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

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

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y^{''}+y^'=sin(x)+cos(x)

y^{^{''}} + y^{^{'}} = sin (x) + cos (x)

y = c_{1} + c_{2} e^{- x} - cos (x)

Solution steps

y^{''} (x) + y^{'} (x) = sin (x) + cos (x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- x} - cos (x)

y = c_{1} + c_{2} e^{- x} - cos (x)

y^{^{'''}} + y^{^{''}} + 4 y^{^{'}} + 4 y = 4 x + 4

y^{^{''}} + 8 y^{^{'}} + 16 = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 e^{2 t}, y^{^{'}} (0) = 5, y (0) = - 3

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

(d^{2} + 1) y = x^{3}

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