[DE] Integrating factor Watch

jayshah31
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#1
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Can someone please tell me what the integrating factor for this question is:

\frac{dy}{dx} + 3x^2y = x^2

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

R = e^{\int 3x^2 \, dx} = e^{x^3}

This gives:

\frac{d}{dx}(e^{x^3}y) = e^{x^3}x^2

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 :p:) tomorrow

Thanks.
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Rocious
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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.
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jayshah31
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(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)
:love:

(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 e^{x^3} 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).

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brooker
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for the RHS use the following substitution u = x^3
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JohnnySPal
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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?
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Oh I Really Don't Care
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(Original post by jayshah31)

///
\frac{d}{dx}(e^{x^3}y) = e^{x^3}x^2

 \frac{d}{dx} (\frac{1}{3} e^{x^3}} = ?


You might find this useful;

Spoiler:
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(Original post by DeanK22)

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

Suppose we have an equation of the form;

 \dispalystyle \frac{dy}{dx} + M(x)y = N(x) \S and M and N are some arbitary functions applied to x. We observe that if;

 \displaystyle Q = f(x)y \Rightarrow \frac{d}{dx}(Q) = \frac{dy}{dx}f(x) + yf'(x)

We wish to transform the LHS of  \S to the above.

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

So  \S becomes  \disaplystyle I(x) \frac{dy}{dx} + I(x)M(x)y = I(x)N(x)

However the LHS of the equation equals  \displaystyle \frac{d}{dx}(f(x)y)

So;

 \displaystyle \frac{dy}{dx}f(x) + f'(x)y = I(x)\frac{dy}{dx} + I(x)M(x)y \Rightarrow f(x) = I(x) \; (A)

and  f'(x) = I(x)M(x) \; (B)

Dividing B by A it is very apparent that;

 \displaystyle \frac{f'(x)}{f(x)}} = M(x) \Rightarrow ln(f(x)) = \int M(x) dx \Rightarrow I(x) = f(x) = e^{\int M(x) dx
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jayshah31
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(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 ... :blushing:.

(Original post by DeanK22)
\frac{d}{dx}(e^{x^3}y) = e^{x^3}x^2

 \frac{d}{dx} (\frac{1}{3} e^{x^3}} = ?

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.
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