First find the homogeneous solution:

yh'' + yh' = 0

The characteristic equation is:

m^2 + m = 0

m(m + 1) = 0

m = 0 or m = -1

Thus

yh(x) = A + Be^{-x}

Now for the particular solution. Since this is a linear equation we may propose a form:

yp = yp1 + yp2

where yp1 produces the -sin(x) term and yp2 produces the e^{x} term.

I would suggest trying:

yp(x) = Csin(x) + Dcos(x) + Ee^{x}

Thus

yp'(x) = Ccos(x) - Dsin(x) + Ee^{x}

yp''(x) = -Csin(x) - Dcos(x) + Ee^{x}

So

yp1'' + yp1' = (C - D)cos(x) + (-C - D)sin(x) + 2Ee^{x} = -sin(x) + e^{x}

Thus

C - D = 0

-C - D = -1

2E = 1

The top equation says D = C, so the middle equation says:

-2C = -1

Thus D = C = 1/2 so

yp(x) = 1/2 * (sin(x) + cos(x)) + e^{x}

Thus the full solution will be:

y(x) = A + Be^{-x} + 1/2 * (sin(x) + cos(x)) + 1/2 * e^{x}

Check:

y'(x) = -Be^{-x} + 1/2 *(cos(x) - sin(x)) + 1/2 * e^{x}

y''(x) = Be^{x} + 1/2 *(-sin(x) - cos(x)) + 1/2 * e^{x}

Adding these up:

y''(x) + y'(x) = -sin(x) + e^{x} (Check!)

-Dan