# [DE] Integrating factorWatch

#1
Can someone please tell me what the integrating factor for this question is:

It's driving me insane. Here is my working:

This gives:

Integration by parts will not aid me for the RHS, I've heard of something about Integration by Reduction, but I'm pretty sure a standard DE question wouldn't require something on an A-Level syllabus. Summary: I have screwed up.

What's my error?

Free rep (worth 18 ) tomorrow

Thanks.
0
9 years ago
#2
You can just jimmy it about and it's a regular seperable ODE. Less of this weird e stuff.

dy/dx = x^2(1 - 3y)

u = x^3 sub will work for the RHS if you must but I don't ever use your method so I shouldn't really advise.

edit: Oh I see so you bring the dx across and the int and d cancel out on the LHS. That's a kind of nice way of doing it and not as weird as it looked to me at first.
0
#3
(Original post by Rocious)
You can just jimmy it about and it's a regular seperable ODE. Less of this weird e stuff.

dy/dx = x^2(1 - 3y)

(Original post by Rocious)
u = x^3 sub will work for the RHS if you must but I don't ever use your method so I shouldn't really advise.
I end up with another thing to integrate by parts (I think), which involves so I'd be stuck in a loop. Anyway, enough of that, thanks for your tip

(Original post by Rocious)
edit: Oh I see so you bring the dx across and the int and d cancel out on the LHS. That's a kind of nice way of doing it and not as weird as it looked to me at first.
Possibly ... this is just a method that we have to learn for the module (DE's where separation of variables cannot be applied).

+rep tomorrow
0
9 years ago
#4
for the RHS use the following substitution u = x^3
0
9 years ago
#5
If you wanted to use the integrating factor method surely you could just use the substitution u=x^3 on the integral you're stuck with?
0
9 years ago
#6
(Original post by jayshah31)

///

You might find this useful;

Spoiler:
Show

(Original post by DeanK22)

A quick and dirty introduction for solving First order linear differential equations;

Suppose we have an equation of the form;

and M and N are some arbitary functions applied to x. We observe that if;

We wish to transform the LHS of to the above.

This implies we require to multiply by a specific function that will allow us to do that. Call this function I(x).

So becomes

However the LHS of the equation equals

So;

and

Dividing B by A it is very apparent that;

0
#7
(Original post by Rocious)
u = x^3 sub will work for the RHS
(Original post by brooker)
for the RHS use the following substitution u = x^3
(Original post by JohnnySPal)
If you wanted to use the integrating factor method surely you could just use the substitution u=x^3 on the integral you're stuck with?
I feel like such a ****. I was doing it by parts, using that same substitution (seeing the product, assuming it could only be done by parts). How embarrassing ... .

(Original post by DeanK22)

You might find this useful;
I believe I'm fine with the Integrating factor thing. It was just the integration on the RHS that I couldn't see, clearly it should have been blatantly obvious.

Thanks all.
0
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