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}))

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))