If y=y(x) solves dydx+4xx^2-1y = x+2(x^2-1)^5/2, x1, with y(2)=29_e(2+√3) and y(√2)=_e(+)…

Question

If $y=y(x)$ solves $\dfrac{dy}{dx}+\dfrac{4x}{x^2-1}y = \dfrac{x+2}{(x^2-1)^{5/2}}$, $x>1$, with $y(2)=\dfrac29\log_e(2+\sqrt3)$ and $y(\sqrt2)=\alpha\log_e(\sqrt\alpha+\beta)+\beta-\sqrt\gamma$, $\alpha,\beta,\gamma\in\mathbb{N}$, then $\alpha\beta\gamma$ is \_\_\_\_\_\_.

Answer: 6

Accepted Answer

I.F. $=(x^2-1)^2$.
$$y(x^2-1)^2 = \int\big[(x+2)\big]dx\ \text{(after simplifying the integrand)} = \sqrt{x^2-1}+2\ln x+\sqrt{x^2-1}+C.$$

Using $y(2)$ condition fixes $C=-\sqrt3$. At $x=\sqrt2$: $y=1+2\ln(\sqrt2+1)-\sqrt3$, giving $\alpha=2,\beta=1,\gamma=3$, so $\alpha\beta\gamma=6$.

Solutions · 1