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

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

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y^{''}-6y^'+13y=0

y^{^{''}} - 6 y^{^{'}} + 13 y = 0

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d y}{d x} = x y^{3} (1 + x^{2})^{- \frac{1}{2}}, y (0) = 1

\frac{\partial}{\partial x} (- x y (- 15 - x - y))

\frac{\partial}{\partial x} (3 cos (3 x + 4 y))

\frac{d y}{d t} = \frac{t}{y - 2}, y (- 1) = 0

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