What is the solution for y^{''}-y^'=-10y ?

The solution for y^{''}-y^'=-10y is y=e^{t/2}(c_{1}cos((sqrt(39)t)/2)+c_{2}sin((sqrt(39)t)/2))

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

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

y = e^{\frac{t}{2}} (c_{1} cos (\frac{39 t}{2}) + c_{2} sin (\frac{39 t}{2}))

Solution steps

y^{''} (t) - y^{'} (t) = - 10 y

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{39 t}{2}) + c_{2} sin (\frac{39 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{39 t}{2}) + c_{2} sin (\frac{39 t}{2}))

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

y^{^{''}} + 5 y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} - 5 y^{^{'}} - 36 y = 0, y^{0} = a, y^{^{'}} 0 = 32

y^{^{''}} - 6 y^{^{'}} + 13 y = 0, y (0) = e^{3} x cos^{2} (x)

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

What is the solution for y^{''}-y^'=-10y ?
The solution for y^{''}-y^'=-10y is y=e^{t/2}(c_{1}cos((sqrt(39)t)/2)+c_{2}sin((sqrt(39)t)/2))