What is the solution for y^{''}-2y+5=0 ?

The solution for y^{''}-2y+5=0 is y=c_{1}e^{sqrt(2)t}+c_{2}e^{-sqrt(2)t}+5/2

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

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

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{5}{2}

Solution steps

y^{''} (t) - 2 y + 5 = 0

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{5}{2}

y = c_{1} e^{2 t} + c_{2} e^{- 2 t} + \frac{5}{2}

y^{^{'''}} - 2 y^{^{''}} + 3 y^{^{'}} - 2 y = 2 e^{x}

y^{^{''}} + y - 5 = e^{2 x} - 3 cos (3 x)

y^{^{'''}} - 18 y^{^{''}} + 864 y = e^{12 x}

(D^{3} - D) y = x

y^{^{''}} + y^{^{'}} = 16 e^{3 x}

What is the solution for y^{''}-2y+5=0 ?
The solution for y^{''}-2y+5=0 is y=c_{1}e^{sqrt(2)t}+c_{2}e^{-sqrt(2)t}+5/2