What is the solution for y^{''}-6y^'+13y=0,y(0)=-3 ?

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

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y^{''}-6y^'+13y=0,y(0)=-3

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = - 3

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

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

5 y^{^{''}} + 45 y = 0, y (0) = - 8, y^{^{'}} (0) = 0

y^{^{''}} + 12 y^{^{'}} + 36 y = 0, y (0) = - 2, y^{^{'}} (0) = 8

3 y^{^{''}} + 12 y^{^{'}} + 12 y = 0

y^{^{''}} + 12 y^{^{'}} + 3 y = 0

4 y^{^{''}} + \frac{16}{0.25} y = 0, y (0) = 0

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