What is the solution for y^{''}+y^'+4.25=0 ?

The solution for y^{''}+y^'+4.25=0 is y=c_{1}+c_{2}e^{-t}-4.25t

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y^{''}+y^'+4.25=0

y^{^{''}} + y^{^{'}} + 4.25 = 0

y = c_{1} + c_{2} e^{- t} - 4.25 t

Solution steps

y^{''} (t) + y^{'} (t) + 4.25 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} - 4.25 t

y = c_{1} + c_{2} e^{- t} - 4.25 t

y^{^{''}} + 5 y^{^{'}} + 6 y = e^{- 5 x} (- 55 + 68 x - 30 x^{2})

2 x^{^{''}} + x = 3 t^{2} + 10 t

y^{^{''}} + 2^{2} y = 2

y^{^{''}} + y^{^{'}} - y = 1

y^{^{'''}} - 5 y^{^{''}} - 4 y^{^{'}} + 20 y = sin (x)

What is the solution for y^{''}+y^'+4.25=0 ?
The solution for y^{''}+y^'+4.25=0 is y=c_{1}+c_{2}e^{-t}-4.25t