What is the solution for 2y^{''}-4y^'+13y=0 ?

The solution for 2y^{''}-4y^'+13y=0 is y=e^t(c_{1}cos(sqrt(11/2)t)+c_{2}sin(sqrt(11/2)t))

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2y^{''}-4y^'+13y=0

2 y^{^{''}} - 4 y^{^{'}} + 13 y = 0

y = e^{t} (c_{1} cos (\frac{11}{2} t) + c_{2} sin (\frac{11}{2} t))

Solution steps

2 y^{''} (t) - 4 y^{'} (t) + 13 y = 0

Solve linear ODE:

y = e^{t} (c_{1} cos (\frac{11}{2} t) + c_{2} sin (\frac{11}{2} t))

y = e^{t} (c_{1} cos (\frac{11}{2} t) + c_{2} sin (\frac{11}{2} t))

20 y^{^{''}} + 3 y^{^{'}} - 2 y = 0

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

x^{^{''}} - 6 x^{^{'}} + 10 x = 0

625 y^{^{''}} - 81 y = 0, y (0) = 9, y^{^{'}} (0) = 6

(D^{3} - D^{2} - 4 D + 4) y = 0

What is the solution for 2y^{''}-4y^'+13y=0 ?
The solution for 2y^{''}-4y^'+13y=0 is y=e^t(c_{1}cos(sqrt(11/2)t)+c_{2}sin(sqrt(11/2)t))