Question

Let $y=y(x)$ be the solution curve of the differential equation $\frac{dy}{dx}+\frac{1}{x^2-1}y={\left(\frac{x-1}{x+1}\right)}^{\frac{1}{2}},x>1$ 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)

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