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^tcos(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)

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} = 6 y

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = - 4, y^{^{'}} (0) = 14

3 y^{^{''}} - 24 y^{^{'}} + 48 y = 0

y^{^{'''}} - 6 y^{^{''}} + 9 y = 0

4 \frac{d ^{2} y}{d t ^{2}} - y = 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^tcos(t)