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17Calculus - Exact Differential Equations and Integrating Factors

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Exact differential equations are first-order differential equations of the form \(\displaystyle{ M(x,y)~dx + N(x,y)~dy = 0 }\) where \(M_y = N_x\). The requirement that \(M_y = N_x\) means the differential equation is what we call exact.

classification: this technique applies to first-order differential equations

\( M(x,y)~dx + N(x,y)~dy = 0 \) and \(M_y = N_x\)

Here are some equivalent ways of writing this differential equation.

\(\displaystyle{ M(x,y)~dx + N(x,y)~dy = 0 }\)

\(\displaystyle{ M(x,y) + N(x,y)y' = 0 }\)

\(\displaystyle{ M(x,y) + N(x,y)\frac{dy}{dx} = 0 }\)

How To Solve

Let's go through an example, step-by-step to see how to solve these problems.

Example

Find the general solution to the exact differential equation \( 2xydx + (x^2-3)dy = 0 \).
Solution - First, let's pull out M and N and make sure this equation is exact.

M(x,y) is the coefficient of the \(dx\), so \( M(x,y) = 2xy \).

N(x,y) is the coefficient of the \(dy\), so \( N(x,y) = x^2-3 \).

To check that this is exact, we need to find \(M_y\) and \(N_x\) and compare them.

\(M_y = 2x\) and \(N_x = 2x\). Since they are equal, this equation is exact.

Now we calculate \(\int{M~dx} = \int{2xy~dx} = x^2 y + f(y) \). Notice that rather than having a constant of integration, we have a constant FUNCTION of integration. This is because when we take the partial derivative of f(y) with respect to x, we get zero.
Now we do a similar integration on N, except with respect to y this time.
\(\int{N~dy} = \int{x^2-3~dy} = x^2 y - 3y + g(x)\) Again we need a FUNCTION of integration, not just a constant.
Now we combine the results, being careful not to duplicate anything. Let's look at the first result of integration, i.e. \( x^2 y + f(y) \). We will keep the \(x^2y\) term. Now we look at the other result. Notice we have another \(x^2y\) term, so we don't need that since we already have one. But we don't have \(-3y\) yet, so we keep that to satisfy the \(f(y)\) requirement in the first result.
So now we have \(x^2y - 3y\) and, since we assume our solution in the form \(F(x,y)=K\), we have \(x^2y-3y=K\) as our final answer.

Here is a great video explaining the technique and showing a couple of examples.

Dr Chris Tisdell - exact equations [1hr-6mins-16secs]

video by Dr Chris Tisdell

In the video above, he does a good job of emphasizing that, before you start actually solving the differential equation, you need to make sure it is exact by calculating \(M_y\) and \(N_x\) and checking that they are equal. Otherwise, this technique will take you down a path that leads nowhere. Most instructors will take off points if you don't show work that you did this. So check with your instructor. But even if they don't take off points, it could save you a lot of time on homework and exams if you do this and it takes only a few seconds. So it is a good investment of your time.

The main difference with these types of problems is that you need to remember the constant of integration is actually a function of integration. This occurs because we are taking the partial integral of one variable while holding the other variable constant. This concept is covered in detail on the multi-variable partial integration page.

Here is another video that is very similar to the first one, with a few different examples. But watching this one also, will help you understand this technique better.

Dr Chris Tisdell - more exact equations [23mins-3secs]

video by Dr Chris Tisdell

Integrating Factors

If we have a first-order equation of the form \( M(x,y)~dx + N(x,y)~dy = 0 \) but \(M_y \neq N_x\), i.e. the equation is not exact (also called inexact), we may be able to convert the equation to exact using integrating factors. You first came across integrating factors when you studied linear, first-order equations. This technique also uses an integrating factor but it is calculated differently.

There are actually two possible integrating factors that convert the differential equation to an exact equation.

\(\displaystyle{ \mu_1 = \exp \int{ \frac{M_y - N_x}{N} dx } }\)

 

\(\displaystyle{ \mu_2 = \exp \int{ \frac{M_y - N_x}{-M} dy } }\)

The trick comes in when you are asked to evaluate these integrals. The evaluation is not always possible and can get quite messy. Also, one integral may yield an integrating factor that is quite complicated while the other one may be much easier to use.

Once we have an integrating factor, we multiply it into the entire differential equation. The result should be an exact equation. (Make sure to check that it is before you finish the problem.) You can then use the technique above to get a solution. Here are a couple of videos with an example.

In this first video, he starts an in-depth example where he calculates an integrating factor for a specific inexact equation. He finishes the example in the next video.

Khan Academy - Integrating Factors (1) [10mins-16secs]

video by Khan Academy

Here is the completion of the example he starts in the previous video. He calculated an integrating factor in the above video and he finishes the problem, which is now exact, in this video.

Khan Academy - Integrating Factors (2) [8mins-25secs]

video by Khan Academy

Okay, time for some practice problems.

Practice

Unless otherwise instructed, solve these differential equations. Make sure to check that the equation is exact before attempting to solve. If the equation is not exact, calculate an integrating factor and use it make the equation exact. If an initial condition is given, find the explicit solution also. Give your answers in exact form.

Basic

\( 2xy~dx + (x^2+3y^2)~dy = 0 \)

Problem Statement

\( 2xy~dx + (x^2+3y^2)~dy = 0 \)

Solution

585 video

video by Dr Chris Tisdell

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\( 2xy~dx + x^2~dy = 0 \)

Problem Statement

\( 2xy~dx + x^2~dy = 0 \)

Solution

587 video

video by Dr Chris Tisdell

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\( 2xy~dx + (x^2-1)~dy = 0 \)

Problem Statement

\( 2xy~dx + (x^2-1)~dy = 0 \)

Final Answer

\( x^2y - y = C \)

Problem Statement

\( 2xy~dx + (x^2-1)~dy = 0 \)

Solution

597 video

video by MIP4U

Final Answer

\( x^2y - y = C \)

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\( [\cos(x)\sin(x)-xy^2]~dx + y(1-x^2)~dy = 0 \)

Problem Statement

\( [\cos(x)\sin(x)-xy^2]~dx + y(1-x^2)~dy = 0 \)

Final Answer

\( (1/2)[y^2(1-x^2) + sin^2(x)] = C \)

Problem Statement

\( [\cos(x)\sin(x)-xy^2]~dx + y(1-x^2)~dy = 0 \)

Solution

598 video

video by MIP4U

Final Answer

\( (1/2)[y^2(1-x^2) + sin^2(x)] = C \)

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\( (y\cos(x)+2xe^y) + (\sin(x)+x^2e^y-1)y' = 0 \)

Problem Statement

\( (y\cos(x)+2xe^y) + (\sin(x)+x^2e^y-1)y' = 0 \)

Final Answer

\( y\sin(x) + x^2e^y - y = C \)

Problem Statement

\( (y\cos(x)+2xe^y) + (\sin(x)+x^2e^y-1)y' = 0 \)

Solution

593 video

video by PatrickJMT

Final Answer

\( y\sin(x) + x^2e^y - y = C \)

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\( (2xy+1) + (x^2+3y^2)y' = 0 \)

Problem Statement

\( (2xy+1) + (x^2+3y^2)y' = 0 \)

Solution

1845 video

video by Dr Chris Tisdell

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\( (e^{x+y}-2x)dx + (e^{x+y}+2y)dy = 0 \), \(y(0)=0\)

Problem Statement

\( (e^{x+y}-2x)dx + (e^{x+y}+2y)dy = 0 \), \(y(0)=0\)

Final Answer

\( 1 = e^{x+y} - x^2 + y^2 \)

Problem Statement

\( (e^{x+y}-2x)dx + (e^{x+y}+2y)dy = 0 \), \(y(0)=0\)

Solution

2086 video

video by Krista King Math

Final Answer

\( 1 = e^{x+y} - x^2 + y^2 \)

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\( xy^3dx + (x^2y^2+1)dy = 0 \)

Problem Statement

Solve \( xy^3dx + (x^2y^2+1)dy = 0 \) using exact integrating factors, if necessary.

Final Answer

\(\displaystyle{ \frac{1}{2}x^2y^2 + \ln|y| = C }\)

Problem Statement

Solve \( xy^3dx + (x^2y^2+1)dy = 0 \) using exact integrating factors, if necessary.

Solution

He makes several mistakes in his solution and answer. First, when calculating the integrating factor, he needs \(dy\) in the integral. Second, since we can't tell if \(y\) is always positive, he needs absolute value signs on his natural log term.

2206 video

video by Engineer4Free

Final Answer

\(\displaystyle{ \frac{1}{2}x^2y^2 + \ln|y| = C }\)

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\( 2x + 3 + (2y-2)y' = 0 \)

Problem Statement

\( 2x + 3 + (2y-2)y' = 0 \)

Final Answer

\( x^2 + 3x + y^2 - 2y = C \)

Problem Statement

\( 2x + 3 + (2y-2)y' = 0 \)

Solution

595 video

video by Khan Academy

Final Answer

\( x^2 + 3x + y^2 - 2y = C \)

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Intermediate

\( (3x+y+1)~dx + (3y+x+1)~dy = 0 \)

Problem Statement

\( (3x+y+1)~dx + (3y+x+1)~dy = 0 \)

Solution

586 video

video by Dr Chris Tisdell

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\( (2x+y+1)~dx + (2y+x+1)~dy = 0 \)

Problem Statement

\( (2x+y+1)~dx + (2y+x+1)~dy = 0 \)

Solution

588 video

video by Dr Chris Tisdell

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\(\displaystyle{ (2xy^{-1}+2ye^{2x}-1) + (e^{2x}+y^2-x^2y^{-2})\frac{dy}{dx} = 0 }\)

Problem Statement

\(\displaystyle{ (2xy^{-1}+2ye^{2x}-1) + (e^{2x}+y^2-x^2y^{-2})\frac{dy}{dx} = 0 }\)

Solution

1822 video

video by Krista King Math

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\( (3x^2-2xy+2)~dx + (6y^2-x^2+3)~dy = 0 \)

Problem Statement

\( (3x^2-2xy+2)~dx + (6y^2-x^2+3)~dy = 0 \)

Solution

596 video

video by Khan Academy

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You CAN Ace Differential Equations

Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

Trig Derivatives

\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\)

 

\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\)

\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\)

 

\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\)

\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\)

 

\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\)

Inverse Trig Derivatives

\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\)

 

\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\)

\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\)

 

\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\)

\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

 

\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\)

Trig Integrals

\(\int{\sin(x)~dx} = -\cos(x)+C\)

 

\(\int{\cos(x)~dx} = \sin(x)+C\)

\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\)

 

\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)

\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\)

 

\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\)

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Topics Listed Alphabetically

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Integrating Factors

Practice

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Practice Instructions

Unless otherwise instructed, solve these differential equations. Make sure to check that the equation is exact before attempting to solve. If the equation is not exact, calculate an integrating factor and use it make the equation exact. If an initial condition is given, find the explicit solution also. Give your answers in exact form.

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