If for the solution curve of dydx+( x)y=2+ x(1+2 x)^2, x(-2,2), f!(3)=√310, then f!(4) is …

Question

If for the solution curve of $\dfrac{dy}{dx}+(	an x)y=\dfrac{2+\sec x}{(1+2\sec x)^2}$, $x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}ight)$, $f\!\left(\dfrac{\pi}{3}ight)=\dfrac{\sqrt3}{10}$, then $f\!\left(\dfrac{\pi}{4}ight)$ is equal to:

Accepted Answer

I.F. $=e^{\int\tan x\,dx}=\sec x$. With Weierstrass substitution $t=\tan(x/2)$, $\cos x=\dfrac{1-t^2}{1+t^2}$, the RHS integral reduces (after substitution $u=t+3/t$) to
$$y\sec x = \frac{2}{t+3/t}+c.$$

Using $x=\pi/3 \Rightarrow t=1/\sqrt3$: $c=0$.

At $x=\pi/4 \Rightarrow t=\sqrt2-1$:
$$y\sqrt2 = \frac{2(\sqrt2-1)}{6-2\sqrt2} \implies y = \frac{4-\sqrt2}{14}.$$

Solutions · 1