What is the solution for y^{''}-6y^'+13y=0,y(0)=e^3xcos(2x) ?

The solution for y^{''}-6y^'+13y=0,y(0)=e^3xcos(2x) is y=e^{3x}(c_{1}cos(2x)+c_{2}sin(2x))

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y^{''}-6y^'+13y=0,y(0)=e^3xcos(2x)

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = e^{3} x cos (2 x)

y = e^{3 x} (c_{1} cos (2 x) + c_{2} sin (2 x))

Solution steps

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

Solve linear ODE:

y = e^{3 x} (c_{1} cos (2 x) + c_{2} sin (2 x))

y = e^{3 x} (c_{1} cos (2 x) + c_{2} sin (2 x))

y^{^{''}} = \frac{( a - 2 )}{a} y

16 y^{^{''}} - 32 y^{^{'}} + 65 y = 0

4 y^{^{''}} + 5 y^{^{'}} - 6 y = 0

\frac{1}{8} y^{^{''}} + 4 y^{^{'}} + 16 y = 0, y (0) = 0, y^{^{'}} (0) = 9

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

What is the solution for y^{''}-6y^'+13y=0,y(0)=e^3xcos(2x) ?
The solution for y^{''}-6y^'+13y=0,y(0)=e^3xcos(2x) is y=e^{3x}(c_{1}cos(2x)+c_{2}sin(2x))