What is the solution for y^{''}+10y^'+41y=0 ?

The solution for y^{''}+10y^'+41y=0 is y=e^{-5t}(c_{1}cos(4t)+c_{2}sin(4t))

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y^{''}+10y^'+41y=0

y^{^{''}} + 10 y^{^{'}} + 41 y = 0

y = e^{- 5 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

y^{''} (t) + 10 y^{'} (t) + 41 y = 0

Solve linear ODE:

y = e^{- 5 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{- 5 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

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

n = 3 \sum \infty 4^{(n + 2)}

\int \frac{x ^{2} + x + 1}{x} d x

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

y = (arctan (5 x))^{2}

What is the solution for y^{''}+10y^'+41y=0 ?
The solution for y^{''}+10y^'+41y=0 is y=e^{-5t}(c_{1}cos(4t)+c_{2}sin(4t))