What is the solution for y^{''}-6y^{''}+12y^'-8y=0 ?

The solution for y^{''}-6y^{''}+12y^'-8y=0 is y=e^{(6t)/5}(c_{1}cos((2t)/5)+c_{2}sin((2t)/5))

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y^{''}-6y^{''}+12y^'-8y=0

y^{^{''}} - 6 y^{^{''}} + 12 y^{^{'}} - 8 y = 0

y = e^{\frac{6 t}{5}} (c_{1} cos (\frac{2 t}{5}) + c_{2} sin (\frac{2 t}{5}))

Solution steps

y^{''} (t) - 6 y^{''} (t) + 12 y^{'} (t) - 8 y = 0

Solve linear ODE:

y = e^{\frac{6 t}{5}} (c_{1} cos (\frac{2 t}{5}) + c_{2} sin (\frac{2 t}{5}))

y = e^{\frac{6 t}{5}} (c_{1} cos (\frac{2 t}{5}) + c_{2} sin (\frac{2 t}{5}))

y^{^{''}} - 36 y = 0, y (0) = 1

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

y^{^{'''}} + 12 y^{^{''}} + 45 y^{^{'}} = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 3, y^{^{'}} (0) = 13

y^{^{''}} + 12 y^{^{'}} + 27 y = 0

What is the solution for y^{''}-6y^{''}+12y^'-8y=0 ?
The solution for y^{''}-6y^{''}+12y^'-8y=0 is y=e^{(6t)/5}(c_{1}cos((2t)/5)+c_{2}sin((2t)/5))