What is the solution for y^{''}-6y^'+13y=0,y(0)=2,y^'(0)=0 ?

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

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y^{''}-6y^'+13y=0,y(0)=2,y^'(0)=0

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = 2, y^{^{'}} (0) = 0

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} + 9 x = 0

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

y^{^{'''}} + 2 y^{^{''}} - 9 y^{^{'}} - 18 y = 0

y^{^{''}} - 9 y^{^{'}} + 8 y = 0

y^{^{''}} + 4 y = 0

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