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

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

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y^{''}-y^'=e^xcos(x)

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + y = e^{- 2 t}

y^{^{''}} + y = - sec^{2} (x)

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

D^{2} (D + 3)^{2} y = 9 e^{- 3 x}

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = e^{6 t}

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