What is the solution for y^{''}-4y^'+13y=0,y(0)=-1,y^'(0)=2 ?

The solution for y^{''}-4y^'+13y=0,y(0)=-1,y^'(0)=2 is y=e^{2t}(-cos(3t)+4/3 sin(3t))

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y^{''}-4y^'+13y=0,y(0)=-1,y^'(0)=2

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 1, y^{^{'}} (0) = 2

y = e^{2 t} (- cos (3 t) + \frac{4}{3} sin (3 t))

Solution steps

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

Solve linear ODE:

y = e^{2 t} (- cos (3 t) + \frac{4}{3} sin (3 t))

y = e^{2 t} (- cos (3 t) + \frac{4}{3} sin (3 t))

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

y^{^{''}} + 6 y^{^{'}} + 13 y = 0, y (0) = 3, y^{^{'}} (0) = - 8

16 y^{^{''}} + 56 y^{^{'}} + 130 y = 0, y (1) = 2, y^{^{'}} (1) = 5

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

y^{^{''}} - 2 y + 10 y = 0

What is the solution for y^{''}-4y^'+13y=0,y(0)=-1,y^'(0)=2 ?
The solution for y^{''}-4y^'+13y=0,y(0)=-1,y^'(0)=2 is y=e^{2t}(-cos(3t)+4/3 sin(3t))