You CAN Ace Differential Equations
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Differential Equations 

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
 
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Exact differential equations are firstorder 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 firstorder 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.
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 multivariable 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.
video by Dr Chris Tisdell
Integrating Factors 

If we have a firstorder 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, firstorder 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 indepth example where he calculates an integrating factor for a specific inexact equation. He finishes the example in the next video.
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.
video by Khan Academy
Okay, time for some practice problems.
Conversion Between ABC Level (or 123) 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. 
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 

video by Dr Chris Tisdell
close solution 
\( 2xy~dx + x^2~dy = 0 \)
Problem Statement 

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

video by Dr Chris Tisdell
close solution 
\( 2x + 3 + (2y2)y' = 0 \)
Problem Statement 

\( 2x + 3 + (2y2)y' = 0 \)
Final Answer 

\( x^2 + 3x + y^2  2y = C \) 
Problem Statement 

\( 2x + 3 + (2y2)y' = 0 \)
Solution 

video by Khan Academy
Final Answer 

\( x^2 + 3x + y^2  2y = C \) 
close solution 
\( 2xy~dx + (x^21)~dy = 0 \)
Problem Statement 

\( 2xy~dx + (x^21)~dy = 0 \)
Final Answer 

\( x^2y  y = C \) 
Problem Statement 

\( 2xy~dx + (x^21)~dy = 0 \)
Solution 

video by MIP4U
Final Answer 

\( x^2y  y = C \) 
close solution 
\( [\cos(x)\sin(x)xy^2]~dx + y(1x^2)~dy = 0 \)
Problem Statement 

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

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

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

video by MIP4U
Final Answer 

\( (1/2)[y^2(1x^2) + sin^2(x)] = C \) 
close solution 
\( (y\cos(x)+2xe^y) + (\sin(x)+x^2e^y1)y' = 0 \)
Problem Statement 

\( (y\cos(x)+2xe^y) + (\sin(x)+x^2e^y1)y' = 0 \)
Final Answer 

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

\( (y\cos(x)+2xe^y) + (\sin(x)+x^2e^y1)y' = 0 \)
Solution 

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 

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 

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 

\(\displaystyle{ \frac{1}{2}x^2y^2 + \lny = C }\) 
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.
video by Engineer4Free
Final Answer 

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

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 

video by Dr Chris Tisdell
close solution 
\( (3x^22xy+2)~dx + (6y^2x^2+3)~dy = 0 \)
Problem Statement 

\( (3x^22xy+2)~dx + (6y^2x^2+3)~dy = 0 \)
Solution 

video by Khan Academy
close solution 
\( (2xy^{1}+2ye^{2x}1) + (e^{2x}+y^2x^2y^{2})\frac{dy}{dx} = 0 \)
Problem Statement 

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

video by Krista King Math
close solution 