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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).

Show all of your work.

(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:

Quadratic Formula

constant coefficients

Ordinary Differential Equation

coefficient equations

non homogeneous cases

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