Let y=y(x) solve (y+1)^2x,dx+ x,dy+y,dx=0, x(0,2).

Question

Let $y=y(x)$ solve $(y+1)	an^2x\,dx+	an x\,dy+y\,dx=0$, $x\in\left(0,\dfrac{\pi}{2}ight)$. If $\lim_{x	o0^+}xy(x)=1$, find $y\left(\dfrac{\pi}{4}ight)$.

Accepted Answer

$$\frac{dy}{dx}+\frac{\sec^2x}{\tan x}y = -\tan x.$$

I.F.$=\tan x$. $y\tan x = x-\tan x+C$.

Using the limit condition: $C=1$. So $y=x\cot x-1+\cot x$. At $x=\pi/4$: $y=\dfrac{\pi}{4}-1+1=\dfrac{\pi}{4}$.

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