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Solutions to Review for Exam III
MATH 2306 (Ritter)
Sections Covered: 8, 9
This review is provided as a courtesy to give some idea of what material is covered. Nothing
else is intended or implied.
(1) Find the general solution of the homogeneous equation.
(a) y 00 −2y 0 +5y = 0
y = c1 ex cos(2x)+c2 ex sin(2x)
(b) y 00 +6y 0 +9y = 0
y = c2 e−3x +c2 xe−3x
(c) y 00 −36y = 0
y = c1 e6x +c2 e−6x
(d) y (4) +3y 00 −4y = 0
y = c1 cos(2x)+c2 sin(2x)+c3 ex +c4 e−x
(e) y 000 +2y 00 +y 0 = 0
y = c1 +c2 e−x +c3 xe−x
(f) 2y 00 −3y 0 −2y = 0
y = c1 e−x/2 +c2 e2x
(2) Solve each IVP
(a) y 00 −3y 0 +2y = 0 y(0) = 0,
(b)
y 00 +2y 0 = 0 y(1) = 0,
y 0 (0) = 2
y 0 (1) = 1
(c) y 00 −2y 0 +5y = 0 y(0) = 0,
y 0 (0) = 2
y = 2e2x −2ex
1 e2
y = − e−2x
2 2
y = ex sin(2x)
(4) Find the general solution of each nonhomogeneous equation
(a) y 00 +6y 0 +9y = ex +3e−3x
(b) y 00 +y 0 −12y = 2x
y = c1 e−3x +c2 xe−3x +
1
1
y = c1 e−4x +c2 e3x − x−
6
72
1 x 3 2 −3x
e + xe
16
2
(c) y 00 +y = 4 cos x
y = c1 cos x+c2 sin x+2x sin x
(5) Determine the form of the particular solution. (Do not bother trying to find any of the
coefficients A, B, etc.)
(a) y 00 −4y 0 +5y = x cos 2x
(b) y 00 +y = x3 +ex
yp = (Ax+B) cos(2x)+(Cx+D) sin(2x)
yp = Ax3 +Bx2 +Cx+D+Eex
(c) y 00 −4y 0 +5y = xe2x sin x
(d)
y 00 −2y 0 +y = 1+ex
yp = (Ax2 +Bx)e2x sin x+(Cx2 +Dx)e2x cos x
yp = A+Bx2 ex
(6) For each homogeneous equation, write out the characteristic equation. If the equation
doesn’t have a characteristic equation, briefly state why.
(a)
3
d3 y dy
d4 y
−2
+ −4y = 0
dx4
dx3 dx
(b) 4y 00 +2xy 0 +ex y = 0
3m4 −2m3 +m−4 = 0
none exists, it’s not constant coefficient
(c) x3 y 000 +2x2 y 00 −4xy 0 +y = 0
(d) y (6) +16y (4) −12y 00 +y = 0
none exists, it’s not constant coefficient
m6 +16m4 −12m2 +1 = 0
(7) For each of the following nonhomogeneous equations, determine whether the method of
undetermined coefficients could be used to determine yp . If not, give a brief explanation.
d4 y
d3 y dy
−2
+ −4y = x3 ex
yes, it could
dx4
dx3 dx
1
(b) 4y 00 +2y 0 +y =
nope, RHS is not of the correct type
1 + x2
(a) 3
(c) x3 y 000 +2x2 y 00 −4xy 0 +y = sin(2x)+x
(d) y (6) +16y (4) −12y 00 +y = x ln x
nope, LHS is not constant coefficient
nope, RHS is not of the correct type