What is the solution for y^{''}+4y^'=4cos(x)+3sin(x)-8 ?

The solution for y^{''}+4y^'=4cos(x)+3sin(x)-8 is y=c_{1}+c_{2}e^{-4x}-2x-16/17 cos(x)+13/17 sin(x)

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y^{''}+4y^'=4cos(x)+3sin(x)-8

y^{^{''}} + 4 y^{^{'}} = 4 cos (x) + 3 sin (x) - 8

y = c_{1} + c_{2} e^{- 4 x} - 2 x - \frac{16}{17} cos (x) + \frac{13}{17} sin (x)

Solution steps

y^{''} (x) + 4 y^{'} (x) = 4 cos (x) + 3 sin (x) - 8

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} - 2 x - \frac{16}{17} cos (x) + \frac{13}{17} sin (x)

y = c_{1} + c_{2} e^{- 4 x} - 2 x - \frac{16}{17} cos (x) + \frac{13}{17} sin (x)

y^{^{''}} - y^{^{'}} - 6 y = e^{- x} + 12 x

5 y^{^{''}} - 15 y = 40 e^{x}

y^{^{''}} + 6 y^{^{'}} + 10 y = e^{- 3 t}, y (0) = 4, y^{^{'}} (0) = 5

y^{^{'''}} = 3 y^{^{''}} + 9 (e^{(5 x)}) + 2 (cos (x))

y^{^{''}} - y^{^{'}} - y = x^{2} - 2

What is the solution for y^{''}+4y^'=4cos(x)+3sin(x)-8 ?
The solution for y^{''}+4y^'=4cos(x)+3sin(x)-8 is y=c_{1}+c_{2}e^{-4x}-2x-16/17 cos(x)+13/17 sin(x)