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

The solution for 2y^{''}-2y^'+y=0,y(0)=-1,y^'(0)=2 is y=e^{t/2}(-cos(t/2)+5sin(t/2))

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

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

y = e^{\frac{t}{2}} (- cos (\frac{t}{2}) + 5 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}) + 5 sin (\frac{t}{2}))

y = e^{\frac{t}{2}} (- cos (\frac{t}{2}) + 5 sin (\frac{t}{2}))

What is the solution for 2y^{''}-2y^'+y=0,y(0)=-1,y^'(0)=2 ?
The solution for 2y^{''}-2y^'+y=0,y(0)=-1,y^'(0)=2 is y=e^{t/2}(-cos(t/2)+5sin(t/2))