What is the solution for y^{''}+4y^'+13y=0 ?

The solution for y^{''}+4y^'+13y=0 is y=e^{-2t}(c_{1}cos(3t)+c_{2}sin(3t))

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y^{''}+4y^'+13y=0

y^{^{''}} + 4 y^{^{'}} + 13 y = 0

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 13 y = 0

Solve linear ODE:

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

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

a re a f (x) = 2 x - 4, 3, 6

\frac{d}{d x} (20 x^{\frac{12}{5}} - 6 x^{\frac{2}{13}})

in v erse l a pl a ce \frac{s + 3}{( s + 3 ) ^{2} + 25}

\int sec^{3} (6 x) d x

a re a - x, x, - 2 x + 6

What is the solution for y^{''}+4y^'+13y=0 ?
The solution for y^{''}+4y^'+13y=0 is y=e^{-2t}(c_{1}cos(3t)+c_{2}sin(3t))