Question

I'm wondering what approach someone would recommend to solve the following ODE; I don't have to solve it in any particular way.

Answer 1

\[(1-x^2)y'=1-x^2+y, \ \ \ |x|<1.\]

Answer 2

substitution?

Answer 3

Alright, cool; I'm reviewing ODE and haven't done this stuff in a while, so I don't remember what that is/how to do it, but I'm going to go read my old text for a moment and then take a shot at it.

Answer 4

ok :) good luck

Answer 5

Is this the thing involving Homogeneous Equations of a certain order?

Answer 6

@iambatman , could you help me out on this? I am stuck.

Answer 7

This ODE is linear

Answer 8

In particular,
\[y'-\frac{1}{1-x^2}y=1\]
with integrating factor
\[\ln\mu(x)=\int\frac{dx}{x^2-1}=-\frac{1}{2}\ln|x+1|+\frac{1}{2}\ln|x-1|\]

Answer 9

\[\mu = e^{\int\limits_{}^{}\frac{1}{1-x^2}dx}=e^{1/2 \ln|x+1|+1/2 \ln|x-1|} = \frac{1}{\sqrt{x+1}}+\sqrt{x-1}\]

Answer 10

Now dealing with that:

\[y=\frac{\int\limits_{}^{}\mu(x)f(x)dx}{\mu(x)}=\frac{\int\limits_{}^{}(1/\sqrt{x+1}+\sqrt{x-1})dx}{1/\sqrt{x+1}+\sqrt{x-1}}\]

Answer 11

\[\int\limits_{}^{}\frac{1}{\sqrt{x+1} \ dx}=2\sqrt{x+1}+C_{1}; \ \int\limits_{}^{}\sqrt{x-1}\ dx=\frac{2(x-1)^{3/2}}{3}+C_{2}.\]

Answer 12

Answer is this, for reference:

Answer 13

With some simplifying,
\[\mu(x)=\sqrt{\frac{x-1}{x+1}}\]
Then distributing gives
\[\sqrt{\frac{x-1}{x+1}}y'-\frac{1}{1-x^2}\sqrt{\frac{x-1}{x+1}}y=\sqrt{\frac{x-1}{x+1}}\]
Noting that
\[\begin{align*}\frac{d}{dx}\sqrt{\frac{x-1}{x+1}}&=\frac{1}{2\sqrt{\frac{x-1}{x+1}}}\frac{x+1-(x-1)}{(x+1)^2}\\\\
&=\frac{(x+1)^{1/2}}{(x-1)^{1/2}(x+1)^2}\\\\
&=\frac{(x-1)^{1/2}}{(x-1)(x+1)^{3/2}}\\\\
&=\frac{(x-1)^{1/2}}{(x^2-1)(x+1)^{1/2}}\\\\
&=-\frac{1}{1-x^2}\sqrt{\frac{x-1}{x+1}}
\end{align*}\]
then gives
\[\frac{d}{dx}\left[\sqrt{\frac{x-1}{x+1}}y\right]=\sqrt{\frac{x-1}{x+1}}\]
Integrating both sides, you have
\[y=\sqrt{\frac{x+1}{x-1}}{\large \int}\sqrt{\frac{x-1}{x+1}}\,dx\]

Answer 14

I'm trying to avoid looking at what you've just written at the moment and work out the kinks in my own algebra without spoiling it, but thank you again for the help; I'm retty certain I'll still need to eventually look at what you've written.

Answer 15

Before you do any looking, there may be a mistake somewhere with my work or with the given solution.

Answer 16

The given solution is right, so it's definitely something I did...

Answer 17

Alright, yeah, I had to take a look, but was in the first place confused about the \[\sqrt{\frac{x-1}{x+1}}\]And how exactly you came to that; the way I was working it out, I would end up, moving from the mu value that I had, with an irreducible collection of terms that resembles a factorable/FOILable expression but isn't, sqrt(x^2-1).

Answer 18

Here's how I'm doing it:

Answer 19

\[\frac{1}{\sqrt{x+1}}+\sqrt{x-1} \cdot \frac{\sqrt{x+1}}{\sqrt{x+1}} = \frac{1+(x-1)^{1/2}(x+1)^{1/2}}{\sqrt{x+1}}\]

Answer 20

\[\frac{1+\sqrt{x^2-1}}{\sqrt{x+1}}\]

Answer 21

From this
\[\ln\mu(x)=-\frac{1}{2}\ln|1+x|+\frac{1}{2}\ln|1-x|\]
We get the IF
\[\begin{align*}\ln\mu(x)&=\frac{1}{2}\ln\left|\frac{1-x}{1+x}\right|\\\\
&=\ln\sqrt{\frac{1-x}{1+x}}\\\\
\mu(x)&=\sqrt{\frac{1-x}{1+x}}
\end{align*}\]
I had to check quickly with Mathematica; for some reason I get \(\color{red}{(1-x)}\) instead of \((x-1)\) in the numerator (I think this result is preferred anyway, since we have the stipulation \(|x|<1\)).

Answer 22

Sorry, just laid down for a bit; so above you just used a property of logarithms to combine the arguments, correct?

Answer 23

This still works though. We have
\[\sqrt{\frac{1-x}{1+x}}y'+\frac{1}{x^2-1}\sqrt{\frac{1-x}{1+x}}y=\sqrt{\frac{1-x}{1+x}}\]
but
\[\begin{align*}\frac{d}{dx}\sqrt{\frac{1-x}{1+x}}&=\frac{1}{2\sqrt{\frac{1-x}{1+x}}}\frac{-(1+x)-(1-x)}{(1+x)^2}\\\\

&=-\frac{(1+x)^{1/2}}{(1-x)^{1/2}(1+x)^2}\\\\

&=\frac{1}{x^2-1}\sqrt{\frac{1-x}{1+x}}
\end{align*}\]
So,
\[\begin{align*}\frac{d}{dx}\left[\sqrt{\frac{1-x}{1+x}}y\right]&=\sqrt{\frac{1-x}{1+x}}\\\\
y&=\sqrt{\frac{1+x}{1-x}}{\large \int}\sqrt{\frac{1-x}{1+x}}\,dx
\end{align*}\]

Answer 24

Yes

Answer 25

I do not see how differentiating \[\sqrt{\frac{1-x}{1+x}}y\]yields no y terms whatsoever, as in the statement in the second to last line, I'm taking a look to try to figure out why that is. I understand everything else, but not that.

Answer 26

In fact, I think it's just the phrasing and order of statements, but I understand almost none of what you just wrote. One moment.

Answer 27

Oh, nevermind, I understand everything else; it's just still that line prior to the last one.

Answer 28

Maybe some color-coding will help?\[\begin{align*}\frac{d}{dx}\left[\color{red}{\sqrt{\frac{1-x}{1+x}}}\color{blue}{y}\right]&=\color{red}{\frac{d}{dx}\left[\sqrt{\frac{1-x}{1+x}}\right]}\color{blue}y+\underbrace{\color{red}{\sqrt{\frac{1-x}{1+x}}}\color{blue}{\frac{dy}{dx}}}\\\\
&=\underbrace{\color{red}{\sqrt{\frac{1-x}{1+x}}}\color{blue}{y'}}+\color{red}{\frac{1}{x^2-1}\sqrt{\frac{1-x}{1+x}}}\color{blue}y\\\\
&=\text{LHS of ODE}\end{align*}\]

Answer 29

I'm not trying to be nitpicky, I swear to god, but it's that the first line in your expression, the RHS of that does not equal the RHS of the second to last line; I get the product rule differentiation and all, just supremely confused about it, it's like you just started writing stuff, and that's clearly not what you did, there was intent behind it, lol.

Answer 30

There are a lot of expressions up there haha, could you be more specific?

Answer 31

Sure thing, doing that right now, one sec.

(1)
(2)

Answer 32

identical LHS, different RHS expressions, primarily the B&W one that makes absolutely no sense to me. At all. Virtually every other expression, gotcha, but differentiating mu*y, I don't understand how differentiating mu*y, you just got a single term, mu.

I'd contend that \[\frac{d}{dx}\Bigg[\sqrt{\frac{1-x}{1+x}}y\Bigg] \neq \sqrt{\frac{1-x}{1+x}}\]As stated in the pictured B&W expression.

Answer 33

\[\frac{d}{dx}(\mu(x)y(x))=\mu(x)f(x)\]

Answer 34

Is that what's going on?

Answer 35

I'm guessing it was the last line of math I wrote; in putting the original problem in the proper form, I forgot that the RHS was multiplied by mu, and since f(x) was just 1, it just meant that the RHS was equal to mu.

Answer 36

Let get back to basics for a moment. A linear ODE has the form
\[y'+f(x)y=g(x)\]
The integrating factor is
\[\mu(x)=\exp\left(\int f(x)\,dx\right)\]
so the ODE can be written as
\[\begin{align*}\mu(x)y'+\mu(x)f(x)y&=\mu(x)g(x)\\\\
\exp\left(\int f(x)\,dx\right)y'+\exp\left(\int f(x)\,dx\right)f(x)y&=\\\\
\mu(x)\frac{dy}{dx}+\frac{d\mu(x)}{dx}y&=\\\\
\frac{d}{dx}\left[\mu(x)y\right]&=
\end{align*}\]
Basically what we have here is a rearrangement of the LHS (the application of the IF) so that we can write the LHS as a derivative. From the beginning, we have that the LHS is equivalent to the RHS by virtue of being an ODE. What we've done is to agebraically manipulate the LHS so that we can isolate the factor of \(y\).

So, while no \(y\) terms appear on the RHS, we still know that LHS = RHS by the transitive property.

Answer 37

Yeah, I get how that works...it wasn't the sheer lack of presence of a y term in and of itself, it's that I thought you were just expressing the LHS in a different manner, not equating the LHS and RHS, which, while identical, have pretty different forms/implications are more important there rather than just screwing around and rearranging identities of a single side.

Answer 38

You were doing \[\frac{d}{dx}\bigg[\mu(x)y(x)\bigg]=\mu(x)f(x),\]not\[\frac{d}{dx}\bigg[\mu(x)y(x)\bigg]=y'(x)\mu(x)+\mu(x)P(x)y(x)\]
Despite all three expressions being equivalent.

Answer 39

Does that make sense?

Answer 40

That's not what I was doing though. What I did was a logical jump from the modified ODE (with the IF distributed) to the fact that the LHS was a derivative:
\[\begin{align*}\sqrt{\frac{1-x}{1+x}}y'+\frac{1}{x^2-1}\sqrt{\frac{1-x}{1+x}}y&=\sqrt{\frac{1-x}{1+x}}\\\\
\frac{d}{dx}\left[\sqrt{\frac{1-x}{1+x}}y\right]&=
\end{align*}\]
In that post, everything in between was just verification, that \(\dfrac{d}{dx}\sqrt{\dfrac{1-x}{1+x}}\) was indeed the \(\mu(x)f(x)\) term (which it was, and this allows me to go from line 1 to line 2).

I hope this addresses your confusion! I have to get to sleep

Answer 41

I understand the math you did. That's all I have to say at this point. I'm 99% sure we're talking about a virtually identical approach in pedantically different ways. Either way, my blood pressure has skyrocketed.

If you think I didn't understand what you did, I don't know what to tell you. Either way, thanks for the help. Goodnight.

Answer 42

All I did was carry out the general procedure for solving a linear ODE. If you understand the reasoning behind the IF approach, then you should have no problem following along. Like you said, if there's still some cause for confusion, it's likely due to either of us having learned to solve linear equations in a slightly (but fundamentally the same) different method.

(If I come across as hostile or condescending, I certainly don't intend to - I usually give askers the benefit of the doubt, but any misunderstanding or confusion puts me on a "let's start from the beginning" track.)

Answer 43

It's fine, I was considerably more upset than I should have been and nonetheless reacted in the wrong manner. Thank you for helping me.

Answer 44

No problem. How's that new question coming along?

Answer 45

I didn't finish it out entirely; due to time constraints, I'm completing most of these problems up to a certain point and keeping a comprehensive record of each problem/how close it is to completion in the form of either embedded pictures of fully worked solutions from my notebook, or links to partial solutions in OpenStudy threads, on some electronic flashcards.

I have a system of flashcards whereby I'm not allowed to put a problem with a given solution in a flashcard until I have at least attempted and completed the majority of that problem and have a record of it somewhere; this encourages me to try at least every single one to completion if I'm going to have it ready in that database to prepare for exams.

I'll come back to it soon, but for now I'm trying to start on at least every review problem to some meaningful degree, before my next class tomorrow night.