What is the solution for y^{''}+6y^'+13y=0 ?

The solution for y^{''}+6y^'+13y=0 is y=e^{-3t}(c_{1}cos(2t)+c_{2}sin(2t))

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y^{''}+6y^'+13y=0

y^{^{''}} + 6 y^{^{'}} + 13 y = 0

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

Solution steps

y^{''} (t) + 6 y^{'} (t) + 13 y = 0

Solve linear ODE:

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

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

\int \frac{x ^{3}}{64} d x

\int (x^{2} - x + 3) d x

\int \frac{2 x}{x ^{2} + y ^{4}} d x

f (x) = \frac{1}{2 x} x^{9}

(- 1, 7), (3, - 1)

What is the solution for y^{''}+6y^'+13y=0 ?
The solution for y^{''}+6y^'+13y=0 is y=e^{-3t}(c_{1}cos(2t)+c_{2}sin(2t))