What is the solution for y^{''}-2y^'+2y=0,y(0)=1,y^'(0)=-1 ?

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

4 y^{^{''}} - y = 0, y (- 9) = 1, y^{^{'}} (- 9) = - 1

\frac{1}{16} y^{^{''}} + y^{^{'}} + 2 y = 0

y^{^{''}} + y^{^{'}} + \frac{1}{3} y = 0

4 y^{^{''}} + 12 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = - 1

x^{^{''}} + 3 x^{^{'}} - 3 x = 0

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