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}

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}