Sample Solutions for Homework 24–26
Winfried Just
Department of Mathematics, Ohio University
February 17, 2017
Winfried Just, Ohio University
MATH3400, Solutions for Homework 24–26
Sample Solution for Homework 24
Consider the DE y 00 + 3y 0 + 2y = e −x + 1.
The complementary solution is yc (x) = c1 e −x + c2 e −2x .
The tentative form of yp (x) is yptentative (x) = Ae −x + B.
Multiplying the term that appears in yc (x) with x k = x we get
yp = Axe −x + B. Differentiate and substitute in the DE:
yp0 = −Axe −x + Ae −x , yp00 = Axe −x − 2Ae −x .
Axe −x − 2Ae −x + 3(−Axe −x + Ae −x ) + 2(Axe −x + B) = e −x + 1.
Match coefficients: (1 − 3 + 2)A = 0, (−2 + 3)A = 1, 2B = 1.
Solve: A = 1, B = 0.5.
The general solution is:
y (x) = c1 e −x + c2 e −2x + xe −x + 0.5.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Solutions for Homework 24–26
Sample Solution for Homework 25
Consider the DE y 00 + 4y 0 + 4y = e −2x .
The complementary solution is yc (x) = c1 e −2x + c2 xe −2x .
Tentatively set yptentative (x) = Ae −2x .
Multiplying the term that appears in yc (x) with x k = x 2 we get
yp (x) = Ax 2 e −2x . Differentiate and substitute in the DE:
yp0 = −2Ax 2 e −2x + 2Axe −2x , yp00 = 4Ax 2 e −2x − 8Axe −2x + 2Ae −2x .
(4Ax 2 −8Ax + 2A)e −x + 4(−2Ax 2 + 2Ax)e −2x + 4Ax 2 e −2x = e −2x .
Match coefficients: A(4 − 8 + 4) = 0, (−8 + 8)A = 0, 2A = 1.
Solve: A = 0.5.
The general solution is:
y (x) = c1 e −2x + c2 xe −2x + 0.5e −2x .
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Solutions for Homework 24–26
Sample Solutions for Homework 26(a)
For each of the following DEs find an appropriate form of yp (x):
(a) y (5) − 3y (4) + 3y 000 − y 00 = 7e x + 7 − 7xe x .
The characteristic polynomial
m5 − 3m4 + 3m3 − m2 = m2 (m − 1) has roots
m1 = 0 of multiplicity k1 = 2 and
m2 = 1 of multiplicity k2 = 3. Thus
yc = c1 + c2 x + c3 e x + c3 xe x + c3 x 2 e x
We set yptentative = A + Be x + Cxe x
Multiply the blue term with x k1 = x 2 and
the red terms with x k2 = x 3 :
yp = Ax 2 + Bx 3 e x + Cx 4 e x
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Solutions for Homework 24–26
Sample Solutions for Homework 26(b)
For each of the following DEs find an appropriate form of yp (x):
(b) y (4) + 2y 00 + y =
7
ex
+ 8 cos x − 9e x sin x.
The characteristic polynomial
m4 + 2m2 + 1 = (m2 + 1)2 has roots
m1 = i, m2 = −i of multiplicity k = 2 each. Thus
yc = c1 sin x + c2 cos x + c3 x sin x + c4 x cos x.
We set
yptentative = Ae x + B sin x + C cos x + De x sin x + Ee x cos x.
Multiply the blue terms with x k = x 2 :
yp = Ae x + Bx 2 sin x + Cx 2 cos x + De x sin x + Ee x cos x.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Solutions for Homework 24–26
Sample Solutions for Homework 26(c)
For each of the following DEs find an appropriate form of yp (x):
(c) y 00 + y = sin x + 3 cos(−x) + 6x sin 6x.
The characteristic polynomial
m2 + 1 has roots
m1 = i, m2 = −i of multiplicity k = 1. Thus
yc = c1 sin x + c2 cos x.
Note that cos(−x) = x. We set yptentative equal to:
A sin x + B cos x + C sin 6x + D sin 6x + Ex sin 6x + Fx cos 6x.
Multiply the blue terms with x k = x:
yp = Ax sin x + Bx cos x + C sin 6x + D sin 6x + Ex sin 6x + Fx cos 6x.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Solutions for Homework 24–26