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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

u^{^{''}} + 0.3 u^{^{'}} + 2.7 H (t) = 0

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

y^{^{''}} - 5 y^{^{'}} - 6 y = 0, y (0) = - 1, y^{^{'}} (0) = 15

y^{^{'''}} + 48 y^{^{''}} + 768 y^{^{'}} + 4096 y = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y^{^{'}} (0) = 1

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