Question

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

• A. $11+6 \log _{ e } 3$
• B. 19
• C. $12-2 \log _e 3$
• D. $19-6 \log _e 3$

Answer 1

Answer: (D)

$ \frac{d y}{d x}+\frac{1}{x^2-1} y=\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}}$
$ \frac{d y}{d x}+P y=Q$
$ \text { I.F. }=e^{\int P d x}=\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}} $
$y\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}}=\int\left(\frac{x-1}{x+1}\right)^1 d x$
$ =x-2 \log _e|x+1|+C$
$ \text { Curve passes through }\left(2, \frac{1}{\sqrt{3}}\right)$
$\Rightarrow C=2 \log _e 3-\frac{5}{3}$
$\text { at } x=8, \sqrt{7} y(8)=19-6 \log _{ e } 3$