General Solution for y"-y'=e^x - sinx

3. Find the general solution for y” + y’=e^ x –sin(x)
Solution:
First find the homogeneous solution:
yh'' + yh' = 0

The characteristic equation is:
m2 + 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 ex term.


yp(x) = Csin(x) + Dcos(x) + Eex
Thus
yp'(x) = Ccos(x) - Dsin(x) + Eex

yp''(x) = -Csin(x) - Dcos(x) + Eex

So
yp1'' + yp1' = (C - D)cos(x) + (-C - D)sin(x) + 2Eex = -sin(x) +ex

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)) + ex

Thus the full solution will be:
y(x) = A + Bex + 1/2 * (sin(x) + cos(x)) + 1/2 * ex

Check:
y'(x) = -Bex + 1/2 *(cos(x) - sin(x)) + 1/2 *ex

y''(x) = Bex + 1/2 *(-sin(x) - cos(x)) + 1/2 * ex

Adding these up:

y''(x) + y'(x) = -sin(x) + ex

You could transform this into a first-order equation by inventing a function u(x) such that u = dy/dx.

So, u' + u = e^x - sin(x)

This can be solved using an integrating factor.