6
$\begingroup$

I am currently solving the following differential equation (link is to another post):

$\dfrac{dr}{d \theta}+r\tan \theta =\frac{1}{\cos \theta}$

The following is in standard form (i.e. $\dfrac{dr}{d\theta}+P(\theta)r=Q(\theta)$). Therefore, I can go and head and solve for the integrating factor:

$\mu(\theta)=e^{\int_{} P(\theta) d\theta}=e^{\int_{} \tan(\theta) d\theta} =e^{-\ln(|\cos(\theta)|)}=|\cos(\theta)|^{-1}$

Multiplying the entire equation by the integration factor allows us to use the "Derivative of a Product" property to yield the following:

$\dfrac{d}{dx}(|\cos(\theta)|^{-1}r)=|\cos(\theta)|^{-1}\sec(\theta)$

Integrating both sides yields a "difficult" integral:

$\int_{} \dfrac{1}{|\cos(\theta)|\cos(\theta)} d\theta$

However, according to solution given here, the absolute value is dropped in the integrating factor (thereby creating an easier problem), meaning $\mu(\theta)=(cos(\theta))^{-1}$. But, why am I allowed to drop the absolute value? Nothing in the problem states the domain of $\theta$ or $r$ and clearly, $|\cos(\theta)|\cos(\theta)\neq \cos^2(\theta)$ for all values of $\theta$.

$\endgroup$
1
  • 3
    $\begingroup$ You can format trig functions like $\cos(\theta)$ or $\tan(\theta)$ or most trig functions, by including immediately before it, a backslash. E.g., \cos(\theta) and \tan(\theta). Similarly, instead of $ln(x)$ we can use a backslash immediately prior to ln, by writing \ln(x). $\endgroup$
    – amWhy
    Aug 27, 2018 at 19:55

3 Answers 3

4
$\begingroup$

$|\cos(\theta)|^{-1}$, 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 $|\cos(\theta)|=\cos(\theta)$.

$\endgroup$
1
  • $\begingroup$ +1 Makholm for pointing this out... $\endgroup$ Aug 27, 2018 at 21:15
2
$\begingroup$

The integrating factor of a differential equation is unique up to a nonzero multiplicative constant.

E.g., the integrating factor of the differential equation $$\dfrac{\mathrm{d}y}{\mathrm{d}x}+g(x)\,y=h(x)$$ is in general $Me^{\int g(x)\,\mathrm{d}x}\,\left(M\neq0\right),$ which attains the solution independently of $M$: \begin{equation} \begin{aligned} \left(Me^{\int g(x)\,\mathrm{d}x}\right)\dfrac{\mathrm{d}y}{\mathrm{d}x}+\left(Me^{\int g(x)\,\mathrm{d}x}\right)g(x)\,y&=\left(Me^{\int g(x)\,\mathrm{d}x}\right)h(x)\\ e^{\int g(x)\,\mathrm{d}x}\dfrac{\mathrm{d}y}{\mathrm{d}x}+e^{\int g(x)\,\mathrm{d}x}g(x)\,y&=e^{\int g(x)\,\mathrm{d}x}h(x)\\ \dfrac{\mathrm{d}}{\mathrm{d}x}\left(e^{\int g(x)\,\mathrm{d}x}\,y\right)&=e^{\int g(x)\,\mathrm{d}x}h(x)\\ y&=\frac1{e^{\int g(x)\,\mathrm{d}x}}\int e^{\int g(x)\,\mathrm{d}x}h(x)\,\mathrm{d}x. \end{aligned} \end{equation}

Dropping any external absolute-value symbol (and the constant of integration) from the integrating factor is equivalent to setting $M=1,$ which is the simplest choice of $M$.

$\endgroup$
0
$\begingroup$

The absolute values originate from logarithmic integrals such as

$$\int\frac{dx}x=\log|x|+C,$$

and often the antilogarithm is taken, giving

$$e^{\log|x|+C}=C'|x|.$$

But $x=0$ corresponds to a singularity and shouldn't be crossed (otherwise the integral is improper). So all $x$'s should have the same sign, so that the correct expressions should be

$$\log x\text{ or }\log(-x)$$

and

$$C'x$$ respectively, where $C'$ can be positive or negative.

And if differentiabiliy is not required at $x=0$, you can have two distinct pieces,

$$\begin{cases}x<0\to C_-x,\\x>0\to C_+x.\end{cases}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.