What is the solution for 6y^{''}-6y^'+39y=0 ?

The solution for 6y^{''}-6y^'+39y=0 is y=e^{t/2}(c_{1}cos((5t)/2)+c_{2}sin((5t)/2))

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6y^{''}-6y^'+39y=0

6 y^{^{''}} - 6 y^{^{'}} + 39 y = 0

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

Solution steps

6 y^{''} (t) - 6 y^{'} (t) + 39 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}))

x \to 0 lim ((1 + \frac{x}{2})^{\frac{3}{x}})

\frac{d}{d x} (ln (x^{2} - 4 x + 5))

x_{1} + x_{2} + x_{3} - 12

\int e^{- x} cos (8 x) d x

\frac{\partial}{\partial x} (12 x^{4} - 6 x^{2} y + 4 y^{3})

What is the solution for 6y^{''}-6y^'+39y=0 ?
The solution for 6y^{''}-6y^'+39y=0 is y=e^{t/2}(c_{1}cos((5t)/2)+c_{2}sin((5t)/2))