# MATH 262 Ordinary Differential Equation with Constant Coefficients Exam Practice

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MATH 262
Problem Set # 3
Assigned: 11 October 2020
Due: 18 October 2020
This assignment explores themes around the topic of constant coeﬃcient second-order equations. The problem set explores second-order and higher-order linear ODEs. Both homogeneous and non-homogeneous cases are considered. You may wish to use the Mathematica
command DSolve on the linear second-order ODEs below to gain further experience with
the software.
1. Solve the following diﬀerential equations for y(x).
(a) y ′′ + 6y ′ + 13y = 0,
y(0) = 1, y ′ (0) = 1.
(b) y ′′ + 2y ′ + 3y = 0
y(0) = 1, y ′ (0) = 0.
(c) y ′′ + 8y ′ + 16y = 0,
y(0) = 0, y ′ (0) = 1. At what x is the function a maximum?
2. Use the method of undetermined coeﬃcients to solve the following ODEs for y(x).
(a) y ′′ − 9y = x2 + 2ex ,
y(0) = 0, y ′ (0) = 1 .
Hint: Using the concept of linearity you need to ﬁnd a distinct particular solution
for each of the terms on the right-hand side.
(b) y ′′ − y = 2e−x .
3. Generalize what you know about constant coeﬃcient equations to ﬁnd the general
solution of the higher-order equations for the function y(x):
(a) Solve y ′′′ − y = 0 (the cubic equation that results is not diﬃcult to factor).
(b) Find the general solution of y IV − 2y ′′ − 8y = 0
(c) Give the general solution of the 4th-order equation y IV − 16y = e−2x .
4. Reduction of order: Given one solution y1 (x) to a linear second-order ODE you can
obtain a second linearly independent solution by writing y(x) = y1 (x)f (x) and ﬁnding
f (x). This approach is eﬀectively the procedure introduced in lecture called variation
of parameters when the equation is non-homogeneous. Traditionally, it is known as
reduction of order for the homogeneous problem, because the substitution reduces the
df
).
second-order ODE to a ﬁrst-order ODE (for dx
Use this idea to obtain the solution to
x(x + 1)y ′′ + (x − 1)y ′ − y = 0 ,
1
y(1) = 0 , y ′ (1) = 1.
(1)
(a) Verify by direct substitution into the equation that one solution to the ODE is
y1 (x) = 1/(1 + x).
(b) Find a second solution to the ODE using the method of reduction of order outlined
above.
(c) Give the general form of the solution and then obtain the constants by satisfying
the initial conditions.
(d) Follow the same procedure (since the intermediate steps are the same so you
can just start with an intermediate step) to obtain the general solution to the
non-homogeneous equation
x(x + 1)y ′′ + (x − 1)y ′ − y = x1/2 .
2
(2)

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Tags:

constant coefficients

Ordinary Differential Equation

coefficient equations

non homogeneous cases

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