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

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

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

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

y = e^{t} (2 cos (t) - 2 sin (t))

Solution steps

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

Solve linear ODE:

y = e^{t} (2 cos (t) - 2 sin (t))

y = e^{t} (2 cos (t) - 2 sin (t))

x^{^{''}} - 9 x^{^{'}} - 14 x = 0

\frac{d ^{2} y}{d t ^{2}} + \frac{d y}{d t} = 0

y^{^{''}} + 10 y^{^{'}} + 25 y = 0, y (- 1) = 2, y^{^{'}} (- 1) = 3

2 y^{^{''}} + 16 y^{^{'}} + 82 y = 0

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

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