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A few small questions about differential equations

battleman13

New member
Jan 31, 2012
6
I am currently taking differential equations in college but I am having trouble understanding a few things.

Our teacher is from Russia so it makes it a bit harder for me to understand him and sometimes for him to understand
our questions....

Currently I have taken Calc 1, 2, 3 and Linear Algebra and understand the topics to an average degree....

diff eq however is giving me wee bits of trouble.


For example I do not know how to tell if an equation is standard form or not?

He gave us the general form for an equation in standard form but I don't "see" how to tell
if an equation is in standard form or not? Some examples he did contradicted my thoughts.

Secondly once you have a linear first order (or I think maybe any order) differential equation in standard form
and it is not seperable (or even if it is) then you can find an integrating factor to make the equation seperable.

I understand the concept, I just for the life of me do not understand what to pick for the different parts that you need
to apply in the formulas for finding said integrating factor?

I will in about two hours post the formulas I have but I think much of my troubles can be helped without them.


Thanks in advance to anyone who can help clear the fog from my problem!
 
Jan 31, 2012
54
our teacher is from russia so it makes it a bit harder for me to understand him and sometimes for him to understand
our questions....
lol

There is a nice(but not to nice) book by Frank Ayres, Theory and Problems of DIFFERENTIAL EQUATIONS. Look for that book.

..
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
I would say that DE's of different types have different "standard forms". You should become adept at transforming DE's from one form to another, in order to recognize the type.

Aside from that, your post is a little bit too general for MHB (we're not a tutoring service). Post a specific question, like you said, and we can help you get unstuck with that particular problem.
 

battleman13

New member
Jan 31, 2012
6
Well the problem is he can show us a thousand different "examples" but its hard for me to see what is "wrong" with an equation that it is not in standard form.

I understand the concepts of calculus decently well but I am lost on these "ordinary differential equations".

I will post a specific example:

This is our "standard form" for 1st order linear equations: $$ \frac{\mathrm{d}y }{\mathrm{d} x} + P(x)y = f(x) $$

He then has it written $$dy + P(x)y = f(x)dx$$

Followed by ( P(x)y - f(x))dx + dy = 0

From there we try to find and integrating factor as a function of two variables mu (x) (greek letter mu)

Ok so say for instance we have this equation: $$ X\frac{\mathrm{d}y }{\mathrm{d} x} - 4y = x^{6}e^{x} $$

This is not in "standard form" but in "standard form" it is written as: $$\frac{\mathrm{d}y }{\mathrm{d} x} - \frac{4}{x} y = x^{5}e^{x}$$

I do not understand how it is determined that the original equation was non standard and now this one is.

For his p(x) he picked the -(4/x) and for his f(x) he picked (x^5)(e^x)

I don't understand why he picked what he did for what... why was the y left out for p(x)?
 
Last edited:
Jan 31, 2012
54
Don't use [tex] and [/tex], we are Capitalists now,- use $$.
 

battleman13

New member
Jan 31, 2012
6
I seem to have got it somewhat now.... at least it's "looking better".
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
I will post a specific example:

This is our "standard form" for 1st order linear equations: $$ \frac{\mathrm{d}y }{\mathrm{d} x} + P(x)y = f(x) $$

He then has it written $$dy + P(x)y = f(x)dx$$
This is technically incorrect. He should have written it $$dy+P(x)y\,dx=f(x)\,dx,$$ but he seems to have corrected the mistake in the next line.

Followed by ( P(x)y - f(x))dx + dy = 0

From there we try to find and integrating factor as a function of two variables mu (x) (greek letter mu)
You can find the integrating factor from the first form. All you need do is identify $P(x)$.

Ok so say for instance we have this equation: $$ X\frac{\mathrm{d}y }{\mathrm{d} x} - 4y = x^{6}e^{x} $$

This is not in "standard form" but in "standard form" it is written as: $$\frac{\mathrm{d}y }{\mathrm{d} x} - \frac{4}{x} y = x^{5}e^{x}$$

I do not understand how it is determined that the original equation was non standard and now this one is.
Because the coefficient of the derivative $dy/dx$ was not 1 before, and now it is.

For his p(x) he picked the -(4/x) and for his f(x) he picked (x^5)(e^x)

I don't understand why he picked what he did for what... why was the y left out for p(x)?
Because when you compare the equation $$\frac{dy}{dx}-\frac{4}{x}\,y=x^{5}e^{x}$$ with $$\frac{dy}{dx}+P(x)\,y=f(x),$$ you find that it is the coefficient of $y$ that is $P(x)$. Does that help?
 

battleman13

New member
Jan 31, 2012
6
It does help actually but I did manage to figure that out late last night! Just in time for our exam today!

So thank you very much!!!!

By the way why how come he is incorrect in your first quote?

Is it because when he multiplied by dx he forgot to multiply it into the P(x)y part of the L.H.S ?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,197
It does help actually but I did manage to figure that out late last night! Just in time for our exam today!

So thank you very much!!!!

By the way why how come he is incorrect in your first quote?

Is it because when he multiplied by dx he forgot to multiply it into the P(x)y part of the L.H.S ?
Yes.

I'm glad, actually, you figured it out yourself, because now you own that result. I'm happy to get you unstuck if you need it, but better yet for you to figure it out.

Cheers.