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Math - College, section 1, Fall 2019
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WebAssign - Diff Eq Tutorial Bank 1/e (Homework)

James Finch

Math - College, section 1, Fall 2019

Instructor: Dr. Friendly

Current Score : 1 / 12

Due : Sunday, January 27, 2030 00:00 EST

Last Saved : n/a Saving... ()

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Question

Points

1	2	3	4
0/1	1/2	–/8	–/1

Total

1/12 (8.3%)

Instructions

Enrich your course with the WebAssign Differential Equations Tutorial Bank. Authored by the WebAssign Community of Teachers, this collection features more than 90 tutorial questions that guide students to a deeper understanding of the skills and concepts. Every question features an algorithmically stepped-out solution for further student support. Additionally, video explanations available at the question level present a variety of learning avenues for tackling tough concepts. Ideal as supplemental problems for your assignments, or as extra practice for students, this free Additional Resource can be added to any WebAssign course.

This sample assignment covers the first few chapters of an introductory course in differential equations.

Question 1 asks the student to solve a first-order linear initial value problem. The tutorial walks the student through the general solution using an integrating factor and finding the particular solution for the given initial condition.

Question 2 covers a first-order equation where the student must first determine that the differential equation is exact, and then find the implicit solution that satisfies the given initial condition. The tutorial guides students through testing the equation for exactness and utilizing the potential function to solve for the implicit solution.

Question 3 provides a second-order homogeneous equation along with two functions, and asks the student to compute the Wronskian and solve the initial value problem. The tutorial takes students through calculating the Wronskian to determine the given functions are linearly independent solutions, and then constructing the particular solution using this information.

Question 4 is an applied problem for a spring-mass system with external forcing. The question asks the student to derive and solve a second-order differential equation that models the system. The tutorial walks the student through the derivation of the non-homogeneous equation and solution using the method of undetermined coefficients. This demo assignment allows many submissions and allows you to try another version of the same question for practice wherever the problem has randomized values.

The answer key and solutions will display after the first submission for demonstration purposes. Instructors can configure these to display after the due date or after a specified number of submissions.

Assignment Submission

For this assignment, you submit answers by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer.

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1. 0/1 points | Previous Answers WADiffEQTutBank1 2.5.004.Tut. My Notes

Solve the following initial value problem.

xy' = x⁻² −

y(1) = 0

y(x) =

x−3−4xx2

$webMathematica generated answer key$

Tutorial

Solution or Explanation

Begin by simplifying the equation to get it in the general form

y' + p(x)y = q(x).

y' + 4x⁻²y = x⁻³

Note that

p(x) = 4x⁻²

and

q(x) = x⁻³.

Since this first-order linear equation is now standard form, we can now compute the integrating factor μ(x) = e∫p(x)dx (note that we disregard the constant of integration).

μ(x) = e∫4x⁻² dx

= e^−4/x

Multiply this function by both sides of the differential equation.

e^−4/xy' + e^−4/x4x⁻²y = e^−4/xx⁻³

Notice that the left-hand side of the equation is the result of the product rule. Rewrite it as the derivative of the appropriate product.

e^−4/xy

e^−4/xx⁻³

e^−4/xy

	e^−4/xx⁻³

Now we have the following.

e^−4/xy =

e^−4/xx⁻³ dx

Compute the integral on the right-hand side of the equation by making a substitution of

u	=	−4x⁻¹
du	=	4x⁻² dx

to get the following.

e^−4/xy = −

ue^u du

Now, for the integral

ue^u du,

we use integration by parts,

wv' = wv −

w'v,

where

v = e^u

and

w = u.

This gives the following.

e^−4/xy

	e^−4/xx⁻³

−

ue^u du

−

ue^u −

e^u du

Perform the integration, put back in terms of x.

e^−4/xy

−

ue^u −

e^u du

−

ue^u − e^u + C

−

e^−4/x − e^−4/x + C

Now solve for the general solution y.

y =

+ Ce^4/x

Finally, use the initial condition,

y(1) = 0,

to determine the constant of integration.

+ Ce⁴

−5

16e⁴

The particular solution that solves the given initial value problem is therefore as follows.

y =

−

16e⁴

e^4/x

This is a first-order, linear differential equation. What methods can be used to solve a linear differential equation of the form

y' + p(x)y = q(x)?

In the given problem, identify the corresponding

p(x)

and

q(x).

How is the initial value used to obtain a particular solution?

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2. 1/2 points | Previous Answers WADiffEQTutBank1 2.6.001.Tut. My Notes

Consider the following initial value problem.

2xe^y + (x²e^y − sin(y))

y(0) = π/3

(a)

Determine if the differential equation is exact.

exact not exact

(b)

Solve for the implicit solution to the initial value problem.

x

$webMathematica generated answer key$

Tutorial

Solution or Explanation

(a)

Recall that a differential equation of the form

M(x, y) dx + N(x, y) dy = 0,

where M and N have continuous first partial derivatives in some rectangular domain D, is exact if

∂M

∂y

∂N

∂x

for all

(x, y)

in D.

To determine whether or not this equation is exact we will identify the functions

M(x, y)

and

N(x, y),

then differentiate them with respect to y and x respectively.

M(x, y)

2xe^y

N(x, y)

x²e^y − sin(y)

∂M

∂y

2xe^y

∂N

∂x

2xe^y

Since

∂M

∂y

∂N

∂x

for all

(x, y),

the given differential equation is exact.

(b)

Now solve the differential equation by first defining the potential function

Φ(x, y),

where

dΦ(x, y) = M(x, y) dx + N(x, y) dy.

Since

dΦ(x, y) = M(x, y) dx + N(x, y) dy = 2xe^ydx + (x²e^y − sin(y)) dy = 0,

then

Φ(x, y) = c,

where c is an arbitrary constant, is an implicit solution to the differential equation and satisfies

∂Φ

∂x

= M(x, y)

and

∂Φ

∂y

(x, y) = N(x, y).

We can determine

Φ(x, y),

by integrating M with respect to x or N with respect to y.

We choose to integrate M with respect to x.

Φ(x, y) =

M(x, y) dx = x²e^y + g(y)

To find

g(y)

we differentiate this result.

∂Φ

∂y

(x, y) =

∂

∂y

(x²e^y + g(y)) = x²e^y + g'(y)

Also note that

∂Φ

∂y

(x, y) = N(x, y) = x²e^y − sin(y)

so we equate terms to get the following.

g'(y) = −sin(y) right double arrow implies

g(y) = cos(y) + c₀

Therefore

Φ(x, y) = x²e^y + cos(y) + c₀.

Our general solution to the given exact differential equation is the following.

Φ(x, y) = x²e^y + cos(y) + c₀ = c₁

We can combine

c₀

and

c₁

to write the solution as the following.

x²e^y + cos(y) = c

Solve for the integration constant using the initial condition,

y(0) = π/3.

Φ(0, π/3) = cos(π/3) = 1/2 right double arrow implies

c = 1/2

Thus, our full solution is defined implicitly to be

x²e^y + cos(y) = 1/2.

A differential equation of the form

M(x, y) dx + N(x, y) dy = 0

where M and N have continuous first partial derivatives in some rectangular domain D is exact if

∂M

∂y

∂N

∂x

for all

(x, y)

in D.

A solution for an exact differential equation involves the potential function

Φ(x, y) = c,

whose total differential is

dΦ(x, y) = M(x, y) dx + N(x, y) dy

for all

(x, y)

in D. What operations must be performed on M and N to determine

Φ(x, y)?

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3. –/8 points WADiffEQTutBank1 3.2.003.Tut. My Notes

Consider the initial value problem.

y'' − 7y' + 6y = 0 y(0) = 0 y'(0) = −5

Also consider the following functions.

y₁(x) = e^x y₂(x) = e^6x

(a)

Show that

y₁(x)

and

y₂(x)

are solutions to the differential equation.

Start with

y₁(x) = e^x.

Compute

y₁'(x) =

n

and

y₁''(x) =

Substitute these values into the given equation to get the following

e^x − 7

+ 6e^x = 0.

Now use

y₂(x) = e^6x

to compute

y₂'(x) =

and

y₂''(x) =

Substitute these values into the given equation to get the following

36e^6x − 7

+ 6e^6x = 0.

(b)

Compute the Wronskian,

W(y₁, y₂).

W(y₁, y₂) =

(c)

Find the solution satisfying the given initial values.

y(x) =

Tutorial

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4. –/1 points WADiffEQTutBank1 3.10.001b.Tut. My Notes

Consider the spring-mass system with external forcing.

Suppose the object attached to the spring has mass of 6 kg and the spring exerts a linear resistive force with spring constant

k = 7 N/m,

and is undamped. The system is subjected to a time-varying periodic force,

f(t) = 5 cos(t).

Derive and solve a differential equation whose solution represents the displacement of the spring from equilibrium at time

t > 0.

x(t) =

Tutorial

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