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  • Find a particular solution satisfying the given condition: x(x^2 - 1) dy/dx = 1; y = 0 when x = 2

12.Find a particular solution satisfying the given condition:

$x\left(x^2-1\right) \frac{d y}{d x}=1 ; y=0$ when $x=2$

Answers (1)

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Given, in the question $x\left(x^2-1\right) \frac{d y}{d x}=1$

$\Longrightarrow \int d y=\int \frac{d x}{x\left(x^2-1\right)}$
$\Longrightarrow \int d y=\int \frac{d x}{x(x-1)(x+1)}$
Let,
$\Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{c}{x-1}$

$\Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{A(x-1)(x+1)+B(x)(x-1)+C(x)(x+1)}{x(x+1)(x-1)}$
$\Rightarrow \frac{1}{x(x+1)(x-1)}=\frac{(A+B+C) x^2+(B-C) x-A}{x(x+1)(x-1)}$

Now comparing the values of $A,B,C$

$A + B + C = 0;  B-C = 0;  A = -1$

Solving these:

$\mathrm{B}=\frac{1}{2} \text { and}\mathrm{C}=\frac{1}{2}$

Now putting the values of $A,B,C$,

$\Rightarrow \frac{1}{x(x+1)(x-1)}=-\frac{1}{x}+\frac{1}{2}\left(\frac{1}{x+1}\right)+\frac{1}{2}\left(\frac{1}{x-1}\right)$
$\Rightarrow \int d y=-\int \frac{1}{x} d x+\frac{1}{2} \int\left(\frac{1}{x+1}\right) d x+\frac{1}{2} \int\left(\frac{1}{x-1}\right) d x$
$\Rightarrow y=-\log x+\frac{1}{2} \log (x+1)+\frac{1}{2} \log (x-1)+\log c$

$\left.\Rightarrow \mathrm{y}=\frac{1}{2} \log \left[\frac{\mathrm{c}^2(\mathrm{x}-1)(\mathrm{x}+1)}{\mathrm{x}^2}\right\}----\mathrm{iii}\right)$
Given, $\mathrm{y}=0$ when $\mathrm{x}=2$
$0=\frac{1}{2} \log \left[\frac{\mathrm{c}^2(2-1)(2+1)}{4}\right\}$
$\Rightarrow \log \frac{3 \mathrm{c}^2}{4}=0$

$\Rightarrow 3 c^2=4$
Therefore,
$\Longrightarrow y=\frac{1}{2} \log \left[\frac{4(x-1)(x+1)}{3 x^2}\right]$
$\Longrightarrow y=\frac{1}{2} \log \left[\frac{4\left(x^2-1\right)}{3 x^2}\right]$

Posted by

HARSH KANKARIA

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