What is the solution for 2y^{''}-4y^'+4y=0 ?

The solution for 2y^{''}-4y^'+4y=0 is y=e^t(c_{1}cos(t)+c_{2}sin(t))

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2y^{''}-4y^'+4y=0

2 y^{^{''}} - 4 y^{^{'}} + 4 y = 0

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

Solution steps

2 y^{''} (t) - 4 y^{'} (t) + 4 y = 0

Solve linear ODE:

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

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

x^{^{''}} + 5 x^{^{'}} - 6 x = 0

y^{^{''}} + 4 y^{^{'}} + 10 y = 0

2 y^{^{''}} + 6 y^{^{'}} + 8 y = 0

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (0) = 0, y^{^{'}} (0) = 1

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

What is the solution for 2y^{''}-4y^'+4y=0 ?
The solution for 2y^{''}-4y^'+4y=0 is y=e^t(c_{1}cos(t)+c_{2}sin(t))