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

The solution for y^{''}-2y+5y=0 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)

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

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

y = c_{1} cos (3 t) + c_{2} sin (3 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t)

y = c_{1} cos (3 t) + c_{2} sin (3 t)

y^{^{''}} - 12 y^{^{'}} + 27 y = 0

y^{^{'''}} - 4 y^{^{''}} + 8 y^{^{'}} = 0

y^{^{''}} - (\frac{8}{3}) y^{^{'}} + α y = 0, y (0) = 1, y^{^{'}} (0) = - 73

6 y^{^{''}} + 11 y^{^{'}} - 6 y = 0

(D^{2} - D - 6) y = 0

What is the solution for y^{''}-2y+5y=0 ?
The solution for y^{''}-2y+5y=0 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)