Nonhomogeneous Second-Order Differential Equations
To solve ay′′ +by′+cy =f(x) we first consider the solution of the form y=yc+ypwhere
ycsolves the differential equaiton ay′′ +by′+cy = 0 and ypsolves the differential equation
ay′′ +by′+cy =f(x).
Since the derivative of the sum equals the sum of the derivatives, we will have a final
solution of 0 + f(x) which gives f(x).
Therefore, we follow a pattern for ypto yield f(x) when each of y,y′and y′′ are substi-
tuted into the above equation.
1. Case I: If f(x) is of the form f(x) = P(x)ekx then we set:
yp=Q(x)ekx where Q(x) is the general polynomial of the same degree as P(x).
For example if the differential equation is set equal to:
(a) f(x) = x2e3x. Set yp= (Ax2+Bx +C)e3x.
(b) f(x) = x3−4x+ 1. Set yp=Ax3+Bx2+C x +D.
(c) f(x) = 6e−2x. Set yp=Ae−2x.
Then we substitute yp,y′
p, and y′′
pand solve for the coefficients.
2. Case II: If f(x) is of the form f(x) = P(x)ekx sin (mx) or f(x) = P(x)ekx cos (mx),
then we set:
yp=Q(x)ekx cos (mx) + R(x)ekx sin (mx) where Q(x) and R(x) are both general
polynomials of the same degree as P(x).
For example if the differential equation is set equal to:
(a) f(x) = 2 cos( 3x). Set yp=Acos (3x) + Bsin (3x)
(b) f(x) = xcos (x). Set yp= (Ax +B) cos (x) + (Cx +D) sin (x)
(c) f(x) = exsin (2x). Set yp=Aexsin (2x) + B excos (2x)
If f(x) is a sum of terms, like f(x) = x2+e−x+ cos (x), do it as separate problems solving
for
yp1=Ax2+Bx +C,
yp2=Ae−xand
yp3=Acos (x) + Bsin (x).
