Question

If the integrating factor of the differential equation is $(x^2-1)\frac{dy}{dx}+2xy=x$ is $f\left(x\right)$ the value of $\frac{f\left(x\right)}{x^2}$ as $x\to \infty$ is

• A. $1$
• B. $2$
• C. $-1$
• D. $0$

Answer 1

Answer: (A)

$1$

$\frac{dy}{dx}+\frac{2x}{x^2-1}y=\frac{x}{x^2-1}⟹f\left(x\right)=e^{\int \frac{2x}{x^2-1}dx}=x^2-1⟹\underset{x\to \infty }{lim}\frac{x^2-1}{x^2}=1$