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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 4 y = x sin (x)

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

4 y^{^{''}} - 4 y^{^{'}} + y = 40 e^{\frac{t}{2}}

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = sec (x)

y^{^{''}} - y^{^{'}} + 225 y = 15 sin (15 t)

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