Question

I'm doing a first order linear ODE using an integrating factor, and while I came to a very nice looking solution, it is not correct, according to my book; where have I made a mistake? Working below momentarily.

Answer 1

\[2(1-x^2)y'-(1+x)y=\sqrt{1-x^2}\]

Answer 2

\[\frac{dy}{dx}-\frac{(1+x)y}{2(1-x^2)}=\frac{\sqrt{1-x^2}}{2(1-x^2)}\]

Answer 3

\[\mu(x)=e^{\int\limits_{}^{}P(x)dx}=e^{\int\limits_{}^{}\frac{(1+x)}{2(1-x^2)}}\]

Answer 4

There's a negative sign that is outside of this integral both in the last expression and in future expressions for some time due to its lack of apparent consequence, but keep in mind that it is there.

Answer 5

\[\int\limits_{}^{}\frac{(1+x)}{2(1-x^2)}dx=\int\limits_{}^{}\frac{1}{2(1-x^2)}dx+\int\limits_{}^{}\frac{x}{2(1-x^2)}dx\]

Answer 6

Factor out the two from the denominator of both expressions as it doesn't affect the integration, but again, keep in mind that it is there.

Answer 7

\[\frac{1}{2}\int\limits_{}^{}\frac{1}{1-x^2}dx=\frac{1}{2}\int\limits_{}^{}\frac{1}{(1+x)(1-x)}dx\]Using partial fractions: \[\frac{1}{2}\int\limits_{}^{}\frac{1}{(1+x)(1-x)}dx=\frac{1}{2}\Bigg( \int\limits_{}^{}\frac{1}{1+x}dx+\int\limits_{}^{}\frac{1}{1-x}dx\Bigg)\]

Answer 8

Supposed to be another one half in there as a result of partial fractions; the outside for that expression is 1/4th, keep in mind.

Answer 9

\[\int\limits_{}^{}\frac{1}{1+x}dx=\ln(1+x); \ \ \ \int\limits_{}^{} \frac{1}{1-x}dx=\ln(1-x)\]

Answer 10

Taking care of the other expression, not yet integrated, \[\frac{1}{2}\int\limits_{}^{}\frac{x}{1-x^2}dx=\frac{1}{2}\ln(1-x^2)\] (All of the constants of integration are eliminated as appropriate)

Answer 11

Thus, transforming that original term, you have:\[\int\limits_{}^{}\frac{(1+x)}{2(1-x^2)}dx=\int\limits_{}^{}\frac{1}{2(1-x^2)}dx+\int\limits_{}^{}\frac{x}{2(1-x^2)}dx\]\[=\frac{1}{4}\bigg[ \ln(1-x)+\ln(1+x)-\ln(1-x^2)\bigg]\]Using properties of logarithms:\[\frac{1}{4}\bigg[\ln\ \frac{(1-x)(1+x)}{(1-x^2)}\bigg]=\frac{1}{4} \bigg[ \ln \frac{(1-x^2)}{(1-x^2)}\bigg] =\frac{1}{4}\ln(1)\]

Answer 12

@wio

Answer 13

\[\mu(x)=e^{\int\limits_{}^{}P(x)dx}=e^{1/4 \ln(1)}=1.\]

Answer 14

Why didn't you factor in the very begining?

Answer 15

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

Answer 16

^That would've been the smarter and significantly less labor intensive thing to do, but either way, we should get the result; I'm doing something wrong in my own stuff, and I want to know what it is.

Answer 17

But yeah, thanks for pointing that out, that would've made this hugely easier, anyways.

Answer 18

\[\int\limits_{}^{} \frac{1}{1-x}dx=\ln(1-x)\]If I let \(u=1-x\), then \(du=-dx\)\[
\int \frac{1}{1-x}dx =- \int \frac 1udu = -\ln(u)=-\ln(1-x)
\]

Answer 19

Ah, I see.

Answer 20

Thus, transforming that original term, you have:\[\int\limits_{}^{}\frac{(1+x)}{2(1-x^2)}dx=\int\limits_{}^{}\frac{1}{2(1-x^2)}dx+\int\limits_{}^{}\frac{x}{2(1-x^2)}dx\]\[=\frac{1}{4}\bigg[ -\ln(1-x)+\ln(1+x)-\ln(1-x^2)\bigg]\]Using properties of logarithms:\[\frac{1}{4}\bigg[-\ln(1-x)+\ln(1+x)-\ln(1-x)-\ln(1+x)\bigg] =-\frac{1}{2}\ln(1-x)\]

Answer 21

This is my attempt to fix, but I'm not sure

Answer 22

Because that logarithm was being subtracted, it would have been in the denominator of the log argument, not the numerator, and everything wouldn't as neatly cancel out as hoped.

Answer 23

Yeah, that totally makes sense. I just wanted to see the error in that. Thank you very much!

Answer 24

\[
e^{-(1/2)\ln(1-x)} =(1-x)^{-1/2}= \frac{1}{\sqrt{1-x}}
\]This is a weird one.