What is the solution for y^{''}+3y^'+2y=e^{-4t},y^'(0)=0,y(0)=1 ?

The solution for y^{''}+3y^'+2y=e^{-4t},y^'(0)=0,y(0)=1 is y=(7e^{-t})/3-3/2 e^{-2t}+1/6 e^{-4t}

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y^{''}+3y^'+2y=e^{-4t},y^'(0)=0,y(0)=1

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- 4 t}, y^{^{'}} (0) = 0, y (0) = 1

y = \frac{7 e ^{- t}}{3} - \frac{3}{2} e^{- 2 t} + \frac{1}{6} e^{- 4 t}

Solution steps

y^{''} (t) + 3 y^{'} (t) + 2 y = e^{- 4 t}

Solve linear ODE:

y = \frac{7 e ^{- t}}{3} - \frac{3}{2} e^{- 2 t} + \frac{1}{6} e^{- 4 t}

y = \frac{7 e ^{- t}}{3} - \frac{3}{2} e^{- 2 t} + \frac{1}{6} e^{- 4 t}

y^{^{''}} + 4 y = 4 t^{2} - 4 t + 10

2 y^{^{''}} - 5 y^{^{'}} + 3 y = (14 x - 10) e^{3 x}

y^{^{''}} + 9 y = 12 cos (3 t), y (0) = - 1, y^{^{'}} (0) = 1

4 x^{^{''}} + 324 x = 8 sin (t)

y^{^{''}} + 5 y^{^{'}} + 4 y = - 2 e^{3 t}

What is the solution for y^{''}+3y^'+2y=e^{-4t},y^'(0)=0,y(0)=1 ?
The solution for y^{''}+3y^'+2y=e^{-4t},y^'(0)=0,y(0)=1 is y=(7e^{-t})/3-3/2 e^{-2t}+1/6 e^{-4t}