What is the solution for y^{''}+100y^'+2500y=30 ?

The solution for y^{''}+100y^'+2500y=30 is y=c_{1}e^{-50t}+c_{2}te^{-50t}+3/250

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y^{''}+100y^'+2500y=30

y^{^{''}} + 100 y^{^{'}} + 2500 y = 30

y = c_{1} e^{- 50 t} + c_{2} t e^{- 50 t} + \frac{3}{250}

Solution steps

y^{''} (t) + 100 y^{'} (t) + 2500 y = 30

Solve linear ODE:

y = c_{1} e^{- 50 t} + c_{2} t e^{- 50 t} + \frac{3}{250}

y = c_{1} e^{- 50 t} + c_{2} t e^{- 50 t} + \frac{3}{250}

3 y^{^{''}} - 5 y^{^{'}} - 2 y = 3 e^{2 x}

4 y^{^{''}} - 8 y^{^{'}} = cos (2 x) + 7 x^{2}

y^{^{''}} - 2 y^{^{'}} + y = \frac{4 e ^{x}}{x ^{2} + 1}

y^{^{''}} + y^{^{'}} = x + e^{x} + x^{4}

\frac{d ^{2} y}{d x ^{2}} + y = csc (x) cot (x)

What is the solution for y^{''}+100y^'+2500y=30 ?
The solution for y^{''}+100y^'+2500y=30 is y=c_{1}e^{-50t}+c_{2}te^{-50t}+3/250