What is the solution for y^{''}-24y^'+12y=0 ?

The solution for y^{''}-24y^'+12y=0 is y=c_{1}e^{2(6+sqrt(33))t}+c_{2}e^{2(6-sqrt(33))t}

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y^{''}-24y^'+12y=0

y^{^{''}} - 24 y^{^{'}} + 12 y = 0

y = c_{1} e^{2 (6 + 33) t} + c_{2} e^{2 (6 - 33) t}

Solution steps

y^{''} (t) - 24 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = c_{1} e^{2 (6 + 33) t} + c_{2} e^{2 (6 - 33) t}

y = c_{1} e^{2 (6 + 33) t} + c_{2} e^{2 (6 - 33) t}

\frac{d ^{2} x}{d t ^{2}} - 4 x = 0

x^{^{''}} - (a - 2) x^{^{'}} - (a - 1) x = 0

3 y^{^{''}} - 6 y^{^{'}} + 9 y = 0

y^{^{''}} + 25 y = 0, y (0) = 4, y^{^{'}} (0) = - 2

x^{''} + x = 0, x (\frac{π}{2}) = 0, x^{^{'}} (\frac{π}{2}) = 1

What is the solution for y^{''}-24y^'+12y=0 ?
The solution for y^{''}-24y^'+12y=0 is y=c_{1}e^{2(6+sqrt(33))t}+c_{2}e^{2(6-sqrt(33))t}