What is the solution for y^{''}+40y^'+625y=0 ?

The solution for y^{''}+40y^'+625y=0 is y=e^{-20t}(c_{1}cos(15t)+c_{2}sin(15t))

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y^{''}+40y^'+625y=0

y^{^{''}} + 40 y^{^{'}} + 625 y = 0

y = e^{- 20 t} (c_{1} cos (15 t) + c_{2} sin (15 t))

Solution steps

y^{''} (t) + 40 y^{'} (t) + 625 y = 0

Solve linear ODE:

y = e^{- 20 t} (c_{1} cos (15 t) + c_{2} sin (15 t))

y = e^{- 20 t} (c_{1} cos (15 t) + c_{2} sin (15 t))

\frac{s ^{2} + 1}{s ^{2} + 4}

\int_{0}^{1} \frac{17}{x ^{5}} d x

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

\int x sin (- 2 x) d x

9 y^{^{''}} + 24 y^{^{'}} + 16 y = 0

What is the solution for y^{''}+40y^'+625y=0 ?
The solution for y^{''}+40y^'+625y=0 is y=e^{-20t}(c_{1}cos(15t)+c_{2}sin(15t))