## 17Calculus Differential Equations - Singular Points

On the page where we discussed solving differential equations using power series, we never really discussed under what conditions that technique could be successfully applied. There are limitations, which we discuss on this page.

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We are going to look at the general differential equation $P(x)y'' + Q(x)y' + R(x)y = 0$ Notice that this is homogeneous since the right side is zero. Also, $$P(x), Q(x)$$ and $$R(x)$$ are functions of $$x$$ only, i.e. there are not y terms or derivatives of x.
Singular points occur where $$P(x) = 0$$. At these points, the $$y''$$ term disappears, which we do not want to happen. Also, we will often write this as $y'' + \frac{Q(x)}{P(x)}y' + \frac{R(x)}{P(x)}y = 0$ and, of course, we do not want zero in the denominators of the second and third terms.

So to find the singular points, it is probably best to write the differential equation in second form above and determine the values of x where the denominators of each of the fractions on the left are zero.

We also require that at least one of $$Q(x)$$ and $$R(x)$$ is not zero at the singular points. Here is how we determine this. Let's call $$x_0$$ a point where $$P(x_0) = 0$$. Using equations, we calculate the two limits

$\lim_{x-x_0}{ (x-x_0) \frac{Q(x)}{P(x)}} ~ \text{ and } ~ \lim_{x-x_0}{ (x-x_0)^2 \frac{R(x)}{P(x)}}$

If both of these limits are finite, then $$x=x_0$$ is a regular singular point of the differential equation.

If one (or both) of these limits is infinite or undefined, then $$x=x_0$$ is an irregular singular point.

Analytic Functions

If the functions, $$P(x), Q(x)$$ and $$R(x)$$ are polynomials, then we don't need any constraints other than the above limits when checking singular points. However, for more complicated functions, we require that both $(x-x_0) \frac{Q(x)}{P(x)} ~ \text{ and } ~ (x-x_0)^2 \frac{R(x)}{P(x)}$ have convergent Taylor Series about $$x_0$$. The term we use to describe this to say that the two functions are analytic.

Fortunately, as long as we have polynomials, exponentials, logarithms, sines or cosines, we are good. They are all analytic and have convergent Taylor Series.

Book Recommendation

If you want more detail on power series solution and radius convergence, this book seems to cover the material in more detail than most other textbooks we have looked at.

You CAN Ace Differential Equations

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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