Differential Equations by substitution

So I'm stuck on this Isaac Physics question -

https://isaacphysics.org/questions/solution_by_substitution2

and I have tried the substitution 4u/v because that differentiates to something that resembles the RHS, although my answer turns out as

u = v^2/2 - 1/2 v or multiples thereof, and I'm unsure as to an alternative way to look at the problem. Should I try to do a reverse substitution, and think about integrating the RHS? I've been stuck for a few hours, and I'd appreciate some help : )

If you move the 4u/v^2 term over, you get (v du/dv - 4u) / v^2.

Any ideas how you might rewrite v du/dv - 4u as a single derivative (times a function of v)?


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I'm getting d/dv(4uv), but this leaves me with 4v du/dv + 4u, and we want v and -4u, so I'm sure I'm not quite right there, I checked the spoiler and I'm confused about the exponent, as surely it would introduce negative powers? The function of y I think is 1/y^2.

edit: negative powers if the exponent is less than 0, which seems to be the only way to get a negative u?

..

Yes, you'll have negative powers - this isn't inherently a problem.

e.g. if you have x dy/dx - 2y = x^4, you can observe

$\dfrac{d}{dx} y/x^2 = \dfrac{1}{x^2}\dfrac{dy}{dx} - \dfrac{2y}{x^3} = \dfrac{1}{x^3}\left(x \dfrac{dy}{dx}-2y\right)$, and from there you can deduce $x^3 \dfrac{d}{dx} y/x^2 = x \dfrac{dy}{dx}-2y$ and you can then substitute that in your original equation to finish.


I agree the natural way to solve this is by IF, but I'm trying to work with what seems to be the intended plan of spotting an appropriate substitution (which is basically the same as "spotting" the IF that works).

Yes, you'll have negative powers - this isn't inherently a problem.

e.g. if you have x dy/dx - 2y = x^4, you can observe

$\dfrac{d}{dx} y/x^2 = \dfrac{1}{x^2}\dfrac{dy}{dx} - \dfrac{2y}{x^3} = \dfrac{1}{x^3}\left(x \dfrac{dy}{dx}-2y\right)$, and from there you can deduce $x^3 \dfrac{d}{dx} y/x^2 = x \dfrac{dy}{dx}-2y$ and you can then substitute that in your original equation to finish.

Rearrange into the form dy/dx + P(x)*y = Q(x) and use the integrating factor. I assume you have heard of integrating factor?

I have heard of it, I haven't actually formally learnt ODEs, but I'm trying to get familiar as I've read that they're pretty important in Physics, but maybe my manipulation earlier today was trash because that lead me to incorrect solutions.

I have found the particular solution (and verified it) using the IF method by first rearranging into the form I mentioned above and spotting that the IF is e^integral(-4/x) dx.

It's easiest to rearrange into that form as you'll spot the IF required.

I'd take the time to learn the concept as it makes these questions sooo simple to solve.

Ive ended up with $\dfrac{d}{dv} (u/v^4) = \frac{1}{v^5} (v \dfrac{du}{dv} - 4u)$, which looks remotely like what we need, and so that whole thing multiplied by 1/v^2?

I can't be bothered continually scrolling up and looking at the attachment in a new tab to be exact about things, but if you know that

$(v \dfrac{du}{dv} - 4u) = f(v)$, then you now know that $\dfrac{d}{dv}(u/v^4) = \frac{1}{v^5} f(v)$, at which point you should have something you can solve by integrating both sides w.r.t. v.

Or to put it another way, you've deduced that w = u/v^4 would be a good substittuion to try.

Thanks, I've just got the correct answer! Thank you for being patient with me

Cool. (The "can't be bothered" was more "I hate having to keep going up and down to refer to and hopefully not misremember part of the equation" impatience than anything with you).

Regarding the IF method: I sometimes find it hard to guess where the Isaac Physics authors are trying to go with certain concepts, but from what I've seen posted here they are trying to get people to think outside the fairly "regimented" box that is taught at A-level (where it's a bit "if you see X, you use technique Y"). So if they're saying "solve this equation by substitution" they probably don't want you to be using the integrating factor method (or at least not in the "rearrange as dy/dx + f(x)y = g(x) and then do these steps" way it's taught at A-level).

I could be completely wrong on that though - as I say, I often find I'm confused by what they're actually after even on questions I'm 100% happy solving on my own terms.