What is the solution for y^{''}-6y^'+13y=0,y(0)=1,y^'(0)=5 ?

The solution for y^{''}-6y^'+13y=0,y(0)=1,y^'(0)=5 is y=e^{3t}(cos(2t)+sin(2t))

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y^{''}-6y^'+13y=0,y(0)=1,y^'(0)=5

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = 1, y^{^{'}} (0) = 5

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

Solution steps

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

Solve linear ODE:

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

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

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

(4 D^{5} - 15 D^{3} - 5 D^{2} + 15 D + 9) y = 0

f^{^{''}} (x) = 6 f^{^{'}} (x) - 6 f (x)

16 y^{^{''}} + 64 y = 0, y (0) = 0.3, y^{^{'}} (0) = - 0.2

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

What is the solution for y^{''}-6y^'+13y=0,y(0)=1,y^'(0)=5 ?
The solution for y^{''}-6y^'+13y=0,y(0)=1,y^'(0)=5 is y=e^{3t}(cos(2t)+sin(2t))