What is the solution for y^{''}-y^'=-y ?

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

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y^{''}-y^'=-y

y^{^{''}} - y^{^{'}} = - y

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

Solution steps

y^{''} (t) - y^{'} (t) = - y

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}))

4 y^{^{''}} - 36 y^{^{'}} + 81 y = 0

4 y^{^{''}} - 4 y^{^{'}} + 37 y = 0, y (0) = 3, y^{^{'}} (0) = \frac{3}{2}

y^{^{''}} + 4 y^{^{'}} + y 8 = 0

y^{^{''}} = 8 y^{^{'}} - 15 y, y (0) = 0, y^{^{'}} (0) = - 2

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

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