What is the solution for y^{''}+6y^'+19y=0 ?

The solution for y^{''}+6y^'+19y=0 is y=e^{-3t}(c_{1}cos(sqrt(10)t)+c_{2}sin(sqrt(10)t))

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y^{''}+6y^'+19y=0

y^{^{''}} + 6 y^{^{'}} + 19 y = 0

y = e^{- 3 t} (c_{1} cos (10 t) + c_{2} sin (10 t))

Solution steps

y^{''} (t) + 6 y^{'} (t) + 19 y = 0

Solve linear ODE:

y = e^{- 3 t} (c_{1} cos (10 t) + c_{2} sin (10 t))

y = e^{- 3 t} (c_{1} cos (10 t) + c_{2} sin (10 t))

y^{^{''}} = 0

\int \frac{1}{25 + ( x - 6 ) ^{2}} d x

\frac{1}{5 s ^{2} + 0.2 s}

\frac{\partial}{\partial y} (e^{5 y} cos (- 5 x))

\int e^{x} sin (4 x) d x

What is the solution for y^{''}+6y^'+19y=0 ?
The solution for y^{''}+6y^'+19y=0 is y=e^{-3t}(c_{1}cos(sqrt(10)t)+c_{2}sin(sqrt(10)t))