What is the solution for 12y^{''}-12y^'+78y=0 ?

The solution for 12y^{''}-12y^'+78y=0 is y=e^{t/2}(c_{1}cos((5t)/2)+c_{2}sin((5t)/2))

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12y^{''}-12y^'+78y=0

12 y^{^{''}} - 12 y^{^{'}} + 78 y = 0

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

Solution steps

12 y^{''} (t) - 12 y^{'} (t) + 78 y = 0

Solve linear ODE:

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

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

\int 5 e^{- 0.5 x} d x

\frac{d}{d x} (65 - 55 cos (\frac{π x}{15}))

\frac{\partial}{\partial x} (\frac{x ^{2} + 4}{y - 3})

y^{^{'}} = \frac{y ^{2} - y}{x}

x + 2, x^{2}, - 1, 2

What is the solution for 12y^{''}-12y^'+78y=0 ?
The solution for 12y^{''}-12y^'+78y=0 is y=e^{t/2}(c_{1}cos((5t)/2)+c_{2}sin((5t)/2))