What is the solution for x^{''}+240x^'-9.8=0 ?

The solution for x^{''}+240x^'-9.8=0 is x=c_{1}+c_{2}e^{-240t}+0.04083…t

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x^{''}+240x^'-9.8=0

x^{^{''}} + 240 x^{^{'}} - 9.8 = 0

x = c_{1} + c_{2} e^{- 240 t} + 0.04083 \dots t

Solution steps

x^{''} (t) + 240 x^{'} (t) - 9.8 = 0

Solve linear ODE:

x = c_{1} + c_{2} e^{- 240 t} + 0.04083 \dots t

x = c_{1} + c_{2} e^{- 240 t} + 0.04083 \dots t

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{2} e^{3} t, y (0) = 2, y^{^{'}} (0) = 17

y^{^{''}} + 6 y^{^{'}} + 13 y = e^{- 3 x} (cos (2 x))

y^{^{''}} + 36 y = sec^{2} (x)

y^{^{''}} + 8 y^{^{'}} + 12 = 1 - H (t - 3)

y^{^{''}} - 2 y^{^{'}} + y = t \cdot e^{t} + 4, y (0) = 1, y^{^{'}} (0) = 1

What is the solution for x^{''}+240x^'-9.8=0 ?
The solution for x^{''}+240x^'-9.8=0 is x=c_{1}+c_{2}e^{-240t}+0.04083…t