What is the solution for x^{''}=ax ?

The solution for x^{''}=ax is x=c_{1}e^{sqrt(a)t}+c_{2}e^{-sqrt(a)t}

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x^{''}=ax

x^{^{''}} = a x

x = c_{1} e^{a t} + c_{2} e^{- a t}

Solution steps

x^{''} (t) = a x (t)

Solve linear ODE:

x = c_{1} e^{a t} + c_{2} e^{- a t}

x = c_{1} e^{a t} + c_{2} e^{- a t}

- 3 y^{^{''}} + \frac{1}{12} y = 0

\frac{1}{8} y^{^{''}} + 4 y^{^{'}} + 28 y = 0, y (0) = 0, y^{^{'}} (0) = 8

\frac{d ^{4} y}{d x ^{4}} - 2 \frac{d ^{2} y}{d x ^{2}} - 8 y = 0

3 y^{^{''}} + 12 y = 0, y (0) = 8, y^{^{'}} (0) = 6

y^{^{''}} + 12 y^{^{'}} + 38 y = 0, y (0) = 2, y^{^{'}} (0) = 0

What is the solution for x^{''}=ax ?
The solution for x^{''}=ax is x=c_{1}e^{sqrt(a)t}+c_{2}e^{-sqrt(a)t}