What is the solution for y^{''}-4y^'+13y=0,y(0)=-5,y^'(0)=-1 ?

The solution for y^{''}-4y^'+13y=0,y(0)=-5,y^'(0)=-1 is y=e^{2t}(-5cos(3t)+3sin(3t))

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y^{''}-4y^'+13y=0,y(0)=-5,y^'(0)=-1

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 5, y^{^{'}} (0) = - 1

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{'''}} = y

x^{^{''}} - 2 x^{^{'}} + x = 0

y^{^{''}} - 4 y^{^{'}} + 4 y = 0, y (0) = 3, y^{^{'}} (0) = 5

3 y^{^{''}} - 4 y = 0

What is the solution for y^{''}-4y^'+13y=0,y(0)=-5,y^'(0)=-1 ?
The solution for y^{''}-4y^'+13y=0,y(0)=-5,y^'(0)=-1 is y=e^{2t}(-5cos(3t)+3sin(3t))