What is the solution for y^{''}-6y^'+12y=0 ?

The solution for y^{''}-6y^'+12y=0 is y=e^{3t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))

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y^{''}-6y^'+12y=0

y^{^{''}} - 6 y^{^{'}} + 12 y = 0

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

Solution steps

y^{''} (t) - 6 y^{'} (t) + 12 y = 0

Solve linear ODE:

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

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

20 x

e^{x - 1}

\frac{d}{d x} (\frac{3 x y}{x ^{2} + y ^{2}})

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

n \to \infty lim (n^{- 2})

What is the solution for y^{''}-6y^'+12y=0 ?
The solution for y^{''}-6y^'+12y=0 is y=e^{3t}(c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t))