Math 3400
Homework 5
Section 3.6
2. Solve by using variation of parameters and the method of undetermined coefficients.
(N H) y 00 − y 0 − 2y = 2e−t .
The associated homogeneous equation is
(H) y 00 − y 0 − 2y = 0.
The characteristic equation is
r2 − r − 2 = (r − 2)(r + 1) = 0,
r = 2, −1.
Take
y2 (t) = e−t .
y1 (t) = e2t ,
The Wronskian for these solutions is
W (y1 , y2 ) = y1 y20 − y10 y2 = −et − 2et = −3et .
The variation of parameters formula give a particular solution
Z t
Z t
y2 (s)g(s)
y1 (s)g(s)
yp (t) = −y1 (t)
ds + y2 (t)
ds.
W (y1 , y2 )(s)
W (y1 , y2 )(s)
In this case we get
2t
yp (t) = −e
2
= e2t
3
Z
t
e−3s
t
Z t 2s −s
2e−s e−s
2e e
−t
ds + e
ds
s
−3e
−3es
Z
2 −t t
2
2
ds − e
1 ds = − e−t − te−t .
3
9
3
Z
Since e−t is a solution of (H), another particular solution is
2
yp (t) = − te−t .
3
The method of undetermined coefficients leads to the third order homogeneous equation
(D + 1)(D − 2)(D + 1)y = 0.
The root r = −1 is repeated, so
y(t) = c1 e2t + c2 e−t + c3 te−t .
Putting this form into (NH) leads to
[−2c3 e−t + c3 te−t ] − [c3 e−t − c3 te−t ] − 2c3 te−t = 2e−t ,
1
or
2
c3 = − .
3
3. Solve by using variation of parameters and the method of undetermined coefficients.
(N H)
y 00 + 2y 0 + y = 3e−t .
The associated homogeneous equation is
(H) y 00 + 2y 0 + y = 0.
The characteristic equation is
r2 + 2r + 1 = (r + 1)2 = 0,
r = 2, −1.
Take
y1 (t) = e−t ,
y2 (t) = te−t .
The Wronskian for these solutions is
W (y1 , y2 ) = y1 y20 − y10 y2 = e−t [e−t − te−t ] + te−t e−t = e−2t .
The variation of parameters formula give a particular solution
Z t
Z t
y2 (s)g(s)
y1 (s)g(s)
yp (t) = −y1 (t)
ds + y2 (t)
ds.
W (y1 , y2 )(s)
W (y1 , y2 )(s)
In this case we get
t
Z t −s −s
3se−s e−s
3e e
−t
yp (t) = −e
ds + te
ds
−2s
e
e−2s
Z t
Z t
3
3
−t
−t
= −3e
s ds + 3te
1 ds = − te−t + 3t2 e−t = t2 e−t .
2
2
The method of undetermined coefficients leads to the third order homogeneous equation
−t
Z
(D + 1)3 y = 0.
The root r = −1 has multiplicity 3, so
y(t) = c1 e−t + c2 te−t + c3 t2 e−t .
Putting this form into (NH) leads to
3
c3 = .
2
4.
2
5. Find the general solution
(N H) y 00 + y = tan(t).
The associated homogeneous equation is
(H) y 00 + y = 0.
The homogeneous equation has a basis of solutions
y1 (t) = cos(t),
y2 (t) = sin(t),
with Wronskian
W (y1 , y2 ) = y1 y20 − y10 y2 = cos2 (t) + sin2 (t) = 1.
The variation of parameters formula give a particular solution
Z t
Z t
y2 (s)g(s)
y1 (s)g(s)
yp (t) = −y1 (t)
ds + y2 (t)
ds.
W (y1 , y2 )(s)
W (y1 , y2 )(s)
In this case we get
Z
t
yp (t) = − cos(t)
Z
t
sin(s) tan(s) ds + sin(t)
cos(s) tan(s) ds
t
Z t
sin2 (s)
= − cos(t)
sin(s) ds.
ds + sin(t)
cos(s)
Consulting a table of integrals leads to the general solution
Z
y(t) = c1 cos(t) + c2 sin(t) − cos(t) log(tan(t) + sec(t)).
14.Solve
t2 y 00 − t(t + 2)y 0 + (t + 2)y = 2t3 .
This is not in standard form. Rewrite it as
(t + 2) 0 (t + 2)
y 00 −
y +
y = 2t.
t
t2
We’re given the solutions of the homogeneous equation
y1 = t,
y2 = tet ,
with Wronskian
W (y1 , y2 ) = y1 y20 − y10 y2 = t[et + tet ] − tet = t2 et .
The variation of parameters formula gives
Z t
Z
y2 (s)g(s)
ds + y2 (t)
yp (t) = −y1 (t)
W (y1 , y2 )(s)
t
y1 (s)g(s)
ds.
W (y1 , y2 )(s)
In this case we get
Z t 2
2s2 es
2s
t
yp (t) = −t
ds + te
ds = −2t2 − 2t.
2
s
2
s
se
se
Since t is a solution of the associated homogeneous equation, another particular solution
Z
t
is
yp (t) = −2t2 .
15.
3