Download Math 285: Differential Equations Quiz 6: Solutions 1. Verify that the

Math 285: Differential Equations
Quiz 6: Solutions
15 Mar 2007: 20 pts
1. Verify that the functions y1 , y2 are solutions to the given the differential equation. Use the Wronskian W (y1 , y2 ) to
show that they form a fundamental set of solutions.
y 00 + 9y = 0;
y1 = cos 3x,
y2 = sin 3x.
We first verify that y1 , y2 are solutions
y1 = cos 3x,
y10 = −3 sin 3x,
y100 = −9 cos 3x,
thus
y100 + 9y1 = −9 cos 3x + 9 cos 3x = 0
y2 = sin 3x,
y20 = 3 cos 3x,
y200 = −9 sin 3x,
thus
y200 + 9y2 = −9 sin 3x + 9 sin 3x = 0.
We now compute the Wronskian
y
W (y1 , y2 ) = 10
y1
sin 3x y2 cos 3x
= 3(cos2 3x + sin2 3x) = 3;
=
y20 −3 sin 3x 3 cos 3x
Since W (y1 , y2 ) = 3 6= 0 for all x, the solutions y1 , y2 form a fundamental set.
2. The indicated function y1 is a solution of the given differential equation on the interval (0, ∞). Use reduction of
order to find a second solution y2 .
x2 y 00 + xy 0 − y = 0; y1 = x.
We look for a second solution of the form y = ux. We have
y 0 = u0 x + u
y 00 = u00 x + 2u0 .
Substituting into the differential equation we obtain:
x2 (u00 x + 2u0 ) + x(u0 x + u) − ux = x2 (u00 x + 3u0 ) = 0.
Dividing through by x2 and substituting w = u0 we obtain the first order homogeneous linear DE:
xw0 + 3w = 0.
We can solve this by the usual method of separation of variables
Z
Z
dw
dx
= −3
w
x
ln |w| = −3 ln |x| + k1
w=
Now since u0 = w we obtain
Z
u=
k2
.
x3
k2
k2
1
dx = − 2 + c1 = c1 + c2 2 .
x3
2x
x
Hence, our solution y becomes
y = ux = c1 x + c2
1
x
and so our second solution is y2 = 1/x.
It is also possible to use formula (5) on page 151 of the text; but first we must divide through by x2 to put the
given differential equation into standard form:
1
1
y 00 + y 0 − 2 y = 0.
x
x
We may now use the formula with P = x1 and y1 = x. We obtain
Z − R P dx
Z − R 1/x dx
Z − ln x+C1
e
e
e
y 2 = y1
dx = x
dx = x
dx
y12
x2
x2
Z
C2
C2
C2
dx = x − 2 + C3 = −
+ C3 x.
=x
3
x
2x
2x
Thus, we can take our second solution to be y2 = 1/x (where C2 = −2 and C3 = 0).