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

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

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y^{''}-6y^'+13y=0,y(0)=e^{3x}cos(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^{^{''''}} + 12 y^{^{''}} + 36 y = 0

2 \frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 3 y = 0

20 y^{^{''}} - 11 y^{^{'}} - 3 y = 0

(D^{3} + 5 D^{2} + 7 D + 3) y = 0

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

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