What is the solution for y^{''}+14y^'+58y=0 ?

The solution for y^{''}+14y^'+58y=0 is y=e^{-7t}(c_{1}cos(3t)+c_{2}sin(3t))

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y^{''}+14y^'+58y=0

y^{^{''}} + 14 y^{^{'}} + 58 y = 0

y = e^{- 7 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

Solution steps

y^{''} (t) + 14 y^{'} (t) + 58 y = 0

Solve linear ODE:

y = e^{- 7 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

y = e^{- 7 t} (c_{1} cos (3 t) + c_{2} sin (3 t))

\frac{1}{2 x} + \frac{1}{7 x ^{\frac{6}{7}}}

\int 2 (x - 2) \cdot ln (x - 2) d x

\int \frac{1}{4 - 6 x} d x

d x + d y + x d y = y d x

f (x) = \frac{2 π - x cos ( x )}{x - 2 π}

What is the solution for y^{''}+14y^'+58y=0 ?
The solution for y^{''}+14y^'+58y=0 is y=e^{-7t}(c_{1}cos(3t)+c_{2}sin(3t))