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

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

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

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

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

Solution steps

y^{''} (t) - 2 y^{'} (t) + 2 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))

y = 4 - x^{2}, (- 1, 3)

\frac{1}{3 s + 1}

y^{^{''}} + 14 y^{^{'}} + 49 y = t^{- 2} e^{- 7 t}

\int (z + 2)^{2} d z

\frac{d}{d x} (cos (2^{x} \cdot x^{5}))

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