We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Differential Equations : Linear & Exact Equations

Study concepts, example questions & explanations for Differential Equations

Create An Account

All Differential Equations Resources

1 Diagnostic Test 29 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : First Order Differential Equations

Find the general solution of the given differential equation and determine if there are any transient terms in the general solution.

$x\frac{dy}{dx}+3y=x^4-x$

Possible Answers:

$\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

$\\y=\frac{1}{7}x^4-\frac{1}{4}x^3+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

$\\y=\frac{1}{7}x^4-\frac{1}{4}x^2+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

$\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{3}$

$\\y=\frac{1}{7}x^3-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

Correct answer:

$\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

Explanation:

First, divide by on both sides of the equation.

$\\\frac{dy}{dx}+\frac{3y}{x}=\frac{x^4}{x}-\frac{x}{x} \\\\\frac{dy}{dx}+\frac{3y}{x}=x^3-1$

Identify the factor p(x) term.

$p(x)=\frac{3}{x}$

Integrate the factor.

$e^{\int p(x)}=e^{\frac{3}{x}}=e^{3\ln x}=e^{\ln x^3}=x^3$

Substitute this value back in and integrate the equation.

$\\\int \frac{d}{dx}[x^3y]=\int x^3(x^3-1)dx \\\\x^3y=\int x^6-x^3dx \\\\x^3y=\frac{x^7}{7}-\frac{x^4}{4}+c$

Now divide by x^3 to get the general solution.

$\\y=\frac{1}{7}\frac{x^7}{x^3}-\frac{1}{4}\frac{x^4}{x^3}+\frac{c}{x^3} \\\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3}$

The transient term means a term that when the values get larger the term itself gets smaller. Therefore the transient term for this function is $cx^{-3}$ .

Report an Error

Example Question #1 : Linear & Exact Equations

Find the solution for the following differential equation:

y' = 3t^2y + t^2 where y(0) = 0 .

Possible Answers:

$y = -\frac{1}{3}$

$y = \frac{e^{t^3} - 1}{3}$

$y = \frac{e^{t^3}}{3}$

$y = e^{t^3}-\frac{1}{3}$

y = -t^3

Correct answer:

$y = \frac{e^{t^3} - 1}{3}$

Explanation:

This equation can be put into the form y' + p(t)y = q(t) as follows:

y' - 3t^2y = t^2 . Differential equations in this form can be solved by use of integrating factor. To solve, take p(t) = -3t^2 and solve for $\mu = e^{\int p(t)dt} = e^{\int -3t^2dt} = e^{-t^3}$

Note, when using integrating factors, the +C constant is irrelevant as we only need one solution, not infinitely many. Thus, we have set C to 0.

Next, note that $(\mu y)' = y'\mu + \mu'y = e^{-t^3}(y' - 3t^2y) = \mu q(t) = e^{-t^3}t^2$

Or more simply, $(\mu y)' = t^2e^{-t^3}$ . Integrating both sides using substitution of variables we find

$\mu y = \int t^2e^{-t^3}dt = -\frac{e^{-t^3}}{3} + C$

Finally dividing by $\mu = e^{-t^3}$ , we see

$y = -\frac{1}{3} + Ce^{t^3}$ . Plugging in our initial condition,

$0 = -\frac{1}{3} + C$

So $C = \frac{1}{3}$

And $y = \frac{e^{t^3} - 1}{3}$ .

Report an Error

Example Question #11 : First Order Differential Equations

Consider the differential equation

y'+y^2=t^4e^t

Which of the terms in the differential equation make the equation nonlinear?

Possible Answers:

The e^t term makes the differential equation nonlinear.

The term makes the differential equation nonlinear

The y^2 term makes the differential equation nonlinear.

The t^4 term makes the differential equation nonlinear

Correct answer:

The y^2 term makes the differential equation nonlinear.

Explanation:

The term y^2 makes the differential equation nonlinear because a linear equation has the form of

y'+f(t)y=g(t)

Report an Error

Example Question #1 : Linear & Exact Equations

Find the general solution of the given differential equation and determine if there are any transient terms in the general solution.

$x\frac{dy}{dx}+3y=x^4-x$

Possible Answers:

$\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{3}$

$\\y=\frac{1}{7}x^4-\frac{1}{4}x^2+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

$\\y=\frac{1}{7}x^3-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

$\\y=\frac{1}{7}x^4-\frac{1}{4}x^3+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

$\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

Correct answer:

$\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3} \\\\\text{Transient Term}=cx^{-3}$

Explanation:

First, divide by on both sides of the equation.

$\\\frac{dy}{dx}+\frac{3y}{x}=\frac{x^4}{x}-\frac{x}{x} \\\\\frac{dy}{dx}+\frac{3y}{x}=x^3-1$

Identify the factor p(x) term.

$p(x)=\frac{3}{x}$

Integrate the factor.

$e^{\int p(x)}=e^{\frac{3}{x}}=e^{3\ln x}=e^{\ln x^3}=x^3$

Substitute this value back in and integrate the equation.

$\\\int \frac{d}{dx}[x^3y]=\int x^3(x^3-1)dx \\\\x^3y=\int x^6-x^3dx \\\\x^3y=\frac{x^7}{7}-\frac{x^4}{4}+c$

Now divide by x^3 to get the general solution.

$\\y=\frac{1}{7}\frac{x^7}{x^3}-\frac{1}{4}\frac{x^4}{x^3}+\frac{c}{x^3} \\\\y=\frac{1}{7}x^4-\frac{1}{4}x+cx^{-3}$

The transient term means a term that when the values get larger the term itself gets smaller. Therefore the transient term for this function is $cx^{-3}$

Report an Error

Example Question #2 : Linear & Exact Equations

Is the following differential equation exact? $2xy^2 + y^2 + 1 + (2x^2y + 2xy + 3)\frac{\mathrm{d} y}{\mathrm{d} x} = 0$

If so, find the general solution.

Possible Answers:

Yes. 2x^2y + 2xy^2 + y = c

Yes. x^2y + xy^2 + 3x + y = c

No. The partial derivatives in the equation do not have the correct relationship.

Yes. x^2y^2 + xy^2 + x + 3y = c

No. The equation does not take the proper form.

Correct answer:

Yes. x^2y^2 + xy^2 + x + 3y = c

Explanation:

For a differential equation to be exact, two things must be true. First, it must take the form $M(x,y) + N(x,y)\frac{\mathrm{d} y}{\mathrm{d} x} = 0$ . In our case, this is true, with M = 2xy^2 + y^2 + 1 and N = 2x^2y + 2xy + 3 . The second condition is that M_y = N_x . Taking the partial derivatives, we find that M_y =4xy + 2y and N_x = 4xy + 2y . As these are equal, we have an exact equation.

Next we find a $\Psi$ such that $\Psi_x = M$ and $\Psi_y = N$ . To do this, we can integrate with respect to or we can integrate with respect to Here, we choose arbitrarily to integrate .

$\Psi = \int \Psi_x dx = \int Mdx = \int(2xy^2 + y^2 + 1)dx =x^2y^2 + xy^2 + x + h(y)$

We aren't quite done yet, because when taking a multivariate integral, the constant of integration can now be a function of y instead of just a constant. However, we know that $\Psi_y = N$ , so taking the partial derivative, we find that 2x^2y + 2xy + h'(y) = 2x^2y + 2xy + 3 and thus that h'(y) = 3 and h(y)=3y .

We now know that $\Psi = x^2y^2 + xy^2 + x + 3y$ , and the point of finding psi was so that we could rewrite $M(x,y) + N(x,y)\frac{\mathrm{d} y}{\mathrm{d} x} = \Psi_x + \Psi_y\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} \Psi}{\mathrm{d} x} = 0$ , and because the derivative of psi is 0, we know it must have been a constant. Thus, our final answer is

$\Psi = x^2y^2 + xy^2 + x + 3y = c$ .

If you have an initial value, you can solve for c and have an implicit solution.

Report an Error

Example Question #3 : Linear & Exact Equations

Is the following differential equation exact? $2x^2y + y + (x^2y + xy)\frac{\mathrm{d} y}{\mathrm{d} x} = 0$ If so, find the general solution.

Possible Answers:

Yes. 4x^2y + 2xy = c

Yes. x^2y + xy^2 + x = c

No. The equation does not take the right form.

Yes. 4x^2y + 2xy^2 + y = c

No. The partial derivatives in the equation do not have the right relationship.

Correct answer:

No. The partial derivatives in the equation do not have the right relationship.

Explanation:

For a differential equation to be exact, two things must be true. First, it must take the form $M(x,y) + N(x,y)\frac{\mathrm{d} y}{\mathrm{d} x} = 0$ . In our case, this is true, with M = 2xy^2 + y and N = x^2y + xy . The second condition is that M_y = N_x . Taking the partial derivatives, we find that M_y =4xy + 1 and N_x = 2xy + y . As these are unequal, we do not have an exact equation.

Report an Error

Example Question #4 : Linear & Exact Equations

Solve the Following Equation

$y' - \frac{y}{x} = xe^x \ \ \ y(1) = e-1$

Possible Answers:

y = e^x - x

y = e^x + x

y = xe^x + x

y = xe^x - x

y = xe^x - 1

Correct answer:

y = xe^x - x

Explanation:

Since this is in the form of a linear equation

y' + P(x)y = Q(x)

we calculate the integration factor

$I(x) = e^{\int P(x)} = e^{\int-\frac{1}{x} \ dx} = e^{-\ln x} = \frac{1}{x}$

Multiplying by I(x) we get

$(y \times \frac{1}{x})' = e^x$

Integrating

$\frac{y}{x} = e^x + C$

y = xe^x + Cx

Plugging in the Initial Condition to solve for the Constant we get

$e - 1 = e + C \implies C = -1$

Our solution is

y = xe^x - x

Report an Error

Example Question #5 : Linear & Exact Equations

Find the general solution of the differential equation

$\frac{dy}{dx} - 5y = -\frac{5}{2}xy^3$

Possible Answers:

$y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)$

$y = \left( \frac{x}{2} + Ce^{-10x} \right)^{-1/2}$

$y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)^{-1/2}$

$y = \left( x - \frac{1}{10} + Ce^{-10x} \right)^{-1/2}$

None of the other answers

Correct answer:

$y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)^{-1/2}$

Explanation:

This is a Bernoulli Equation of the form

$\frac{dy}{dx} + P(x) = Q(x)y^n$

which requires a substitution

$v = y^{1-n}$

to transform it into a linear equation

Rearranging our equation gives us

$y^{-3} \frac{dy}{dx} - 5y^{-2} = -\frac{5}{2}x$

Substituting $v = y^{-2}$

$-\frac{1}{2} \frac{dv}{dx} - 5v = -\frac{5}{2}x$

$\frac{dv}{dx} + 10v = 5x$

Solving the linear ODE gives us

$v = \frac{x}{2} - \frac{1}{20} + Ce^{-10x}$

Substituting in $v = y^{-2}$ and solving for

$y = \left( \frac{x}{2} - \frac{1}{20} + Ce^{-10x} \right)^{-1/2}$

Report an Error

Example Question #6 : Linear & Exact Equations

Solve the differential equation

$\frac{dy}{dx} = -\frac{2xy^2+1}{2x^2y}$

Possible Answers:

xy^2+x = C

x^2y^2+x = C

x^2y+x = C

xy+x = C

None of the other answers

Correct answer:

x^2y^2+x = C

Explanation:

Rearranging the following equation

$(2xy^2 + 1)\ dx +2x^2y \ dy = 0$

This satisfies the test of exactness, so integrating we have

x^2y^2+x = C

Report an Error

View Differential Equations Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

View Differential Equations Tutors

Omar
Certified Tutor

Arizona State University, Bachelors, Civil Engineering.

View Differential Equations Tutors

Yoonsik
Certified Tutor

Seoul National University, Bachelor of Science, Physics. University of Pennsylvania, Doctor of Philosophy, Atomic and Molecul...

All Differential Equations Resources

1 Diagnostic Test 29 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

French Tutors in Houston, MCAT Tutors in San Diego, Spanish Tutors in Washington DC, Reading Tutors in Washington DC, Chemistry Tutors in San Diego, Computer Science Tutors in Phoenix, MCAT Tutors in Houston, Reading Tutors in New York City, Reading Tutors in Houston, Physics Tutors in Phoenix

Popular Courses & Classes

ISEE Courses & Classes in Boston, ISEE Courses & Classes in New York City, LSAT Courses & Classes in Chicago, SSAT Courses & Classes in Los Angeles, ACT Courses & Classes in San Diego, MCAT Courses & Classes in Seattle, MCAT Courses & Classes in Boston, GMAT Courses & Classes in Denver, GMAT Courses & Classes in Chicago, LSAT Courses & Classes in Houston

Popular Test Prep

SAT Test Prep in Seattle, GRE Test Prep in New York City, GMAT Test Prep in Atlanta, MCAT Test Prep in Miami, LSAT Test Prep in Dallas Fort Worth, ISEE Test Prep in Houston, LSAT Test Prep in San Diego, ACT Test Prep in San Diego, LSAT Test Prep in Miami, ACT Test Prep in Chicago

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Differential Equations : Linear & Exact Equations

Study concepts, example questions & explanations for Differential Equations

All Differential Equations Resources

Example Questions

Example Question #1 : First Order Differential Equations

Example Question #1 : Linear & Exact Equations

Example Question #11 : First Order Differential Equations

Example Question #1 : Linear & Exact Equations

Example Question #2 : Linear & Exact Equations

Example Question #3 : Linear & Exact Equations

Example Question #4 : Linear & Exact Equations

Example Question #5 : Linear & Exact Equations

Example Question #6 : Linear & Exact Equations

All Differential Equations Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: