What is the solution for y^{''}-y^'+6y=0 ?

The solution for y^{''}-y^'+6y=0 is y=e^{t/2}(c_{1}cos((sqrt(23)t)/2)+c_{2}sin((sqrt(23)t)/2))

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y^{''}-y^'+6y=0

y^{^{''}} - y^{^{'}} + 6 y = 0

y = e^{\frac{t}{2}} (c_{1} cos (\frac{23 t}{2}) + c_{2} sin (\frac{23 t}{2}))

Solution steps

y^{''} (t) - y^{'} (t) + 6 y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{23 t}{2}) + c_{2} sin (\frac{23 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{23 t}{2}) + c_{2} sin (\frac{23 t}{2}))

\frac{d}{d θ} (\frac{4}{2 cos ( θ ) - sin ( θ )})

f (x) = \frac{8 x}{x ^{2} + 1}, (1, 4)

\frac{d}{d x} (\frac{4 x}{3 - tan ( x )})

x \to 0 lim (sin (a x))

\frac{10 s}{s ^{2} + 16}

What is the solution for y^{''}-y^'+6y=0 ?
The solution for y^{''}-y^'+6y=0 is y=e^{t/2}(c_{1}cos((sqrt(23)t)/2)+c_{2}sin((sqrt(23)t)/2))