Let y=y(x) solve dydx+1x^2-1y = (x-1x+1)^1/2, x1, passing through (2,13).

Question

Let $y=y(x)$ solve $\dfrac{dy}{dx}+\dfrac{1}{x^2-1}y = \left(\dfrac{x-1}{x+1}\right)^{1/2}$, $x>1$, passing through $\left(2,\dfrac13\right)$. Then $\sqrt7\,y(8)$ is:

Accepted Answer

I.F. $=\sqrt{\dfrac{x-1}{x+1}}$.
$$y\sqrt{\frac{x-1}{x+1}} = \int\frac{x-1}{x+1}\,dx = x-2\ln|x+1|+C.$$

Using $\left(2,\dfrac13\right)$: $\dfrac13\cdot\dfrac{1}{\sqrt3} = 2-2\ln3+C \implies C=2\ln3-\dfrac53$.

At $x=8$: $y\cdot\dfrac{\sqrt7}{3} = 8-2\ln9+2\ln3-\dfrac53 \implies \sqrt7\,y(8)=19-6\ln3$.

*(Answer key value: B = 19.)*

Solutions · 1