What is the solution for y^{''}-6y^'+13y=0,y(0)=-3,y^'(0)=-17 ?

The solution for y^{''}-6y^'+13y=0,y(0)=-3,y^'(0)=-17 is y=e^{3t}(-3cos(2t)-4sin(2t))

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y^{''}-6y^'+13y=0,y(0)=-3,y^'(0)=-17

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = - 3, y^{^{'}} (0) = - 17

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = - 3, y^{^{'}} (0) = - \frac{68}{5}

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

y^{^{'''}} - 5 y^{^{''}} - 33 y^{^{'}} - 27 y = 0

- \frac{( y ^{^{''}} )}{( y ^{^{'}} )} = 4

49 y^{^{''}} - 70 y^{^{'}} + 25 y = 0

What is the solution for y^{''}-6y^'+13y=0,y(0)=-3,y^'(0)=-17 ?
The solution for y^{''}-6y^'+13y=0,y(0)=-3,y^'(0)=-17 is y=e^{3t}(-3cos(2t)-4sin(2t))