What is the solution for y^{''}+8y^'+41y=0 ?

The solution for y^{''}+8y^'+41y=0 is y=e^{-4t}(c_{1}cos(5t)+c_{2}sin(5t))

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y^{''}+8y^'+41y=0

y^{^{''}} + 8 y^{^{'}} + 41 y = 0

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

Solution steps

y^{''} (t) + 8 y^{'} (t) + 41 y = 0

Solve linear ODE:

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

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

arccos (1 - x^{2})

\frac{d}{d x} (\frac{( x - 1 )}{2})

\int \frac{1}{7 - 4 x - x ^{2}} d x

\frac{( s + 1 )}{( s ^{2} + 2 s + 9 )}

x = 3 y, x = 2 + y^{2}

What is the solution for y^{''}+8y^'+41y=0 ?
The solution for y^{''}+8y^'+41y=0 is y=e^{-4t}(c_{1}cos(5t)+c_{2}sin(5t))