Let y = y(x) be the solution curve of the differential equation dydx+1x2−1y=(x−1x+1)1/2, x>1 passing through the point (2,13). Then, 7y(8) is

Given , diferential equation dydx+1x2−1y=(x−1x+1)12,x>1Which is of first order linear differential equations .Here,p= p=1x2−1,Q=X−1X+1 IF = e∫pdx= e ∫1x2−1dx=e12log|x−1x+1|=x−1x+1Now ,solution is given by y.IF= ∫Q.IF dx⇒ yx−1x+1=∫x−1x+1x−1x+1dx ⇒ yx−1x+1=∫x−1x+1dx ⇒ yx−1x+1=∫(1−2x+1)dx ⇒ yx−1x+1=x−2log(x+1)+c∴ The curve passes through (2,13)∴ 1313=2−2log3+C ⇒ 13−2+2log3=C ⇒ C=2 log 3−53 ∴ yx−1x+1=x−2log(x+1)+2log 3−53 y= x+1x−1[x−2 log (x+1)+2log3−53] y(8)=37[8−2log9+2log3−53] 7y(8)=3[193−2log32+2log 3] =3[19−12log3+6log33]=19−6log3

Let y = y(x) be the solution curve of the differential equation dydx+1x2−1y=(x−1x+1)1/2, x>1 passing through the point (2,13). Then, 7y(8) is

Given , diferential equation dydx+1x2−1y=(x−1x+1)12,x>1Which is of first order linear differential equations .Here,p= p=1x2−1,Q=X−1X+1 IF = e∫pdx= e ∫1x2−1dx=e12log|x−1x+1|=x−1x+1Now ,solution is given by y.IF= ∫Q.IF dx⇒ yx−1x+1=∫x−1x+1x−1x+1dx ⇒ yx−1x+1=∫x−1x+1dx ⇒ yx−1x+1=∫(1−2x+1)dx ⇒ yx−1x+1=x−2log(x+1)+c∴ The curve passes through (2,13)∴ 1313=2−2log3+C ⇒ 13−2+2log3=C ⇒ C=2 log 3−53 ∴ yx−1x+1=x−2log(x+1)+2log 3−53 y= x+1x−1[x−2 log (x+1)+2log3−53] y(8)=37[8−2log9+2log3−53] 7y(8)=3[193−2log32+2log 3] =3[19−12log3+6log33]=19−6log3

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Let y = y(x) be the solution curve of the differential equation $\frac{d y}{d x} + \frac{1}{x^{2} - 1} y = {(\frac{x - 1}{x + 1})}^{1 / 2}, x > 1$ passing through the point $(2, \sqrt{\frac{1}{3}}) .$ Then, $\sqrt{7} y (8)$ is

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$12 - 2 \log_{e} 3$

$11 + 6 \log_{e} 3$

$19 - 6 \log_{e} 3$

answer is D.

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Detailed Solution

Given , diferential equation
$\frac{d y}{d x} + \frac{1}{x^{2} - 1} y = {(\frac{x - 1}{x + 1})}^{\frac{1}{2}}, x > 1$
Which is of first order linear differential equations .
Here,p= $p = \frac{1}{x^{2} - 1}, Q = \sqrt{\frac{X - 1}{X + 1}}$
IF = $e^{\int p d x} = e^{\int \frac{1}{x^{2} - 1} d x} = e^{\frac{1}{2} \log | \frac{x - 1}{x + 1} |} = \sqrt{\frac{x - 1}{x + 1}}$
Now ,solution is given by
y.IF= $\int Q . I F d x$
$\Rightarrow y \sqrt{\frac{x - 1}{x + 1}} = \int \sqrt{\frac{x - 1}{x + 1}} \sqrt{\frac{x - 1}{x + 1}} d x \Rightarrow y \sqrt{\frac{x - 1}{x + 1}} = \int \frac{x - 1}{x + 1} d x \Rightarrow y \sqrt{\frac{x - 1}{x + 1}} = \int (1 - \frac{2}{x + 1}) d x \Rightarrow y \sqrt{\frac{x - 1}{x + 1}} = x - 2 \log (x + 1) + c$
∴ The curve passes through $(2, \sqrt{\frac{1}{3}})$
$∴ \sqrt{\frac{1}{3}} \sqrt{\frac{1}{3}} = 2 - 2 \log 3 + C \Rightarrow \frac{1}{3} - 2 + 2 \log 3 = C \Rightarrow C = 2 \log 3 - \frac{5}{3} ∴ y \sqrt{\frac{x - 1}{x + 1}} = x - 2 \log (x + 1) + 2 \log 3 - \frac{5}{3} y = \sqrt{\frac{x + 1}{x - 1}} [x - 2 \log (x + 1) + 2 \log 3 - \frac{5}{3}] y (8) = \frac{3}{\sqrt{7}} [8 - 2 \log 9 + 2 \log 3 - \frac{5}{3}] \sqrt{7} y (8) = 3 [\frac{19}{3} - 2 \log 3^{2} + 2 \log 3] = 3 [\frac{19 - 12 \log 3 + 6 \log 3}{3}] = 19 - 6 \log 3$