What is the solution for y^{''}+50y^'+100=0 ?

The solution for y^{''}+50y^'+100=0 is y=c_{1}+c_{2}e^{-50t}-2t

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y^{''}+50y^'+100=0

y^{^{''}} + 50 y^{^{'}} + 100 = 0

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

Solution steps

y^{''} (t) + 50 y^{'} (t) + 100 = 0

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} = - 18 x, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} - 8 y^{^{'}} + 17 y = x^{2} e^{4 x}

y^{^{''}} + 25 y^{^{'}} = 6 sin (x)

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = \frac{e ^{x}}{x ^{3}}

y^{^{''}} - y^{^{'}} = t^{2}

What is the solution for y^{''}+50y^'+100=0 ?
The solution for y^{''}+50y^'+100=0 is y=c_{1}+c_{2}e^{-50t}-2t