The point (1,−1) satisfies the inequality 40x−22y≤50.

Determine the singular points of the given differential equation. Classify each singular point as regular or irregular. (Enter y

ludmilkaskok [199]

Answer:

Step-by-step explanation:

Given that:

The differential equation; $(x^2-4)^2y'' + (x + 2)y' + 7y = 0$

The above equation can be better expressed as:

$y'' + \dfrac{(x+2)}{(x^2-4)^2} \ y'+ \dfrac{7}{(x^2- 4)^2} \ y=0$

The pattern of the normalized differential equation can be represented as:

y'' + p(x)y' + q(x) y = 0

This implies that:

$p(x) = \dfrac{(x+2)}{(x^2-4)^2} \$

$p(x) = \dfrac{(x+2)}{(x+2)^2 (x-2)^2} \$

$p(x) = \dfrac{1}{(x+2)(x-2)^2}$

Also;

$q(x) = \dfrac{7}{(x^2-4)^2}$

$q(x) = \dfrac{7}{(x+2)^2(x-2)^2}$

From p(x) and q(x); we will realize that the zeroes of (x+2)(x-2)² = ±2

When x = - 2

$\lim \limits_{x \to-2} (x+ 2) p(x) = \lim \limits_{x \to2} (x+ 2) \dfrac{1}{(x+2)(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{1}{(x-2)^2}$

$\implies \dfrac{1}{16}$

$\lim \limits_{x \to-2} (x+ 2)^2 q(x) = \lim \limits_{x \to2} (x+ 2)^2 \dfrac{7}{(x+2)^2(x-2)^2}$

$\implies \lim \limits_{x \to2} \dfrac{7}{(x-2)^2}$

$\implies \dfrac{7}{16}$

Hence, one (1) of them is non-analytical at x = 2.

Thus, x = 2 is an irregular singular point.

5 0

3 years ago

A rhombus has diagonals of 10cm and 6cm. Find the angles of the rhombus and its area.

Marrrta [24]

Answer:

area =length ×breath =10×6=60

3 0

3 years ago

Can you guys please help. I really don't know how to solve this

Nonamiya [84]

The inequalities that can be used to represent this problem are x < 0 or x > 5. On the number line, you will want to place an open circle on both 0 and 5. The open circle signifies that x is not equal to these values, but is greater or less than. On the open circle on 0, draw an arrow pointing left, toward the negative values. On the open circle on 5, draw an arrow pointing right, toward the positive values.

Hope this helps!! :)

3 0

3 years ago

Find the difference between points M(6,16) and Z(-1,14) to the nearest tenth.

svlad2 [7]

I'm not sure what you mean by difference, so I am guessing you meant to write "distance". If this is not the case, then feel free to report me. Anyway, to find the distance between two points you would use the distance formula.

Distance Formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .

Let's substitute our point coordinates into the formula. We'll substitute -1 and 6 for x_2 and x_1, and 14 and 16 for y_2 and y_1. After substituting, your formula will now look like: $d=\sqrt{(-1-6)^2+(14-16)^2}$ .