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12.Find a particular solution satisfying the given condition:

$x(x^2 -1)\frac{dy}{dx} =1;\ y = 0\ \textup{when} \ x = 2$

Answers (1)

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Given, in the question

$x(x^2 -1)\frac{dy}{dx} =1$

$\\ \implies \int dy=\int \frac{dx}{x(x^2 -1)} \\ \implies \int dy=\int \frac{dx}{x(x -1)(x+1)}$

Let,

Now comparing the values of A,B,C

A + B + C = 0; B-C = 0; A = -1

Solving these:

Now putting the values of A,B,C

Given, y =0 when x =2

Therefore,

$\\ \implies y = \frac{1}{2}\log[\frac{4(x-1)(x+1)}{3x^2}]$

$\\ \implies y = \frac{1}{2}\log[\frac{4(x^2-1)}{3x^2}]$

Posted by

HARSH KANKARIA

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