What is the solution for y^{''}-5y^'+12y=0 ?

The solution for y^{''}-5y^'+12y=0 is y=e^{(5t)/2}(c_{1}cos((sqrt(23)t)/2)+c_{2}sin((sqrt(23)t)/2))

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y^{''}-5y^'+12y=0

y^{^{''}} - 5 y^{^{'}} + 12 y = 0

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{23 t}{2}) + c_{2} sin (\frac{23 t}{2}))

Solution steps

y^{''} (t) - 5 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{23 t}{2}) + c_{2} sin (\frac{23 t}{2}))

y = e^{\frac{5 t}{2}} (c_{1} cos (\frac{23 t}{2}) + c_{2} sin (\frac{23 t}{2}))

25 y^{^{''}} + 70 y^{^{'}} + 49 y = 0

y^{^{''}} - y^{^{'}} - 20 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

6 f^{^{''}} (t) = 3 f^{^{''}} (t) - 5 f (t)

y^{^{''}} - 6 y^{^{'}} + 16 y = 0

y^{^{''}} - 4 y^{^{'}} + 13 y = 0, y (0) = - 1, y^{^{'}} (0) = 5

What is the solution for y^{''}-5y^'+12y=0 ?
The solution for y^{''}-5y^'+12y=0 is y=e^{(5t)/2}(c_{1}cos((sqrt(23)t)/2)+c_{2}sin((sqrt(23)t)/2))