Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : Differential Equations

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #71 : Differential Equations

Find the solution to the differential equation

$\frac{dy}{dx}=y\cos x$

with initial value $y(\pi/2)=2$ .

Possible Answers:

$y(x)=e^{\sin x}$

$y(x) = 2e\cdot e^{\sin x}$

$y(x) = 2e^{\sin x}$

$y(x)=\frac{2}{e}e^{\sin x}$

Correct answer:

$y(x)=\frac{2}{e}e^{\sin x}$

Explanation:

We can solve this by method of separating variables:

$\frac{dy}{dx}=y\cos x$

$\frac{1}{y}dy = \cos x dx$

So we integrate to get

$\int \frac{1}{y}dy = \int \cos x dx$

$\iff \ln y = \sin x+c$

$\iff e^{\ln y}=y=e^{\sin x+c}=e^{\sin x}e^c$

$\Rightarrow y(x)=e^{\sin x}e^c$

So now we plug in the initial value $y(\pi/2)=2$ to get the specific solution:

$\small y(\pi/2)=2=e^{\sin \pi/2}e^c$

$\small \iff 2 = e^1e^c=e^{1+c}$

So we solve for $\small c$ to get $\small \small c=\ln 2-1$ . We plug this $\small c$ into $\small y(x)$ to get

$\small y(x)=e^{\sin x}e^{\ln 2-1}$

which becomes

$\small y(x)=\frac{2}{e}e^{\sin x}$

after simplifying.

Report an Error

Example Question #315 : Equations

Find the general solution to the differential equation

$\small \frac{dy}{dt}=yt^3$ .

Possible Answers:

$\small y=e^{\frac{t^4}{4}}e^c$

$\small \small \small y=e^{\frac{t^4}{4}}+e^c$

$\small \small \small y=e^{3t^2}+e^c$

$\small \small y=e^{3t^2}e^c$

Correct answer:

$\small y=e^{\frac{t^4}{4}}e^c$

Explanation:

We can use separation of variables to solve the differential equation

$\small \frac{dy}{dt}=yt^3$

We have

$\small \frac{1}{y}dy=t^3dt$

$\small \small \int\frac{1}{y}dy=\int t^3dt$

$\small \small \implies \ln y=\frac{t^4}{4}+c$

$\small \small \small \iff e^{\ln y}=y=e^{\frac{t^4}{4}+c}=e^{\frac{t^4}{4}}e^{c}$

So the general solution is $\small y=e^{\frac{t^4}{4}}e^c$ , or $\small \small y=Ke^{\frac{t^4}{4}}$ for $\small K=e^c$ .

Report an Error

Example Question #63 : How To Find Solutions To Differential Equations

What is the solution to the differential equation

$\frac{df}{dx}=sec^3(x)tan(x)?$

Possible Answers:

f(x)=3sec^3(x)tan^2(x)+sec^5(x)+c

$f(x)=\frac{sec^3(x)}{3}+c$

$f(x)=\frac{sec^4(x)tan^2(x)}{8}+c$

$f(x)=\frac{sec^4(x)}{4}+c$

Correct answer:

$f(x)=\frac{sec^3(x)}{3}+c$

Explanation:

First, multiply both sides of the equation by dx to find

df=sec^3(x)tan(x)dx

Second, integrate both sides of the equation

$\int df=\int sec^3(x)tan(x)dx$

To integrate the right side, we can use u-substitution. Let u=sec(x). Then du=sec(x)tan(x)dx. Therefore we can rewrite the integral in terms of u as:

$\int df=\int u^2du$

Next take the antiderivative of both sides

$f=\frac{u^3}{3}+c$

Substitute sec(x) for u to find the final solution in terms of x.

$f(x)=\frac{sec^3(x)}{3}+c$

Report an Error

Example Question #72 : Differential Equations

Find the solution to the separable differential equation.

$\frac{dy}{dx}=xy+2y$

Possible Answers:

$y=\frac{{}x^{2}y^{2}}{4}+y^2+c$

$y=\frac{x^2}{2}+x+c$

$y=ln\left(\frac{x^2}{2}+x\right)+c$

$y=ce^{\frac{x^2}{2}+2x}$

Correct answer:

$y=ce^{\frac{x^2}{2}+2x}$

Explanation:

To solve this separable differential equation, we first need to rewrite it so that the left side is expressed entirely in terms of y and the right side in terms of x.

First factor the right side of the equation:

$\frac{dy}{dx}=xy+2y \rightarrow \ \frac{dy}{dx}=y(x+2)$

Next, divide both sides of the equation by y and multiply both sides of the equation by dx, which results in:

$\frac{dy}{y}=(x+2)dx$

Integrate both sides of the equation:

$\int\frac{dy}{y}=\int(x+2)dx$

$ln(y)=\frac{x^2}{2}+2x+c$

Finally, exponentiate both sides of the equation to solve for y:

$e^{ln(y)}=e^{\frac{x^2}{2}+2x+c}$

$y=e^{\frac{x^2}{2}+x+c}=e^{c}e^{\frac{x^2}{2}+2x}$

Finally, e^c is a constant, so just write it as 'c'.

$y=ce^{\frac{x^2}{2}+2x}$

Report an Error

Example Question #65 : How To Find Solutions To Differential Equations

Find the derivative of the following function:

$y=(3x+7)(4x-3)^{3}$

Possible Answers:

$\frac{dy}{dx}=(12x+28)(4x-3)^{2}+3(4x-3)^3$

$\frac{dy}{dx}=(9x+21)(4x-3)^{2}+3(4x-3)^{3}$

$\frac{dy}{dx}=(12x-28)(4x-3)^2+3(4x-3)^3$

$\frac{dy}{dx}=(36x+84)(4x-3)^{2}+3(4x-3)^{3}$

$\frac{dy}{dx}=(21x+28)+3(4x-3)^{3}$

Correct answer:

$\frac{dy}{dx}=(36x+84)(4x-3)^{2}+3(4x-3)^{3}$

Explanation:

Computation of the derivative requires use of the Product Rule and the Chain Rule.

$y=(3x+7)(4x-3)^{3}$

$\frac{dy}{dx}=\frac{d}{dx}[(3x+7)(4x-3)^3]$

A good way to remember the Product Rule is by memorizing this saying: "First Times the Derivative of the Second, Plus the Second Times the Derivative of the First." Or if that doesn't help then you can just write out the formula:

For:

y = f(x)*g(x) Where f(x) and g(x) are differentiable functions

$\frac{dy}{dx}=f(x)g'(x)+g(x)f'(x)$

As you can see, the "saying" from above matches the formula.

In this case:

f(x)=3x+7 , g(x)=(4x-3)^3

Applying the Product Rule:

$\frac{dy}{dx}=(3x+7)\frac{d}{dx}[(4x-3)^3]+(4x-3)^3\frac{d}{dx}[(3x+7)]$

To compute the derivatives of (4x-3)^3 and (3x+7) , simply apply the chain rule:

For:

y=(f(x))^a

$\frac{dy}{dx}=a*(f(x))^{a-1}*f'(x)$

Where a is an integer and f(x) is a differentiable function.

Applying the Chain Rule:

$\frac{dy}{dx}=(3x+7)[3(4x-3)^2(4)]+(4x-3)^3(3)$

Simplify the expression to match one of the answer choices:

$\frac{dy}{dx}=(36x+84)(4x-3)^2+3(4x-3)^3$

Report an Error

Example Question #66 : How To Find Solutions To Differential Equations

Find the derivative with respect to for the following function:

$y=\frac{3x^2-5}{(2x^2-7)^3}$

Possible Answers:

$\frac{dy}{dx}=\frac{-24x^3+18x}{(2x^2-7)^4}$

$\frac{12x^3-36x^2+12x}{(2x^2-7)^5}$

$\frac{12x^3-78x^2-60x}{(2x^2-7)^3}$

$\frac{12x^3-36x^2-102x}{(2x^2-7)^5}$

$\frac{24x^3+18x}{(2x-7)^4}$

Correct answer:

$\frac{dy}{dx}=\frac{-24x^3+18x}{(2x^2-7)^4}$

Explanation:

Computation of the derivative requires use of the Quotient Rule and Chain Rule.

$y=\frac{3x^2-5}{(2x^2-7)^3}$

$\frac{dy}{dx}=\frac{d}{dx}\left[\frac{3x^2-5}{(2x^2-7)^3}\right]$

A good way to memorize the Quotient Rule is by memorizing this saying: "Bottom Times the Derivative of the Top, Minus the Top Times the Derivative of the Bottom, All Over the Bottom Squared." Or if that doesn't help then you can just write out the formula:

For:

$y=\frac{f(x)}{g(x)}$

Where f(x) and g(x) are differentiable functions where g(x) Does Not Equal 0!

$\frac{dy}{dx}=\frac{g(x)*f'(x)-f(x)*g'(x)}{(g(x))^2}$

As you can see, the "saying" from above matches the formula.

In this case:

f(x)=3x^2-5 , g(x)=(2x^2-7)^3

Applying the Quotient Rule:

$\frac{dy}{dx}=\frac{(2x^2-7)^3\frac{d}{dx}[3x^2-5]-(3x^2-5)\frac{d}{dx}[(2x^2-7)^3]}{((2x^2-7)^3)^2}$

To compute the derivatives of 3x^2-5 and (2x^2-7)^3 , simply apply the chain rule:

For:

y=(f(x))^a

$\frac{dy}{dx}=a*(f(x))^{a-1}*f'(x)$

Where a is an integer and f(x) is a differentiable function.

Applying the Chain Rule:

$\frac{dy}{dx}=\frac{(2x^2-7)^3(6x)-(3x^2-5)[3(2x^2-7)^2(4x)]}{(2x^2-7)^6}$

We can simplify this equation further by factoring out a common (2x^2-7)^2 from the numerator.

$\frac{dy}{dx}=\frac{(2x^2-7)^2[6x(2x^2-7)-(3x^2-5)(12x)]}{(2x^2-7)^6}$

You will notice that the can cancel out with some terms from the denominator resulting in:

$\frac{dy}{dx}=\frac{(6x)(2x^2-7)-(3x^2-5)(12x)}{(2x^2-7)^4}$

Distributing the terms out in the numerator and simplifying the expression results in the final answer, which matches one of the answer choices:

$\frac{dy}{dx}=\frac{-24x^3+18x}{(2x^2-7)^4}$

Report an Error

Example Question #73 : Differential Equations

Find the derivative of the following function.

$f(x)=(3x^3+5x^2-7)^{^{10}}$

Possible Answers:

$(90x^2+100x)(3x^3+5x^2-7)^{9}$

$(9x^2+10x)(3x^3+5x^2-7)^{9}$

$(9x^2+10x)(300x^3+50x^2-70)^{9}$

$10(3x^3+5x^2-7)^{9}$

$(90x^2-100x)(3x^3+5x^2-7)^{9}$

Correct answer:

$(90x^2+100x)(3x^3+5x^2-7)^{9}$

Explanation:

$D_x[(3x^3+5x^2-7)^{10}]$ Finding the derivative requires use of the chain rule.

$D_x[(f(x))^a]=a*f(x)^{a-1}*f'(x)$ Where f(x) is a differentiable function and a is an integer.

From the problem statement:

f(x)=3x^3+5x^2-7

Computing the derivative:

f'(x)=9x^2+10x

Applying the Chain Rule:

$D_x[(3x^3+5x^2-7)^{10}]=10*(3x^3+5x^2-7)^{9}*(9x^2+10x)$

Simplify to match the answer choice:

$(90x^2+100x)(3x^3+5x^2-7)^{9}$

Report an Error

Example Question #68 : How To Find Solutions To Differential Equations

For:

y=sin(x^2+2)cos(x-3)

Find $\frac{dy}{dx}$ :

Possible Answers:

2xsin(x-3)cos(x^2+2)-cos(x-3)sin(x^2+2)

2xcos(x-3)cos(x^2+2)-sin(x-3)sin(x^2+2)

2xsin(x-3)sin(x^2+2)-cos(x-3)cos(x^2+2)

2xcos(x^2+2)-sin(x^2+2)

2xcos(x+3)cos(x^2+2)-sin(x+3)sin(x^2+2)

Correct answer:

2xcos(x-3)cos(x^2+2)-sin(x-3)sin(x^2+2)

Explanation:

Computation of the derivative requires use of the Product Rule and Chain Rule.

y=sin(x^2+2)cos(x-3)

$\frac{dy}{dx}=\frac{d}{dx}[sin(x^2+2)cos(x-3)]$

For:

y=f(x)*g(x) Where f(x) and g(x) are differentiable functions

As you can see, the "saying" from above matches the formula.

In this case:

f(x)=sin(x^2+2) , g(x)=cos(x-3)

Applying the Product Rule:

$\frac{dy}{dx}=sin(x^2+2)\frac{d}{dx}[cos(x-3)]+cos(x-3)\frac{d}{dx}[sin(x^2+2)]$

To compute the derivatives of cos(x-3) and sin(x^2+2) , simply apply the chain rule:

For:

y=sin(u)

$\frac{dy}{dx}=cos(u)\frac{du}{dx}$ , Where u is a differentiable function

For:

y=cos(u)

$\frac{dy}{dx}=-sin(u)\frac{du}{dx}$ , Where u is a differentiable function

Applying the Chain Rule:

$\frac{dy}{dx}=sin(x^2+2)(-sin(x-3)(1))+cos(x-3)(cos(x^2+2)(2x))$

Simplify the expression to match one of the answer choices:

$\frac{dy}{dx}=2xcos(x-3)cos(x^2+2)-sin(x-3)sin(x^2+2)$

Report an Error

Example Question #69 : How To Find Solutions To Differential Equations

Find the derivative with respect to x for the following function:

$y=e^{sin(3x)}+(x^5-3x)^3$

Possible Answers:

$\frac{dy}{dx}=3sin(3x)e^{sin(3x)-1}+(12x^4-9)(x^5-3x)^2$

$\frac{dy}{dx}=-3cos(3x)e^{sin(3x)}+(12x^4-9)(x^5-3x)^2$

$\frac{dy}{dx}=3cos(3x)e^{sin(3x)}+(15x^4-9)(x^5-3x)^2$

$\frac{dy}{dx}=-3sin(3x)e^{sin(3x)-1}+(12x^4-9)(x^5-3x^2)^2$

$\frac{dy}{dx}=3cos(3x)e^{sin(x)}+3(x^5-3x)^2$

Correct answer:

$\frac{dy}{dx}=3cos(3x)e^{sin(3x)}+(15x^4-9)(x^5-3x)^2$

Explanation:

Computation of the derivative requires use of the Chain Rule.

$y=e^{sin(3x)}+(x^5-3x)^3$

$\frac{dy}{dx}=\frac{d}{dx}[e^{sin(3x)}+(x^5-3x)^3)]$

For:

$y=e^{f(u)}$

$\frac{dy}{dx}=e^{f(u)}*f'(u)$

Here f(u)=sin(3x) , therefore, f'(u)=3cos(3x)

For:

y=(f(x))^a

$\frac{dy}{dx}=a*(f(x))^{a-1}*f'(x)$

Applying these two rules results in:

$\frac{dy}{dx}=e^{sin(3x)}(3cos(3x))+3(x^5-3x)^2(5x^4-3)$

Simplifying this equation results in one of the answer choices:

$\frac{dy}{dx}=3cos(3x)e^{sin(3x)}+(15x^4-9)(x^5-3x)^2$

Report an Error

Example Question #1371 : Functions

Solve the differential equation:

$\frac{dy}{dx}=\frac{2sinx+cosx}{y}$

Possible Answers:

$y=\sqrt{2sinx-4cosx+c}$

$y=\sqrt{4cosx-2sinx+c}$

$y=\sqrt{8cosx-4sinx+c}$

y=4cosx-2sinx+c

Correct answer:

$y=\sqrt{2sinx-4cosx+c}$

Explanation:

This is a separable equation, meaning to solve we want to separate, then integrate.

Step 1: Separate by putting the y components on one side of the equation, and x components on the other.

ydy=(2sinx+cosx)dx

Step 2: Integrate both sides of the equation.

$\int ydy= \int (2sinx+cosx)dx$

This gives us the following:

$\frac{1}{2}y^{2}=-2cosx+sinx+c$

Step 3: Now to simplify we can solve for y.

$y^{2}=2sinx-4cosx+c$

$y=\sqrt{2sinx-4cosx+c}$

Report an Error

View Calculus Tutors

Gregory
Certified Tutor

Pennsylvania State University-Main Campus, Bachelor of Science, Chemical Engineering. Carnegie Mellon University, Doctor of P...

View Calculus Tutors

Tracy
Certified Tutor

Clarion University of Pennsylvania, Bachelor of Science, Physics.

View Calculus Tutors

Yaakoub
Certified Tutor

CUNY Queens College, Bachelor of Science, Finance. Vaughn College of Aeronautics and Technology, Bachelor of Science, Electri...

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Biology Tutors in Phoenix, LSAT Tutors in Philadelphia, GRE Tutors in Boston, Chemistry Tutors in Boston, GMAT Tutors in Phoenix, Reading Tutors in New York City, Spanish Tutors in Los Angeles, French Tutors in Seattle, Calculus Tutors in Phoenix, French Tutors in Boston

Popular Courses & Classes

ISEE Courses & Classes in San Diego, MCAT Courses & Classes in Los Angeles, LSAT Courses & Classes in San Diego, ISEE Courses & Classes in Boston, MCAT Courses & Classes in Washington DC, ISEE Courses & Classes in Philadelphia, ACT Courses & Classes in Boston, SAT Courses & Classes in Atlanta, ISEE Courses & Classes in Phoenix, MCAT Courses & Classes in Denver

Popular Test Prep

LSAT Test Prep in Seattle, GRE Test Prep in San Francisco-Bay Area, MCAT Test Prep in Houston, GMAT Test Prep in New York City, ACT Test Prep in Boston, SSAT Test Prep in Los Angeles, SSAT Test Prep in Seattle, SSAT Test Prep in Houston, SSAT Test Prep in Atlanta, SAT Test Prep in Atlanta

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

Calculus 1 : Differential Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #71 : Differential Equations

Example Question #315 : Equations

Example Question #63 : How To Find Solutions To Differential Equations

Example Question #72 : Differential Equations

Example Question #65 : How To Find Solutions To Differential Equations

Example Question #66 : How To Find Solutions To Differential Equations

Example Question #73 : Differential Equations

Example Question #68 : How To Find Solutions To Differential Equations

Example Question #69 : How To Find Solutions To Differential Equations

Example Question #1371 : Functions

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: