What is the solution for y^{''}+8/100 y^'+12/10000 y=0 ?

The solution for y^{''}+8/100 y^'+12/10000 y=0 is y=c_{1}e^{-t/(50)}+c_{2}e^{-(3t)/(50)}

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y^{''}+8/100 y^'+12/10000 y=0

y^{^{''}} + \frac{8}{100} y^{^{'}} + \frac{12}{10000} y = 0

y = c_{1} e^{- \frac{t}{50}} + c_{2} e^{- \frac{3 t}{50}}

Solution steps

y^{''} (t) + \frac{8}{100} y^{'} (t) + \frac{12}{10000} y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{t}{50}} + c_{2} e^{- \frac{3 t}{50}}

y = c_{1} e^{- \frac{t}{50}} + c_{2} e^{- \frac{3 t}{50}}

y^{^{''}} - y = 0, y (ln (2)) = 5, y (0) = 1

\frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 2 x = 0, x (0) = 0

y^{^{''''}} + 2 y^{^{'''}} + 2 y^{^{''}} + 2 y^{^{'}} + y = 0

2 \cdot y^{^{''}} + 8 \cdot y^{^{'}} + 8 y = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 1

What is the solution for y^{''}+8/100 y^'+12/10000 y=0 ?
The solution for y^{''}+8/100 y^'+12/10000 y=0 is y=c_{1}e^{-t/(50)}+c_{2}e^{-(3t)/(50)}