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^{^{''}} + 8 y^{^{'}} + 41 y = 0, y (0) = 3, y^{^{'}} (0) = - 11

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

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

2 x^{^{''}} + 2 x^{^{'}} + 7 x = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 20, y^{^{'}} (0) = 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)