What is the solution for 20y^{''}+4y^'+y=0 ?

The solution for 20y^{''}+4y^'+y=0 is y=e^{-t/(10)}(c_{1}cos(t/5)+c_{2}sin(t/5))

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20y^{''}+4y^'+y=0

20 y^{^{''}} + 4 y^{^{'}} + y = 0

y = e^{- \frac{t}{10}} (c_{1} cos (\frac{t}{5}) + c_{2} sin (\frac{t}{5}))

Solution steps

20 y^{''} (t) + 4 y^{'} (t) + y = 0

Solve linear ODE:

y = e^{- \frac{t}{10}} (c_{1} cos (\frac{t}{5}) + c_{2} sin (\frac{t}{5}))

y = e^{- \frac{t}{10}} (c_{1} cos (\frac{t}{5}) + c_{2} sin (\frac{t}{5}))

(x + x)^{^{'}}

\int \frac{8}{( x - 2 ) ^{2}} d x

7 x (x + y) y^{^{'}} = 7 y (x - y)

\frac{\partial}{\partial x} (8 x^{0.5} y^{0.5})

\frac{d}{d x} (- \frac{2 ( - 3 x ^{2} + 1 )}{( x ^{2} + 1 ) ^{3}})

What is the solution for 20y^{''}+4y^'+y=0 ?
The solution for 20y^{''}+4y^'+y=0 is y=e^{-t/(10)}(c_{1}cos(t/5)+c_{2}sin(t/5))