What is the solution for x^{''}+100x^'+1/(0.0004)x=30 ?

The solution for x^{''}+100x^'+1/(0.0004)x=30 is x=c_{1}e^{-50t}+c_{2}te^{-50t}+0.012

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x^{''}+100x^'+1/(0.0004)x=30

x^{^{''}} + 100 x^{^{'}} + \frac{1}{0.0004} x = 30

x = c_{1} e^{- 50 t} + c_{2} t e^{- 50 t} + 0.012

Solution steps

x^{''} (t) + 100 x^{'} (t) + \frac{1}{0.0004} x = 30

Solve linear ODE:

x = c_{1} e^{- 50 t} + c_{2} t e^{- 50 t} + 0.012

x = c_{1} e^{- 50 t} + c_{2} t e^{- 50 t} + 0.012

y^{^{''}} - y = 1 + x^{2} e^{x}

y^{^{''}} + 9 y = cos (3 t), y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 4 y^{^{'}} - 5 y = 12, 5 y (0) = 0

y^{^{''}} - y^{^{'}} + y = 6 sin (x) - 3 e^{2 x}

\frac{1}{16} \frac{d ^{2} x}{d t ^{2}} + x = 10 sin (4 t)

What is the solution for x^{''}+100x^'+1/(0.0004)x=30 ?
The solution for x^{''}+100x^'+1/(0.0004)x=30 is x=c_{1}e^{-50t}+c_{2}te^{-50t}+0.012