Let y=y(x) solve dydx+√2,y2^4x-2x = xe^^-1(√22x), 0

Question

Let $y=y(x)$ solve $\dfrac{dy}{dx}+\dfrac{\sqrt2\,y}{2\cos^4x-\cos2x} = xe^{	an^{-1}(\sqrt2\cot2x)}$, $0<x<\dfrac{\pi}{2}$, with $y\left(\dfrac{\pi}{4}ight)=\dfrac{\pi^2}{32}$. If $y\left(\dfrac{\pi}{3}ight)=\dfrac{\pi^2}{18}e^{-	an^{-1}(\alpha)}$, then $3\alpha^2$ is \_\_\_\_\_\_.

Answer: 2

Accepted Answer

$$\int\frac{\sqrt2\,dx}{2\cos^4x-\cos2x} = -\frac{1}{\sqrt2}\tan^{-1}(\sqrt2\cot2x).$$

So I.F. $=e^{-\tan^{-1}(\sqrt2\cot2x)}$, and $ye^{-\tan^{-1}(\sqrt2\cot2x)} = \dfrac{x^2}{2}+c$.

Using $y(\pi/4)=\pi^2/32 \Rightarrow c=0$. So $y = \dfrac{x^2}{2}e^{\tan^{-1}(\sqrt2\cot2x)}$. At $x=\pi/3$:
$$y = \frac{\pi^2}{18}e^{-\tan^{-1}(2/3)} \implies \alpha=\frac23 \implies 3\alpha^2=2.$$

Solutions · 1