What is the solution for linear y^{''}-6y^'+13y=0 ?

The solution for linear y^{''}-6y^'+13y=0 is y=e^{3t}(c_{1}cos(2t)+c_{2}sin(2t))

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linear y^{''}-6y^'+13y=0

linear

y^{^{''}} - 6 y^{^{'}} + 13 y = 0

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

Solution steps

y^{''} (t) - 6 y^{'} (t) + 13 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 6 y^{^{'}} + 7 y = 0

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

2 y^{^{''}} - y^{^{'}} - 3 y = 0, y (2) = 5, y^{^{'}} (2) = 0

y^{^{''}} + (- 2) y^{^{'}} + y = 0

What is the solution for linear y^{''}-6y^'+13y=0 ?
The solution for linear y^{''}-6y^'+13y=0 is y=e^{3t}(c_{1}cos(2t)+c_{2}sin(2t))