EXAMPLE: PROBLEM 3.6.17 OF THE TEXT
110.302 DIFFERENTIAL EQUATIONS
PROFESSOR RICHARD BROWN
Problem. Find the general solution to
x2 y 00 − 3xy 0 + 4y = x2 ln x,
x > 0,
given that y1 (x) = x2 and y2 (x) = x2 ln x form a fundamental set of solutions
to the homogeneous version x2 y 00 − 3xy 0 + 4y = 0.
Strategy. We use the method of Variation of Parameters with Y (x) = u1 (x)x2 + u2 (x)x2 ln x. To
do this, we will need to place the ODE in its standard form to retrieve the non-homogeneous part,
g(t). Here, the standard form is y 00 − x3 y 0 + x42 y = ln x. So g(x) = ln x.
Solution. With the assumed form for Y (x), we get the system
u01 x2 + u02 x2 ln x = 0
u01 + u02 ln x = 0
(1)
ln x
u01 2x + u02 (2x ln x + x) = ln x
or
2u01 + u02 (1 + 2 ln x) =
.
(2)
x
Using the simplified system at right, we multiply Equation 1 by −2 and add to Equation 2 and get
ln x
ln x
1
u02 (−2 ln x + 1 + 2 ln x) =
, or u02 =
, so u2 = (ln x)2 .
x
x
2
Substituting this back into Equation 1, we get
(ln x)2
1
ln x
0
0
0
u1 + u2 ln x = 0 = u1 +
ln x, or u01 = −
, so u1 = − (ln x)3 .
x
x
3
or
Thus our particular solution to the ODE is
Y (x) = u1 (x)y1 (x) + u2 (x)y2 (x)
1
1
= − (ln x)3 x2 + (ln x)2 x2 ln x
3
2
1 2
= x (ln x)3 .
6
And our general solution to the ODE is
1
y(x) = c1 x2 + c2 x2 ln x + x2 (ln x)3 .
6
We can also have done this directly using the integrals constructed in the book. To see this, first
we calculate the Wronskian of the two homogeneous solutions:
2
x x2 ln x
2 2
= 2x3 ln x + x3 − 2x3 ln x = x3 ,
W (y1 , y2 ) = W (x , x ln x) = 2x 2x ln x + x which is never zero on the interval x > 0.
1
2
110.302 DIFFERENTIAL EQUATIONS PROFESSOR RICHARD BROWN
Then
Z
u1 (x) =
−y2 g
dx = −
W (y1 , y2 )
Z
(x2 ln x)(ln x)
dx = −
x3
and
Z
y1 g
dx = −
W (y1 , y2 )
The rest of the result follows.
u2 (x) =
Z
x2 ln x
dx =
x3
Z
Z
(ln x)2
1
dx = − (ln x)3 ,
x
3
ln x
1
dx = (ln x)2 .
x
2