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}(-cos(t/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) = - 1, y^{^{'}} (0) = 0

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

Solution steps

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

Solve linear ODE:

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

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

4 y^{^{''}} + 4 y^{^{'}} - 3 y = 0

y^{^{''}} + 4 y = 0, y (π) = 0, y (0) = 0

x^{^{''}} - 2 x^{^{'}} = 0

y^{^{''}} - 25 y^{^{'}} + 144 y = 0, y (0) = 2

y^{^{''}} + 2 y^{^{'}} - 8 y = 0, y^{^{'}} (0) = 14

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}(-cos(t/2)+sin(t/2))