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

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

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x^{''}+x^'+1/2 x=0,x(0)=5,x^'(0)=0

x^{^{''}} + x^{^{'}} + \frac{1}{2} x = 0, x (0) = 5, x^{^{'}} (0) = 0

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

Solution steps

x^{''} (t) + x^{'} (t) + \frac{1}{2} x = 0

Solve linear ODE:

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

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

y^{^{'''}} - 19 y^{^{''}} + 80 y^{^{'}} + 100 y = 0

y^{^{''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y = 0

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

8 y^{^{''}} - 8 y^{^{'}} + 100 y = 0

y^{^{'''}} - 9 y^{^{''}} + 18 y = 0, y (0) = 9

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