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

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

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x^{''}=bx

x^{^{''}} = b x

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + \frac{16}{3} y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = - \frac{4}{3}

2 y^{^{''}} + 5 y^{^{'}} = 0

y^{^{'''}} + 2 y^{^{''}} + 5 y^{^{'}} - 26 y = 0, y (0) = e^{2 x}

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

9 \frac{d ^{2} y}{d x ^{2}} - 49 y = 0, y (0) = 1, y^{^{'}} (0) = 0

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