What is the solution for y^{''}+y^'+9.25=0 ?

The solution for y^{''}+y^'+9.25=0 is y=c_{1}+c_{2}e^{-t}-9.25t

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y^{''}+y^'+9.25=0

y^{^{''}} + y^{^{'}} + 9.25 = 0

y = c_{1} + c_{2} e^{- t} - 9.25 t

Solution steps

y^{''} (t) + y^{'} (t) + 9.25 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} - 9.25 t

y = c_{1} + c_{2} e^{- t} - 9.25 t

x^{^{''}} + 49 x = 65 cos (6 t)

y^{^{''}} - 8 y^{^{'}} + 16 y = 4 e^{4 x}

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

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

y^{^{''}} + y^{^{'}} = n

What is the solution for y^{''}+y^'+9.25=0 ?
The solution for y^{''}+y^'+9.25=0 is y=c_{1}+c_{2}e^{-t}-9.25t