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

The solution for y^{''}-4y^'+3y=3e^{-2t},y(0)=0,y^'(0)=0 is y=(1/2-1/5)e^{3t}-1/2 e^t+1/5 e^{-2t}

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + y = 20 sin (x)

y^{^{''}} + 4 y = cos (3 x)

2 y^{^{''}} - 16 y + 4 = 0

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = e^{x} + 1

y^{^{''}} + y = 5 t^{2} - 5, y (0) = 0, y^{^{'}} (0) = - 6

What is the solution for y^{''}-4y^'+3y=3e^{-2t},y(0)=0,y^'(0)=0 ?
The solution for y^{''}-4y^'+3y=3e^{-2t},y(0)=0,y^'(0)=0 is y=(1/2-1/5)e^{3t}-1/2 e^t+1/5 e^{-2t}