What is the solution for 100y^{''}+20y^'+y=0 ?

The solution for 100y^{''}+20y^'+y=0 is y=c_{1}e^{-t/(10)}+c_{2}e^{-t/(10)}t

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100y^{''}+20y^'+y=0

100 y^{^{''}} + 20 y^{^{'}} + y = 0

y = c_{1} e^{- \frac{t}{10}} + c_{2} e^{- \frac{t}{10}} t

Solution steps

100 y^{''} (t) + 20 y^{'} (t) + y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{t}{10}} + c_{2} e^{- \frac{t}{10}} t

y = c_{1} e^{- \frac{t}{10}} + c_{2} e^{- \frac{t}{10}} t

\int - x^{- 3} d x

x \to 1 + lim (x - 2)

\frac{\partial}{\partial x} (x y^{2} e^{x^{2}})

t sin (0.89 t)

\frac{5}{8 t}

What is the solution for 100y^{''}+20y^'+y=0 ?
The solution for 100y^{''}+20y^'+y=0 is y=c_{1}e^{-t/(10)}+c_{2}e^{-t/(10)}t