2) y = 2 + e^{\frac{- t}{2}}

x^{^{''}} + x^{^{'}} = 2 t + 1

y^{^{''''}} + 2 y^{^{''}} + y = 3 t + 8

y^{^{''}} - 9 y^{^{'}} + 18 = 0

What is the solution for y^{''}+5y^'+4y=2e^{-5x},y(0)=4,y^'(0)=3 ?
The solution for y^{''}+5y^'+4y=2e^{-5x},y(0)=4,y^'(0)=3 is y=(3+7/2)e^{-x}-3e^{-4x}+1/2 e^{-5x}