What is the solution for 4y^{''}-16y=1 ?

The solution for 4y^{''}-16y=1 is y=c_{1}e^{2t}+c_{2}e^{-2t}-1/16

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4y^{''}-16y=1

4 y^{^{''}} - 16 y = 1

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{1}{16}

Solution steps

4 y^{''} (t) - 16 y = 1

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{1}{16}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} - \frac{1}{16}

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

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

y^{^{''''}} + 2 y^{^{''}} + y = 5 + cos (5 t)

y^{^{''}} - 2 y^{^{'}} - 48 y = - 136 t + 96 t^{2}

y^{^{''}} + 15 y^{^{'}} + 56 y = e^{- 6 x}

What is the solution for 4y^{''}-16y=1 ?
The solution for 4y^{''}-16y=1 is y=c_{1}e^{2t}+c_{2}e^{-2t}-1/16