Question

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

Answer 1

$\frac{dy}{dx}+\frac{1}{x^2-1}y=\sqrt{\frac{x-1}{x+1}},x>1$
Integrating factor
I.F. $=e^{\int \frac{1}{x^2-1}dx}=e^{\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|}$
$=\sqrt{\frac{x-1}{x+1}}$
Solution of differential equation
$y\sqrt{\frac{x-1}{x+1}}=\int \frac{x-1}{x+1}dx=\int (1-\frac{2}{x+1})dx$
$y\sqrt{\frac{x-1}{x+1}}=x-2\ln |x+1|+C$
Curve passes through $(2,\sqrt{\frac{1}{3}})$
$\frac{1}{\sqrt{3}}\times \frac{1}{\sqrt{3}}=2-2\ln 3+C$
$C=2\ln 3-\frac{5}{3}$
Now$y\sqrt{\frac{x-1}{x+1}}=x-2\ln |x+1|+2\ln 3-\frac{5}{3}$
$y\left(8\right)\times \frac{\sqrt{7}}{3}=8-2\ln 9+2\ln 3-\frac{5}{3}$
$\sqrt{7}\cdot y\left(8\right)=19-6\ln 3$