[Math] Absolute value in integrating factor of First-Order Linear Differential Equation

integration

Question states:
$$ y' + \frac{y}{x} = 6x+2$$
Obviously x cannot be zero. If we assume that $x$ is positive (i.e. $x>0$), we find the integrating factor as $$u(x)=e^{\int \frac{1}{x} dx}$$ which is equal to $x$. Then the solution is $$y(x)= \frac{1}{u(x)} \int (6x+2)(u(x)) dx = \frac {1}{x} \int 6x^2+2x \ dx = 2x^2+x+\frac{C}{x}.$$
Now, we assumed that $x$ is positive. But I couldn't get the same answer when I didn't make this assumption; that is, the integrating factor is $$u(x)=e^{\int \frac{1}{x} dx} = e^{ \ln \lvert x\rvert} = \lvert x\rvert.$$
Then this problem gets way more complicated, as the solution becomes
$$y(x)= \frac{1}{u(x)} \int (6x+2)(u(x)) dx = \frac {1}{\lvert x\rvert} \int 6x \lvert x\rvert +2\lvert x\rvert \ dx.$$
My calculus textbook omitted the absolute value altogether; that is, the textbook indicated that the integrating factor was just $x$. Because the textbook is written by quite reputable and trustworthy authors (Ron Larson and Bruce Edwards), I was wondering (1) if treating the integrating factor as just $x$ is acceptable, and/or (2) How the solution is still correct if we must use the integrating factor as $\lvert x\rvert$. If we can omit the absolute value sometimes, how do we know when we can omit the absolute value sign and when we shouldn't? (As a side note, I fully understand why there's absolute value sign for the antidervative of $ \frac{1}{x} $).

Best Answer

You see that in your last equation attain the same result for $x>0$ and $x<0$. So absolute sign is omitted.

Related Solutions

Integration – Absolute Value in Integrals

In calculus courses one deals with the function $\ln\colon(0,+\infty)\to\mathbb R$. For $x>0$ it holds that $$ \frac{d}{dx}\ln x=\frac{1}{x}. $$ Note that the function $x\mapsto 1/x$ is defined for $x\in\mathbb R$ with $x\neq 0$. Integrating, one has $$ \int \frac{1}{x}\,dx= \begin{cases} \ln x+C_1 & x>0\\ \ln (-x)+C_2 & x<0. \end{cases} \tag{*} $$ This is sometimes written $$ \int\frac{1}{x}\,dx=\ln |x|+C $$ but in applications one consider either $x>0$ or $x<0$.

In complex analysis one also deals with a logarithm, taking complex numbers as arguments (I will not dig into the subject of branches here). One common definition is $$ \log z=\ln |z|+i\arg z. $$ Here $\ln$ is the logarithm defined in calculus. $\arg$ is the argument function. As it happens, also $$ \frac{d}{dz}\log z=\frac{1}{z}. $$ Here $\frac{d}{dz}$ is the complex derivative. Thus, in the complex setting $$ \int\frac{1}{z}\,dz=\log z+C, $$ i.e. the logarithm is still an antiderivative of $1/z$. Here, again, one should be more careful with domains, but since I understand it as OP is not into complex analysis yet (correct me if wrong), I do not dwell about that.

Finally, maple has no reason to expect the $x$ in $1/x$ to be real, so when you hit

 int(1/x,x);

maple makes no assumption on $x$, but return

$\log x$.

Note that, if $x$ happens to be real and positive, and one assumes $\arg x=0$ (which is natural), then $$ \log x=\ln|x|+i\arg x=\ln x $$ so the different logarithms really coincide.

What should you do?

I suggest that you, every time you must integrate something that will return a logarithm, pause and think if you know anything about the variable. If you know that $x$ is positive, then $\int 1/x\,dx=\ln x+C$. If you know that $x$ is negative, then $\int 1/x\,dx=\ln(-x)+C=\ln|x|+C$ and if you do not know, then use $(*)$ above.

[Math] Deciding when to drop the absolute values in differential equation

$|\cos(\theta)|^{-1}$, or for that matter $\int \frac{d\theta}{|\cos\theta|\cos \theta}$, diverges to infinity for $\theta\to\pm\pi/2$ -- so if you're interested only in the connected component of the solution that contains $\theta=0$, it will only be defined on the open interval $(-\pi/2,\pi/2)$ anyway. In this interval $\cos(\theta)$ is always positive, and therefore $|\cos(\theta)|=\cos(\theta)$.

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