What is the solution for y^{''}-14y^'+53y=0 ?

The solution for y^{''}-14y^'+53y=0 is y=e^{7t}(c_{1}cos(2t)+c_{2}sin(2t))

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y^{''}-14y^'+53y=0

y^{^{''}} - 14 y^{^{'}} + 53 y = 0

y = e^{7 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

Solution steps

y^{''} (t) - 14 y^{'} (t) + 53 y = 0

Solve linear ODE:

y = e^{7 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

y = e^{7 t} (c_{1} cos (2 t) + c_{2} sin (2 t))

\int 20 cos (10 t + \frac{π}{6}) + 2 d t

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f (x) = - 6 x

\frac{d}{d x} (6 (x^{2} + 4)^{8})

\frac{d}{d x} (\frac{π x}{3})

What is the solution for y^{''}-14y^'+53y=0 ?
The solution for y^{''}-14y^'+53y=0 is y=e^{7t}(c_{1}cos(2t)+c_{2}sin(2t))