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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} - 6 y = 0

\frac{d ^{4} y}{d x ^{4}} + 2 y^{^{'''}} - 2 y^{^{'}} - y = 0

y^{^{''}} - 6 y^{^{'}} + 13 y = 0

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 6 y^{^{'}} + 7 y = 0

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