What is the solution for y^{''}+14y^'+65y=0 ?

The solution for y^{''}+14y^'+65y=0 is y=e^{-7t}(c_{1}cos(4t)+c_{2}sin(4t))

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y^{''}+14y^'+65y=0

y^{^{''}} + 14 y^{^{'}} + 65 y = 0

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

Solution steps

y^{''} (t) + 14 y^{'} (t) + 65 y = 0

Solve linear ODE:

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

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

f (x) = 3 x + 10, (2, 4)

\int \frac{26}{( x - 1 ) ( x ^{2} + 25 )} d x

(x^{2} - 7 x + 4) 25 - x^{2}

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

x y^{2} d x + (1 + x^{2} y) d y = 0

What is the solution for y^{''}+14y^'+65y=0 ?
The solution for y^{''}+14y^'+65y=0 is y=e^{-7t}(c_{1}cos(4t)+c_{2}sin(4t))