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=e^{t/2}(1/2 cos(t)+3/4 sin(t))

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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 = e^{\frac{t}{2}} (\frac{1}{2} cos (t) + \frac{3}{4} sin (t))

Solution steps

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

Solve linear ODE:

y = e^{\frac{t}{2}} (\frac{1}{2} cos (t) + \frac{3}{4} sin (t))

y = e^{\frac{t}{2}} (\frac{1}{2} cos (t) + \frac{3}{4} sin (t))

y^{^{''}} + y = 0, y (\frac{π}{3}) = 2

9 y^{^{''}} + 24 y^{^{'}} - 33 y = 0

\frac{d ^{2} y}{d x ^{2}} = (4 y)

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

y^{^{'''}} + 3 y^{^{''}} + 4 y^{^{'}} + 12 y = 0

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=e^{t/2}(1/2 cos(t)+3/4 sin(t))