What is the solution for 4y^{''}-4y+5y=0,y(0)= 1/2 ,y^'(0)=1 ?

The solution for 4y^{''}-4y+5y=0,y(0)= 1/2 ,y^'(0)=1 is y= 1/2 cos(t/2)+2sin(t/2)

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4y^{''}-4y+5y=0,y(0)= 1/2 ,y^'(0)=1

4 y^{^{''}} - 4 y + 5 y = 0, y (0) = \frac{1}{2}, y^{^{'}} (0) = 1

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

Solution steps

4 y^{''} (t) - 4 y + 5 y = 0

Solve linear ODE:

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

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

y^{^{''}} - 10 y^{^{'}} - 16 y = 0

y^{^{'}} + y = y^{^{''}}

\frac{5}{6} y^{^{''}} + 10 y^{^{'}} + 60 y = 0, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 8 y^{^{'}} + 9 y = 0, y (0) = 5, y^{^{'}} (0) = 0

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

What is the solution for 4y^{''}-4y+5y=0,y(0)= 1/2 ,y^'(0)=1 ?
The solution for 4y^{''}-4y+5y=0,y(0)= 1/2 ,y^'(0)=1 is y= 1/2 cos(t/2)+2sin(t/2)