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

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

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-y+y^'-y=e^xcos(x)

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}}, x y + 7 ey (x) = 7 e

f (x) = tan (x) - 1 sec (x)

\frac{\partial}{\partial x} (ln (13 x^{2} - 10 y^{2}))

\frac{\partial}{\partial x} (x^{2} y^{3} sin (x - 2 y))

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