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Particular Solution
Consider the nonhomogeneous differential equation y(n)+an−1(x)y(n−1)+…+a1(x)y′+a0(x)y=g(x).
Recall from the section about linear independence that the solution of the corresponding homogeneous equation y(n)+an−1(x)y(n−1)+…+a1(x)y′+a0(x)y=0 is given as yh=c1y1(x)+c2y2(x)+…+cnyn(x).
If yp is the solution of the nonhomogeneous equation, the general solution of the nonhomogeneous equation is y=yh+yp.
Proof.
Plug y=yh+yp into the differential equation:
(yh+yp)(n)+an−1(x)(yh+yp)(n−1)+…+a1(x)(yh+yp)′+a0(x)y=
=(yh(n)+an−1(x)yh(n−1)+…+a1(x)yh′+a0(x)yh)+(yp(n)+an−1(x)yp(n−1)+…+a1(x)yp′+a0(x)yp)=
=0+g(x)=g(x)
So, indeed, y=yp+yh is the solution of the nonhomogeneous differential equation.
Now, let's consider an example.
Example. The solution of the homogeneous equation y′′+y′−2y=0 is yh=c1e−2x+c2ex. For the nonhomogeneous equation y′′+y′−2y=2(1+x−x2), the particular solution is yp=x2. So, the general solution of the nonhomogeneous differential equation is y′′+y′−2y=2(1+x−x2) is y=yh+yp=c1e−2x+c2ex+x2.