What is the solution for 6y^{''}-6y^'+15y=0 ?

The solution for 6y^{''}-6y^'+15y=0 is y=e^{t/2}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))

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6y^{''}-6y^'+15y=0

6 y^{^{''}} - 6 y^{^{'}} + 15 y = 0

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

Solution steps

6 y^{''} (t) - 6 y^{'} (t) + 15 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}))

\frac{d}{d x} (xL (x) - x)

\int \frac{7 x}{x ^{3} + 1} d x

y = (x + 7) (7 x + 2)

\frac{d}{d x} (x^{3} (8 - 3 x))

\frac{d}{d x} ((1 + 6 x^{2}) (x - x^{2}))

What is the solution for 6y^{''}-6y^'+15y=0 ?
The solution for 6y^{''}-6y^'+15y=0 is y=e^{t/2}(c_{1}cos((3t)/2)+c_{2}sin((3t)/2))