Let y=y(x) be the solution curve of the differential equation dy dx+ 1 x^2-1y=( x-1 x+1)^ 1 2, x>1 passing through the point (2,√ 1 3). Then √7y(8)…

Given,

$\frac{dy}{dx}+\frac{y}{{x}^{2}-1}={(\frac{x-1}{x+1})}^{\frac{1}{2}}$dxdy​+x2−1y​=(x+1x−1​)21​ is a linear differential equation.

Here $\displaystyle I.F.={e}^{\int \frac{dx}{{x}^{2}-1}}={e}^{\frac{1}{2}\mathrm{ln}(\frac{x-1}{x+1})}=\sqrt{\frac{x-1}{x+1}}$I.F.=e∫x2−1dx​=e21​ln(x+1x−1​)=x+1x−1​​

General solution will be $\displaystyle y\sqrt{\frac{x-1}{x+1}}=\int \frac{x-1}{x+1}dx+C$yx+1x−1​​=∫x+1x−1​dx+C

$$\Rightarrow y\sqrt{\frac{x-1}{x+1}}=x-2\mathrm{ln}(x+1)+c$$⇒yx+1x−1​​=x−2ln(x+1)+c

Given $y(2)=\frac{1}{\sqrt{3}}$y(2)=3​1​

$$\Rightarrow c=2\mathrm{ln}3-\frac{5}{3}$$⇒c=2ln3−35​

So, the equation of the curve is $y\sqrt{\frac{x-1}{x+1}}=x-2\mathrm{ln}(x+1)+2\mathrm{ln}3-\frac{5}{3}$yx+1x−1​​=x−2ln(x+1)+2ln3−35​

Now putting $x=8$x=8, we get

$$\frac{\sqrt{7}y(8)}{3}=8-4\mathrm{ln}3+2\mathrm{ln}3-\frac{5}{3}$$37​y(8)​=8−4ln3+2ln3−35​

$$\Rightarrow \sqrt{7}y(8)=19-6\mathrm{ln}3$$⇒7​y(8)=19−6ln3