Question

Note: This is NOT a question. This is a tutorial.

How to solve non-exact differential equations? Suppose you are given

\((4xy + 3y^2 - x)\text dx + x(x+2y)\text dy = 0\). See comment below to see how!

Answer 1

The first step is to put the Differential Equation in the form Mdx +

Ndy = 0 where M and N are functions of x and/or y. In this problem, the

equation is already in this form. M is a function of x and/or y which

is \(4xy + 3y^2 - x\) and N is also a function of x and/or y but we

need to simplify it and turn it into \((x^2 + 2xy)dy\) for convenience

purposes.

The next step is to find the partial derivative of M with respect to y,

and the partial derivative of N with respect to x. In this problem, we

take \(\frac{\partial (4xy + 3y^2 - x)}{\partial y}\) and

\(\frac{\partial (x^2 + 2x)}{\partial x}\)

That will give us \(4x + 6y\) for \(\frac{\partial M}{\partial y}\) and

\(2x + 2y\) for \(\frac{\partial N}{\partial x}\). As you notice,

\(\frac{\partial M}{\partial y} \ne \frac{\partial N}{\partial x}\).

Therefore, it is NOT an exact differential equation, thus, we introduce

an integrating factor to make it exact. But first, a step must be made

before using the Integrating Factor formula. What is this step? There

are two cases:

Case I \(\frac{1}{N} (\frac{\partial M}{\partial y} - \frac{\partial

N}{\partial x}) = f(x)\)

Case II \(\frac{1}{M} (\frac{\partial M}{\partial y} - \frac{\partial

N}{\partial x}) = g(y)\)

What do these mean? Basically, you first subtract \(\frac{\partial

N}{\partial x}\) from \(\frac{\partial M}{\partial y}\). Then, you

simplify and factor out the difference. Then compare it with M and N.

If you divide the difference by N, the quotient MUST be a function of x

and x alone. If the quotient does not result in a function of x, then

you divide by M, and the quotient MUST be a function of y and y alone.

Let's look at our problem \[\frac{\partial M}{\partial y} -

\frac{\partial N}{\partial x} = 4x + 6y - (2x + 2y) = 4x + 6y - 2x - 2y

= 2x + 4y\]let's factor out 2. This gives us \(2(x+2y)\). Now, let's

compare with M and N. If we divide this by M, we would be getting a

function of x AND y, so we cannot use it. Therefore, we divide by N,

which is \(x^2 + 2xy\). \[\frac{2(x+2y)}{x^2 + 2xy} =

\frac{2\cancel{(x+2y)}}{x\cancel{(x+2y)}} = \frac{2}{x}\] this is

clearly a function of x and x alone.

Therefore, now we can use the formula for Integrating Factor which is

\(e^{\int f(x)dx}\) or \(e^{-\int g(y)dy}\), depending on the quotient

we got from the previous step. Since our quotient in the previous step

was a function of x, we use the first formula. \[e^{\int f(x)dx} =

e^{\int \frac{2}{x} dx} = e^{2\int \frac{1}{x}dx} = e^{2\ln x} = e^{\ln

x^2} = x^2\] Therefore, \(x^2\) is our Integrating Factor. What do we

do with it? We multiply it by M and N, and \(\textit{should}\) make it

an exact differential equation.

\[x^2(4xy + 3y^2 - x)dx + x^3 (x+2y)dy = 0\] Let's distribute. \[(4x^3

y + 3x^2 y^2 - x^3)dx + (x^4 + 2x^3 y)dy = 0\] Now, we take the partial

derivative of M with respect to y, and the partial derivative of N with

respect to x. \[\frac{\partial M}{\partial y} = 4x^3 + 6x^2 y \]

\[\frac{\partial N}{\partial x} = 4x^3 + 6x^2y\] Therefore,

\(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\) thus,

it is now an exact Differential Equation!! So, now, we just proceed on solving the new equation via the Exact Differential Equation solution approach.

Answer 2

I had no idea. Thanks, for reminding me why I like OS so much :D

Answer 3

Nice of you to do these tutorials! :)
You're really good with LaTeX! (I seen the color latex that you did the other day...)

Answer 4

And if the IF is function of both x and y
http://math.stackexchange.com/questions/37641/finding-integrating-factor-when-if-will-be-a-function-of-x-and-y

Answer 5

that's a huge trouble ... i guess

Answer 6

@lgbasallote must appreciate you ,, great work
I prefer to say u as Great work SIR :)
gr8 and thanks again . these type of teachers are needed in os :) keep it up SIR

Answer 7

give me some topics and i will keep it up :p lol

Answer 8

I will request you to post of these topics :
1) complex numbers
2) kinematics in 1g , 2g , 3g
3) circular motion ( dynamics and kinematics )
4) Limits and differential calculus

:-P

Answer 9

i have a derivative tutorial here http://openstudy.com/study#/updates/4f9cb59de4b000ae9ed1a3e8 :D

Answer 10

number 2 and 3 are physics though -__-

Answer 11

:D that is why i typed there :-P i was joking :-P

Answer 12

Excellent tutorial. I have one question: where doe the e^∫f(x)dx come from?

Answer 13

assume IF is only function of y or x, use the condition of exactness ... you will get linear DE ... whose solution is that factor.

Answer 14

Thanks @lgbasallote , love me some ODEs.

Answer 15

To add to mathslover's suggestions, complex calculus might be an interesting one (how feasible fitting it into a single tutorial I don't know, however).
Great tutorials you're doing.

Answer 16

Complex integration only might be feasible to fit within one post.

Answer 17

I don't think it's possible at all to fit conformal mapping and analytics within the same post.

Answer 18

Actually, I'm reviewing my real analysis textbook (at a leisurely pace, since it's my favorite math subject), and I'm willing to create these DIY math posts concurrent to me reading. If anyone's interested... This helps me remember and learn, too.

It might start off slow, though. My real analysis text covers some basic theory first, and I like reading that stuff.

Answer 19

I, for one, may be interested. I've got to have something interesting to do in between the near-perpetual revising for my history test.

Answer 20

@henpen here is a derivation of the expression for the integrating factor in a linear DE, I'm not sure how that proof might extend to this case though. Might be a good exercise to unite the two.

Answer 21

http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

Answer 22

Thanks.

Answer 23

You are so amazingly smart! I envy yew ;D

Answer 24

oh stop it you :""> im not *that* smart :">

Answer 25

Excellent in terms of fundamenetals , :D
For those experienced with differential equations, the integrating factor x^2 is observable ...

Answer 26

goddarn you math people.