Question

General solution of (1+x^2)dy/dx + 2xy = 3x^2?

Answer 1

What are you solving for x, dy/dx?

Answer 2

y

Answer 3

I don't get it. Is this an algebra exercise? Solve for y?

Answer 4

y in terms of x (you need some kind of integration to get rid of the dy/dx).

Answer 5

wait

Answer 6

Everybody seems very hot on integration methods so I thought we would step up a little:-)

Answer 7

divide by (1 + x^2) on both sides

Answer 8

this is homogenous equation and should be solved by putting y = vx

Answer 9

dy/dx + 2xy/(1+x^2) = 3x^2 / (1+x^2)

Answer 10

Integrating Factor = \[e^{\int\limits_{}^{}[2xdx/(1+x^2)]}\]

Answer 11

multiply by IF on both sides

Answer 12

(1 + x^2)y = 3x^2dx

Answer 13

Making progress...2x is derivative of x^2+1 which should remind you of the Product Rule.

Answer 14

ur testing me? i already know how to do this

Answer 15

y = x^3/ (1+x^2) + c

Answer 16

If you know, let others try...

Answer 17

u should put up a signpost saying "TRAINING EXERCISE" then, coz i thot u needed the explanation

Answer 18

Maybe the others would like to see the explanation?

Answer 19

i did it on top...its a linear differential eqn..

Answer 20

once you divide by 1+x^2 on both sides

Answer 21

Why does it work?

Answer 22

why does what work?

Answer 23

he divided it by 1 + x^2 to get it in linear form, which is:

dy/dx + P(x)*y = Q(x)

Answer 24

now u got it?

Answer 25

whers he?

Answer 26

Sorry, system went down...

Answer 27

Yes, I have contrived this example in order to show how it works...

Answer 28

I do apologize for any confusion...

Answer 29

D(1+x^2)y = (1+x^2)dy/dx + 2xy (product Rule)
and the RHS is same as LHS of original eq.

So D(1+x^2)y) = 3x^2

Now lhs is just the derivative of (1+x^2)y so you can apply direct integration..

OK, contrived, so how to find the IF when it is not so convenient?

Answer 30

I can type out the diffeq solution if you want. Using the method of integrating factors.

Answer 31

Well thanks for that, following the method is OK of course, I guess I am just one of those that likes to see the "why" of the method and I was wondering if others thought the same.

Perhaps unfashionable? Or maybe this is just not the right sort of forum for that sort of thing as I haven't seen anyone put up a "TRAINING EXERCISE" notice to their questions.

Answer 32

I can get out my diffeq book and derive for you the method of integrating factors :P But you can look that up of course :)

Answer 33

We seem to be talking slightly at cross purposes, I already have the derivation, the question was designed to lead up to it (but we never got that far.-)

Answer 34

are you a teacher, estudier?

Answer 35

I was, once upon a time, I am retired now.

Answer 36

much respect!

Answer 37

Are there other forums maybe like this but with the possibility for more of a discussion?

Answer 38

u should ve specified sir, here we think ur a student looking fr answers

Answer 39

sorry for being rude earlier

Answer 40

No offence taken, I am learning as well:-)

Answer 41

r u lookin for the derivation of the integrating factor

Answer 42

I had in mind to explain it to some of the students but I didn't really know if there were any that wanted to know it, so I thought to come at it indirectly via a series of integration questions.

Answer 43

The level here seems to be something like A-level/first year Uni and with many below that, is that right?

Answer 44

yes - thats about right - theres a wide range of levels.
By the way i tried solving that diff. equation by putting y = vx and separate the variables but the algebra was heavy - i gave up - might be my age i suppose!

Answer 45

estudier, thank you for teaching me that the integrating factgor method is derived from the product rule

Answer 46

jimmy, the equation is not separable, even with y = vx

Answer 47

yep - u must be right - I mean -even I failed!! lol!

Answer 48

@jamesm That is the key point, sep var relies on integration by substitution which is equivalent to the chain rule.

So it is natural to ask whether you can solve fode deriving from integration by parts (ie the product rule for derivatives).

Answer 49

I will write something up and leave it here for anyone to look at if they want to.

Answer 50

great idea

Answer 51

is the equation is homogenous estuder? seems so to me it is. I guess that not all of them can be solved with y=vx

Answer 52

Linear fode if:- dy/dx + g(x) y = h(x)

Linear fode homogenous if h(x) = 0 for all x (else inhomogenous).

Answer 53

Let me do writeup, then questions (if any).

Answer 54

To recap, we got the answer for the original question via seeing that lhs was like

p dy/dx + dp/dx y (p = 1 + x^2) and reexpressing it with the Product rule as d p(y)/dx.

If we rewrite the original as dy/dx + 2xy/(1+x^2) = 3x^2/(1+x^2), you can see it is the multiplying through (which was already done in the original question) by (1+x^2) that allowed the quick result because it allowed direct integration.

So how to find this factor usually?

Well, it has to be such that when you multiply on the lhs it looks like

p dy/dx + p(2x/(1+x^2))y

and also like p dy/dx + dp/dx y

So if you compare these two then you can see that p is a particular solution of

dp/dx = (2x/(1+x^2)) p which is homogeneous linear fode solvable by sep var to give

|p| = exp (int(2x/(1+x^2)) dx).

This can be generalized for any linear fode of form dy/dx + g(x)y = h(x) whereby g(x) replaces the

2x/(1-x^2) of our original question.

And that is where the method comes from, exp your g(x), multiply through by p(x), rearrange and direct integrate (although just as with sep var, it might not be possible to perform the final integration).

Answer 55

If people get it, we can go on to to sode's...

Answer 56

yes - I follow you ok . I didn't rewalise how rusty I was with fodes!