What is the solution for y^{''}+8y^'+65y=0 ?

The solution for y^{''}+8y^'+65y=0 is y=e^{-4t}(c_{1}cos(7t)+c_{2}sin(7t))

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y^{''}+8y^'+65y=0

y^{^{''}} + 8 y^{^{'}} + 65 y = 0

y = e^{- 4 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

Solution steps

y^{''} (t) + 8 y^{'} (t) + 65 y = 0

Solve linear ODE:

y = e^{- 4 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

y = e^{- 4 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

x \to 0 - lim (\frac{x}{13 x ^{4} - 6 x})

\frac{\partial}{\partial y} (x yz)

\int \frac{2 x}{4 - x ^{2}} d x

y^{^{''}} + 2 y = - 18 x^{2} e^{2} x

\frac{d}{d x} (- \frac{9}{( 1 + 3 x ) ^{2}})

What is the solution for y^{''}+8y^'+65y=0 ?
The solution for y^{''}+8y^'+65y=0 is y=e^{-4t}(c_{1}cos(7t)+c_{2}sin(7t))