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nordsb [41]
3 years ago
8

(3a^2+4a-6)-(6a^3-3a+8)Please Explain 15 points :)

Mathematics
2 answers:
Kobotan [32]3 years ago
6 0
Distribute and add like terms
-(6a³-3a+8)=-6a³+3a-8

now add
3a²+4a-6-6a³+3a-8
regroup
-6a³+3a²+4a+3a-6-8
-6a³+3a²+7a-14
dat is expanded and compacted
lions [1.4K]3 years ago
4 0
I think this is what your doing am assuming . forgot but just put -3a+1a solve it and then do -1+2 and you got your anwser

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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.

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