What is the solution for y^{''}+1y^'+y=0 ?

The solution for y^{''}+1y^'+y=0 is y=e^{-t/2}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))

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y^{''}+1y^'+y=0

y^{^{''}} + 1 y^{^{'}} + y = 0

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

Solution steps

y^{''} (t) + 1 y^{'} (t) + y = 0

Solve linear ODE:

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

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

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

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

y^{^{''}} - y^{^{'}} - 30 y = 0, y (0) = x, y^{^{'}} (0) = 5

x^{^{''}} = a x

- 3 y^{^{''}} + \frac{1}{12} y = 0

What is the solution for y^{''}+1y^'+y=0 ?
The solution for y^{''}+1y^'+y=0 is y=e^{-t/2}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))