MATH 1005A - Notes 1
Reduction of Order
Consider a linear, second-order, homogeneous equation in standard form, y 00 + p(x)y 0 + q(x)y 0 = 0:
Suppose that one solution y1 is known. Then a second, independent solution y2 is obtained by letting y2 (x) = u(x)y1 (x); u(x) to be determined.
0
0
0
00 y2 = uy1 ) y2 = u0 y1 + uy1 ; y 00 = u00 y1 + 2u0 y1 + uy1 ;

so y2 is a solution if and only if
0
00
0
[u00 y1 + 2u0 y1 + uy1 ] + p(x)[u0 y1 + uy1 ] + q(x)[uy1 ] = 0;

i.e.,
00
0
0
u[y1 + p(x)y1 + q(x)y1 ] + u00 y1 + 2u0 y1 + p(x)u0 y1 = 0:
00
0
Since y1 is a solution, y1 + p(x)y1 + q(x)y1 = 0. Thus, y2 is a solution if and only if
0
u00 y1 + u0 [2y1 + p(x)y1 ] = 0;

i.e.,

u00
2y 0 + p(x)y1 y0 =¡ 1
= ¡2 1 ¡ p(x): u0 y1 y1 Integration with respect to x then gives
0

ln ju j = ¡2 ln jy1 j ¡

Z

p(x) dx:

Taking the exponential of both sides and using the fact that e¡2 ln jy1 j = eln jy1 j obtain ju0 j =

¡2

=

1
, we
2
y1

1 ¡ R p(x) dx
1 R e ; or u0 = § 2 e¡ p(x) dx:
2
y1 y1 Taking the plus sign and integrating once more, we obtain
Z
1 ¡ R p(x) dx u(x) = e dx:
2
y1
Since

1 ¡ R p(x) dx e 6= 0;
2
y1

u(x) is not a constant, hence y1 = erx and y2 = u(x)erx are linearly independent.
For the equation ay 00 + by 0 + cy = 0 with b2 ¡ 4ac = 0; y = erx ) ar2 + br + c = 0 ) p b b ¡b § b2 ¡ 4ac
=¡
) y1 = erx = e¡ 2a x : r= 2a
2a

2
In standard form, the equation is b c b y 00 + y 0 + y =