What is the solution for y^{''}-30y^'+255y=0 ?

The solution for y^{''}-30y^'+255y=0 is y=e^{15t}(c_{1}cos(sqrt(30)t)+c_{2}sin(sqrt(30)t))

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y^{''}-30y^'+255y=0

y^{^{''}} - 30 y^{^{'}} + 255 y = 0

y = e^{15 t} (c_{1} cos (30 t) + c_{2} sin (30 t))

Solution steps

y^{''} (t) - 30 y^{'} (t) + 255 y = 0

Solve linear ODE:

y = e^{15 t} (c_{1} cos (30 t) + c_{2} sin (30 t))

y = e^{15 t} (c_{1} cos (30 t) + c_{2} sin (30 t))

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

\frac{\partial}{\partial x} (\frac{( x - y )}{( x + y )})

\frac{\partial}{\partial x} (e^{x + y + 3})

(\frac{x ^{3}}{3})^{^{'}}

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

What is the solution for y^{''}-30y^'+255y=0 ?
The solution for y^{''}-30y^'+255y=0 is y=e^{15t}(c_{1}cos(sqrt(30)t)+c_{2}sin(sqrt(30)t))