What is the solution for y^{''}+20y^'+500=12 ?

The solution for y^{''}+20y^'+500=12 is y=c_{1}+c_{2}e^{-20t}-(122t)/5

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y^{''}+20y^'+500=12

y^{^{''}} + 20 y^{^{'}} + 500 = 12

y = c_{1} + c_{2} e^{- 20 t} - \frac{122 t}{5}

Solution steps

y^{''} (t) + 20 y^{'} (t) + 500 = 12

Solve linear ODE:

y = c_{1} + c_{2} e^{- 20 t} - \frac{122 t}{5}

y = c_{1} + c_{2} e^{- 20 t} - \frac{122 t}{5}

\int \frac{1}{x - 1 + 1 + x} d x

y^{^{'}} = 2 + y - 2 x + 3

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

y

2

What is the solution for y^{''}+20y^'+500=12 ?
The solution for y^{''}+20y^'+500=12 is y=c_{1}+c_{2}e^{-20t}-(122t)/5