What is the solution for y^{''}+y=2e^{-8t},y(0)=0,y^'(0)=0 ?

The solution for y^{''}+y=2e^{-8t},y(0)=0,y^'(0)=0 is y=-2/65 cos(t)+16/65 sin(t)+2/65 e^{-8t}

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y^{''}+y=2e^{-8t},y(0)=0,y^'(0)=0

y^{^{''}} + y = 2 e^{- 8 t}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{2}{65} cos (t) + \frac{16}{65} sin (t) + \frac{2}{65} e^{- 8 t}

Solution steps

y^{''} (t) + y = 2 e^{- 8 t}

Solve linear ODE:

y = - \frac{2}{65} cos (t) + \frac{16}{65} sin (t) + \frac{2}{65} e^{- 8 t}

y = - \frac{2}{65} cos (t) + \frac{16}{65} sin (t) + \frac{2}{65} e^{- 8 t}

y^{^{'''}} - 6 y^{^{'}} - 7 = 0

\frac{d ^{2} y}{d x ^{2}} + 9 y = 15 x^{2} - 9

y^{^{''}} + 6 y^{^{'}} + 8 y = 2 t + e^{t}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{x}, y (1) = 0, y^{^{'}} (1) = 1

y^{^{''}} + y^{^{'}} + 4 y = e^{- 1.3 t}

What is the solution for y^{''}+y=2e^{-8t},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+y=2e^{-8t},y(0)=0,y^'(0)=0 is y=-2/65 cos(t)+16/65 sin(t)+2/65 e^{-8t}