What is the solution for 12y^{''}-12y^'+30y=0 ?

The solution for 12y^{''}-12y^'+30y=0 is y=e^{t/2}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))

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12y^{''}-12y^'+30y=0

12 y^{^{''}} - 12 y^{^{'}} + 30 y = 0

y = e^{\frac{t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

Solution steps

12 y^{''} (t) - 12 y^{'} (t) + 30 y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{3 t}{2}) + c_{2} sin (\frac{3 t}{2}))

\int \frac{3 x ^{3}}{4 + x ^{8}} d x

\frac{d x}{d y} = 3 x^{2} y^{3}

\int 3 x^{3} - 4 x^{2} + \frac{7}{x} - \frac{1}{x ^{2}} + 2 e^{x} d x

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 x} (x + 1)

f (x) = 9 x - 8 x, (1, 1)

What is the solution for 12y^{''}-12y^'+30y=0 ?
The solution for 12y^{''}-12y^'+30y=0 is y=e^{t/2}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))