Download Math 285: Differential Equations Quiz 7: Solutions 1. Solve the given

Math 285: Differential Equations
Quiz 7: Solutions
29 Mar 2007: 20 pts
1. Solve the given differential equation (i.e, find the general solution) by the method of undetermined coefficients:
y 00 − 2y 0 − 3y = 6 − 2e−x
First we consider the associated homogeneous equation. The auxiliary equation is m2 − 2m − 3 = 0. The roots are
m = 3, −1 and therefore the complementary solution is yc = c1 e3x + c2 e−x . To find a particular solution to the
nonhomogeneous equation using the method of undetermined coefficients we apply the superposition principle. First
we look for a particular solution yp1 to the equation:
y 00 − 2y 0 − 3y = 6
We try a solution of the form yp1 = A where A is a constant. Since yp0 1 = yp001 = 0, we get upon substition −3A = 6
so yp1 = A = −2. Next we look for a particular solution yp2 to the equation:
y 00 − 2y 0 − 3y = −2e−x
We try a solution of the form yp2 = Bxe−x since e−x is already a solution of the homogeneous equation. We have
yp0 2 = B(1 − x)e−x
and yp002 = B(x − 2)e−x .
Substitution into the equation gives
B(x − 2)e−x − 2B(1 − x)e−x − 3Bxe−x = −2e−x
−4Bxe−x = −2e−x .
So B = 1/2 and yp2 = 21 xe−x and so
1
yp = yp1 + yp2 = −2 + xe−x .
2
Thus the general solution is:
1
y = yp + yc = −2 + xe−x + c1 e3x + c2 e−x .
2
2. Solve the given differential equation by the method of variation of parameters. Hint: Look for solutions of the form
f (x)
f (x)
y = u1 y1 + u2 y2 , where u01 = − y2W
and u02 = y1W
(note that W = W (y1 , y2 ) is the Wronskian).
y 00 + 4y = 4 sec 2x
First we consider the associated homogeneous equation. The auxiliary equation is m2 +4 = 0. The roots are m = ±2i
and therefore the complementary solution is yc = c1 cos 2x + c2 sin 2x. Thus y1 = cos 2x and y2 = sin 2x and the
Wronskian is
y1 y2 cos 2x
sin 2x = 2(cos2 2x + sin2 2x) = 2.
=
W = W (y1 , y2 ) = 0
y1 y20 −2 sin 2x cos 2x
Hence we look for solutions of the form y = u1 y1 + u2 y2 , where
y2 f (x)
sin 2x · 4 sec 2x
u01 = −
=−
= −2 tan 2x
W
2
y1 f (x)
cos 2x · 4 sec 2x
u02 =
=
= 2.
W
2
Integrating we obtain:
Z
Z
−2 sin 2x
u1 = −2 tan 2x dx =
dx = ln | cos 2x| + C1
cos 2x
Z
u2 = 2 dx = 2x + C2 .
Hence the general solution is
y = u1 y1 + u2 y2 = (ln | cos 2x| + C1 ) cos 2x + (2x + C2 ) sin 2x.