What is the solution for y^{''}+10y^'+16y=40cos(4t) ?

The solution for y^{''}+10y^'+16y=40cos(4t) is y=c_{1}e^{-2t}+c_{2}e^{-8t}+sin(4t)

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y^{''}+10y^'+16y=40cos(4t)

y^{^{''}} + 10 y^{^{'}} + 16 y = 40 cos (4 t)

y = c_{1} e^{- 2 t} + c_{2} e^{- 8 t} + sin (4 t)

Solution steps

y^{''} (t) + 10 y^{'} (t) + 16 y = 40 cos (4 t)

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{- 8 t} + sin (4 t)

y = c_{1} e^{- 2 t} + c_{2} e^{- 8 t} + sin (4 t)

y^{^{''}} - 2 y^{^{'}} + y = \frac{13 e ^{t}}{1 + t ^{2}}

y^{^{''}} + 36 y = 7 sec (6 x)

y^{^{'''}} - y^{^{'}} = 2 sin (t)

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{3} e^{2} t, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{x} sin (2 x)

What is the solution for y^{''}+10y^'+16y=40cos(4t) ?
The solution for y^{''}+10y^'+16y=40cos(4t) is y=c_{1}e^{-2t}+c_{2}e^{-8t}+sin(4t)