Second-Order Linear Equations
Example 1 (section 7.8, pg. 1172, problem 10). Solve the initial-value problem using
the method of undetermined coefficients:
y00 +y0
−2y=x+ sin 2x, y(0) = 1, y0(0) = 0
Solution:
1) First, we solve the corresponding homogeneous equation y00+y0
−2y= 0. The characteristic
equation is r2+r−2=(r−1)(r+ 2) = 0 with roots r= 1,−2. So the general solution of
the homogeneous equation is
yc(x) = C1ex+C2e−2x
2) Second, we look for a particular solution of the given differential equation. Since G(x) =
x+ sin 2x, we seek a particular solution of the form
yp(x) = Ax +B+Ccos 2x+Dsin 2x
Then
y0
p(x) = A−2Csin 2x+ 2Dcos 2x, y00
p=−4Ccos 2x−4Dsin 2x
So, substituting into the given differential equation, we have
−4Ccos 2x−4Dsin 2x+A−2Csin 2x+2Dcos 2x−2 (Ax +B+Ccos 2x+Dsin 2x) = x+sin 2x
or
−2Ax + (A−2B)+(−4C+ 2D−2C) cos 2x+ (−4D−2C−2D) sin 2x=x+ sin 2x
or
−2Ax + (A−2B)+(−6C+ 2D) cos 2x+ (−6D−2C) sin 2x=x+ sin 2x
This is true if
−2A= 1 A−2B= 0 −6C+ 2D= 0 −6D−2C= 1
The solution of the system is
A=−
1
2B=−
1
4C=−
1
20 D=−
3
20
A particular solution is therefore
yp(x) = −
1
2x−
1
4
−
1
20 cos 2x−
3
20 sin 2x
3) By the superposition principle, the general solution is
y(x) = yc(x) + yp(x) = C1ex+C2e−2x
−
1
2x−
1
4
−
1
20 cos 2x−
3
20 sin 2x
1
