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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

4 y^{^{'''}} - 7 y^{^{''}} - 34 y^{^{'}} - 8 y = 0

y^{^{''}} - 5 y^{^{''}} + 6 y = 0

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = 1, y^{^{'}} (0) = 3

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = 1, y^{^{'}} (0) = 5

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

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