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

The solution for y^{''}+2y^'-3y=e^{-3t},y(0)=0,y^'(0)=0 is y= 1/16 e^t-1/16 e^{-3t}-(e^{-3t}t)/4

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

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

y = \frac{1}{16} e^{t} - \frac{1}{16} e^{- 3 t} - \frac{e ^{- 3 t} t}{4}

Solution steps

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

Solve linear ODE:

y = \frac{1}{16} e^{t} - \frac{1}{16} e^{- 3 t} - \frac{e ^{- 3 t} t}{4}

y = \frac{1}{16} e^{t} - \frac{1}{16} e^{- 3 t} - \frac{e ^{- 3 t} t}{4}

x^{^{''}} + x = 4 e^{t}

y^{^{''}} + 9 y = t, y (0) = 0, y^{^{'}} (0) = 0

D u = 50 - \frac{1}{500}

d (n) = 8 - 6 (n - 1)

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

What is the solution for y^{''}+2y^'-3y=e^{-3t},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+2y^'-3y=e^{-3t},y(0)=0,y^'(0)=0 is y= 1/16 e^t-1/16 e^{-3t}-(e^{-3t}t)/4