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

The solution for y^{''}+4y=e^{-4t},y(0)=2 is y= 39/20 cos(2t)+c_{2}sin(2t)+1/20 e^{-4t}

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

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

y = \frac{39}{20} cos (2 t) + c_{2} sin (2 t) + \frac{1}{20} e^{- 4 t}

Solution steps

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

Solve linear ODE:

y = \frac{39}{20} cos (2 t) + c_{2} sin (2 t) + \frac{1}{20} e^{- 4 t}

y = \frac{39}{20} cos (2 t) + c_{2} sin (2 t) + \frac{1}{20} e^{- 4 t}

y^{^{''}} + 25 y = 40 cos (5 t), y (0) = y

y^{^{''}} - 4 y^{^{'}} + 8 y = e^{2 x} + sin (2 x)

y^{^{''}} + 25 y = sec^{3} (5 t)

y^{^{''}} + 4 y^{^{'}} - 21 y = 5 e^{2 x} + 4

y^{^{''}} + 4 y = 1 - t^{3}

What is the solution for y^{''}+4y=e^{-4t},y(0)=2 ?
The solution for y^{''}+4y=e^{-4t},y(0)=2 is y= 39/20 cos(2t)+c_{2}sin(2t)+1/20 e^{-4t}