What is the solution for y^{''}-y^'+9y=0 ?

The solution for y^{''}-y^'+9y=0 is y=e^{t/2}(c_{1}cos((sqrt(35)t)/2)+c_{2}sin((sqrt(35)t)/2))

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y^{''}-y^'+9y=0

y^{^{''}} - y^{^{'}} + 9 y = 0

y = e^{\frac{t}{2}} (c_{1} cos (\frac{35 t}{2}) + c_{2} sin (\frac{35 t}{2}))

Solution steps

y^{''} (t) - y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{35 t}{2}) + c_{2} sin (\frac{35 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{35 t}{2}) + c_{2} sin (\frac{35 t}{2}))

y^{^{''}} + 25 y = 0, y (0) = 6, y^{^{'}} (0) = - 3

y^{^{'''}} + y^{^{''}} - 72 y^{^{'}} = 0

u^{^{''}} + 16 H (t) = 0

y^{^{'''}} + 4 y^{^{''}} + 13 y^{^{'}} = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = - 3, y^{^{'}} (0) = 2

What is the solution for y^{''}-y^'+9y=0 ?
The solution for y^{''}-y^'+9y=0 is y=e^{t/2}(c_{1}cos((sqrt(35)t)/2)+c_{2}sin((sqrt(35)t)/2))