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The solution of differential equation $(x^2-1) \large\frac{dy}{dx}$$+2xy =\large\frac{1}{x^2-1}$

$(a)\;y(x^2-1)=\log | \frac{1+x}{1-x}|+c \\ (b)\;y(x^2-1)=\log | \frac{1-x}{1+x}|+c \\ (c)\;y(x^2-1)=\log | \frac{x-1}{x+1}|+c \\ (d)\;y(x^2-1)=\frac{1}{2}\log | \frac{x-1}{x+1}|+c$
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$\large\frac{dy}{dx}+\frac{2x}{x^2-1}$$y=\large\frac{1}{(x^2-1)^2}$-------(1)
This is linear differential equation
Integrating factor $=e^{\int \large\frac{2x}{(x^2-1)}}$$dx=e^{\log (x^2-1)}=x^2-1$
multiplying both sides of (i) by $I.F =x^2-1$
we get $(x^2-1)\large\frac{dy}{dx}$$+2xy=\large \frac{1}{x^2}$$-1$
$y(x^2-1) -\int\large\frac{1}{x^2-1} dx+c$
=> $y(x^2-1)=\large\frac{1}{2} $$\log |\large\frac{x-1}{x+1}|+c$
Hence d is the correct answer.
answered Feb 5, 2014 by meena.p
 

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