You CAN Ace Differential Equations

17Calculus Subjects Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

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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

Lets jump right into a video to see how we solve these. Here is a great video clip 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. Regardless, these integrating factors can, at times, be quite useful. Let's look at an example in this next video.

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

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

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. Give your answers in exact form.

Basic Problems

\( 2xy^2 + 4x^3 + (2x^2y +4y^3)y' = 0 \)

Problem Statement

\( 2xy^2 + 4x^3 + (2x^2y +4y^3)y' = 0 \)

Final Answer

\( y = x^2y^2 + x^4 + y^4 + C \)

Problem Statement

\( 2xy^2 + 4x^3 + (2x^2y +4y^3)y' = 0 \)

Solution

First, check for exactness. Rewriting the equation in differential form, we have \( (2xy^2 + 4x^3)dx + (2x^2y +4y^3)dy = 0 \)
\( M(x,y) = 2xy^2 + 4x^3 \) \( \to \) \( M_y = (2x)(2y) + 0 = 4xy \)
\( N(x,y) = 2x^2y+4y^3 \) \( \to \) \( N_x = 2y(2x)+ 0 = 4xy \)
Since \( M_y = N_x \), the differential equation is exact and so we can solve it.
\( \psi_x = M(x,y) = 2xy^2+4x^3 \) \( \to \) \( \psi = 2y^2 (x^2/2) + (4x^4)/4 + h(y) = \) \( x^2y^2 + x^4 + h(y) \)
\( \psi_y = N(x,y) = 2x^2y+4y^3 \) \( \to \) \( \psi = 2x^2 (y^2/2) + (4y^4)/4 + g(x) = \) \( x^2y^2 + y^4 + g(x) \)
Equating these results, we get \( \psi = x^2y^2 + x^4 + y^4 + C \)

Final Answer

\( y = x^2y^2 + x^4 + y^4 + C \)

close solution

\( y+2e^x + (1+e^{-x})y' = 0 \)

Problem Statement

\( y+2e^x + (1+e^{-x})y' = 0 \)

Final Answer

\( y = ye^x +y + e^{2x} + C \)

Problem Statement

\( y+2e^x + (1+e^{-x})y' = 0 \)

Solution

First we check for exactness. Rewriting the differential equation, we have \( (y+2e^x)dx + (1+e^{-x})dy = 0 \).
\( M(x,y) = y+2e^x \) \( \to \) \( M_y = 1 \)
\( N(x,y) = 1+e^{-x} \) \( \to \) \( N_x = -e^{-x} \)
Since \( M_y \neq N_x \), this equation is not exact. We need to find an integrating factor.
\(\begin{array}{rcl} \displaystyle{\frac{d\mu}{dx}} & = & \displaystyle{\frac{M_y - N_x}{N}\mu} \\ & = & \displaystyle{\frac{1+e^{-x}}{1+e^{-x}}\mu} \\ & = & \mu \\ \displaystyle{\frac{d\mu}{\mu}} & = & dx \\ ln|\mu| & = & x + k ~~ \text{(let k=0)}\\ \mu & = & e^x \end{array}\)
So our integrating factor is \( \mu=e^x \). Multiplying the differential equation by this factor, we get
\( e^x[t+2e^x] + e^x[1+e^{-x}]y' = 0 \) \( \to \) \( (ye^x+2e^{2x})dx + (e^x+1)dy = 0 \)
Let's check that we now have an exact equation.
\( M(x,y) = ye^x + 2e^{2x} \) \( \to \) \( M_y = e^x \)
\( N(x,y) = e^x+1 \) \( \to \) \( N_x = e^x \)
Since we now have an exact equation, we can solve it.
\( \psi_x = M(x,y) = ye^x + 2e^{2x} \) \( \to \) \( \psi = ye^x + e^{2x} + h(y) \)
\( \psi_y = N(x,y) = e^x+1 \) \( \to \) \( \psi = ye^x + y + g(x) \)
Equating we get \( \psi = ye^x +y + e^{2x} + C \)

Final Answer

\( y = ye^x +y + e^{2x} + C \)

close solution

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

Problem Statement

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

Solution

585 solution video

video by Dr Chris Tisdell

close solution

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

Problem Statement

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

Solution

587 solution video

video by Dr Chris Tisdell

close solution

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

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 solution video

video by Khan Academy

Final Answer

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

close solution

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

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 solution video

video by MIP4U

Final Answer

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

close solution

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

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 solution video

video by MIP4U

Final Answer

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

close solution

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

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 solution video

video by PatrickJMT

Final Answer

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

close solution

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

Problem Statement

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

Solution

1845 solution video

video by Dr Chris Tisdell

close solution

Find the explicit solution for the exact differential equation \( (e^{x+y}-2x)dx + (e^{x+y}+2y)dy = 0 \), \(y(0)=0\).

Problem Statement

Find the explicit solution for the exact differential equation \( (e^{x+y}-2x)dx + (e^{x+y}+2y)dy = 0 \), \(y(0)=0\).

Final Answer

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

Problem Statement

Find the explicit solution for the exact differential equation \( (e^{x+y}-2x)dx + (e^{x+y}+2y)dy = 0 \), \(y(0)=0\).

Solution

2086 solution video

video by Krista King Math

Final Answer

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

close solution

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

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 solution video

video by Engineer4Free

Final Answer

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

close solution

Intermediate Problems

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

Problem Statement

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

Solution

586 solution video

video by Dr Chris Tisdell

close solution

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

Problem Statement

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

Solution

588 solution video

video by Dr Chris Tisdell

close solution

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

Problem Statement

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

Solution

596 solution video

video by Khan Academy

close solution

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

Problem Statement

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

Solution

1822 solution video

video by Krista King Math

close solution
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