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3 years ago

10

Someone please help me

1 answer:

Mars2501 [29]3 years ago

4 0

Answer:

x=-1,-2

Step-by-step explanation:

Multiply the given equation with (x+1) on both sides. We'll get,

x+2/x=-3

or,x^2+2=-3x

or,x^2+3x+2=0

or,x^2+2x+x+2=0

or,x(x+2)+1(x+2)=0

or,(x+2)(x+1)=0

or,x=-2,-1

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ludmilkaskok [199]

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Step-by-step explanation:

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

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

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

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

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

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

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

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

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

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

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

Thus, x = 2 is an irregular singular point.

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