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

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

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

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

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

y^{^{''}} + 14 y^{^{'}} + 53 y = 0

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

(d^{3} - 3 d^{2} - 10 d) y = 0

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