What is the solution for y^{''}+10y^'+26y=0 ?

The solution for y^{''}+10y^'+26y=0 is y=e^{-5t}(c_{1}cos(t)+c_{2}sin(t))

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y^{''}+10y^'+26y=0

y^{^{''}} + 10 y^{^{'}} + 26 y = 0

y = e^{- 5 t} (c_{1} cos (t) + c_{2} sin (t))

Solution steps

y^{''} (t) + 10 y^{'} (t) + 26 y = 0

Solve linear ODE:

y = e^{- 5 t} (c_{1} cos (t) + c_{2} sin (t))

y = e^{- 5 t} (c_{1} cos (t) + c_{2} sin (t))

f (x) = x^{2} 9 - x^{2}

\frac{d}{d x} ((sin (\frac{x}{2}) - cos (\frac{x}{2}))^{2})

cos (x), sin (x), 0, \frac{π}{2}

\frac{d y}{d x} = 10 y \cdot e^{x + 7}

\int (x + 1) 3 x + 2 d x

What is the solution for y^{''}+10y^'+26y=0 ?
The solution for y^{''}+10y^'+26y=0 is y=e^{-5t}(c_{1}cos(t)+c_{2}sin(t))