What is the solution for 12y^{''}-12y^'+6y=0 ?

The solution for 12y^{''}-12y^'+6y=0 is y=e^{t/2}(c_{1}cos(t/2)+c_{2}sin(t/2))

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12y^{''}-12y^'+6y=0

12 y^{^{''}} - 12 y^{^{'}} + 6 y = 0

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

Solution steps

12 y^{''} (t) - 12 y^{'} (t) + 6 y = 0

Solve linear ODE:

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

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

\frac{\partial}{\partial s} (\frac{( s - r )}{t})

s

\int \frac{x + 4}{2 x + 6} d x

\int r e^{r} d r

2 x e^{x - 1} + e^{x - 1} (x^{2} + 1)

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