What is the solution for y^{''}+10y^'+16y=40cos(4x) ?

The solution for y^{''}+10y^'+16y=40cos(4x) is y=c_{1}e^{-2x}+c_{2}e^{-8x}+sin(4x)

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y^{''}+10y^'+16y=40cos(4x)

y^{^{''}} + 10 y^{^{'}} + 16 y = 40 cos (4 x)

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

Solution steps

y^{''} (x) + 10 y^{'} (x) + 16 y = 40 cos (4 x)

Solve linear ODE:

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

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

y^{^{''}} - 8 y^{^{'}} + 6 y = 3 x^{2} - 8

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

x^{^{''}} - 25 x = e^{5 t}

y^{^{'''}} + y^{^{''}} - 2 y = - e^{3 x} (9 + 67 x + 17 x^{2})

y^{^{'''}} + 3 y^{^{''}} + 3 y^{^{'}} + y = x + 6 e^{- x}

What is the solution for y^{''}+10y^'+16y=40cos(4x) ?
The solution for y^{''}+10y^'+16y=40cos(4x) is y=c_{1}e^{-2x}+c_{2}e^{-8x}+sin(4x)