What is the solution for y^{''}-4y^'+13y=0,y(0)=-5 ?

The solution for y^{''}-4y^'+13y=0,y(0)=-5 is y=e^{2t}(-5cos(3t)+c_{2}sin(3t))

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y^{''}-4y^'+13y=0,y(0)=-5

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 5

y = e^{2 t} (- 5 cos (3 t) + c_{2} sin (3 t))

Solution steps

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

Solve linear ODE:

y = e^{2 t} (- 5 cos (3 t) + c_{2} sin (3 t))

y = e^{2 t} (- 5 cos (3 t) + c_{2} sin (3 t))

y^{^{''}} + 5 y^{^{'}} + 4 y = 0, y (0) = 1, y (0) = 0

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

y^{^{''}} + 5 y^{^{'}} + 4 y = 0, y^{0} = 1

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

36 y^{^{''}} - 36 y^{^{'}} + 13 y = 0

What is the solution for y^{''}-4y^'+13y=0,y(0)=-5 ?
The solution for y^{''}-4y^'+13y=0,y(0)=-5 is y=e^{2t}(-5cos(3t)+c_{2}sin(3t))