Let $$y = y(x)$$ be a solution curve of the differential equation $$(y+1)\tan^2 x \, dx + \tan x \, dy + y \, dx = 0$$, $$x \in \left(0, \frac{\pi}{2}\right)$$. If $$\lim_{x \to 0^+} xy(x) = 1$$, then the value of $$y\left(\frac{\pi}{4}\right)$$ is:

We are given the differential equation

$$ (y+1)\tan^2 x \;dx+\tan x \;dy + y \;dx =0 ,\qquad 0<x<\dfrac{\pi}{2}. $$

First we collect the differentials so that it looks like the standard exact-differential form $$M(x,y)\,dx+N(x,y)\,dy=0.$$

$$\begin{aligned} (y+1)\tan^2 x \;dx+y\;dx+\tan x\;dy &=0\\ \bigl((y+1)\tan^2 x+y\bigr)\,dx+\tan x\,dy &=0. \end{aligned}$$

Thus we have

$$M(x,y)=y\tan^2 x+\tan^2 x+y, \qquad N(x,y)=\tan x.$$

For exactness we must have $$\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}.$$ Let us check:

$$\frac{\partial M}{\partial y}= \tan^2 x+1,\qquad\frac{\partial N}{\partial x}= \sec^2 x.$$

Remembering the trigonometric identity $$\tan^2 x+1=\sec^2 x,$$ we find that the two partial derivatives are equal, so the differential equation is indeed exact.

Therefore there exists a potential function $$F(x,y)$$ such that

$$\frac{\partial F}{\partial x}=M,\qquad\frac{\partial F}{\partial y}=N.$$

Integrating $$M$$ with respect to $$x$$ while treating $$y$$ as a constant, we write

$$F(x,y)=\int\Bigl((y+1)\tan^2 x+y\Bigr)\,dx+g(y),$$

where $$g(y)$$ is an arbitrary function of $$y$$ (because the partial derivative with respect to $$x$$ “misses” such a term).

We now evaluate the integral. First recall the identity $$\tan^2 x=\sec^2 x-1.$$ Also recall the elementary antiderivative

$$\int \sec^2 x\,dx=\tan x.$$

Hence

$$\int\tan^2 x\,dx=\int(\sec^2 x-1)\,dx=\tan x - x +C.$$

Using this result we get

$$\begin{aligned} F(x,y)=&(y+1)\bigl(\tan x-x\bigr)+y x+g(y). \end{aligned}$$

Next we enforce the second requirement $$\dfrac{\partial F}{\partial y}=N=\tan x.$$ Differentiating $$F$$ with respect to $$y$$ gives

$$\frac{\partial F}{\partial y}=\bigl(\tan x-x\bigr)+x+g'(y)=\tan x+g'(y).$$

Setting this equal to $$N=\tan x$$ immediately forces $$g'(y)=0$$, so $$g(y)$$ must be a constant, which we absorb into the constant of the final solution. Consequently the exact integral of the differential equation is

$$F(x,y)=(y+1)\bigl(\tan x-x\bigr)+y x=C,$$

where $$C$$ is a constant determined by the initial/limit condition.

We are told that the solution satisfies the limiting condition

$$\lim_{x\to0^+}x\,y(x)=1.$$

To exploit this, let us study the behaviour of $$F(x,y)$$ as $$x\to0^+$$. For small $$x$$ we use the Maclaurin expansion $$\tan x = x+\dfrac{x^3}{3}+O(x^5).$$ Therefore

$$\tan x - x = \frac{x^3}{3}+O(x^5).$$

Substituting into $$F(x,y)$$ we write, for small $$x,$$

$$F(x,y)= (y+1)\Bigl(\tfrac{x^3}{3}+O(x^5)\Bigr)+y x.$$

The limit information tells us that $$y x\to 1$$ as $$x\to0^+,$$ hence near the origin $$y\approx\dfrac{1}{x}.$$ Then

$$ (y+1)\frac{x^3}{3} \approx \frac{1}{x}\cdot \frac{x^3}{3}= \frac{x^2}{3}\xrightarrow[x\to0]{}0.$$

So the first bracketed term vanishes in the limit, while $$y x\to1.$$ Consequently

$$C=\lim_{x\to0^+}F(x,y)=1.$$

Thus the explicit implicit equation of the required solution curve is

$$ (y+1)\bigl(\tan x-x\bigr)+y x = 1. $$

We are asked to find $$y\!\left(\dfrac{\pi}{4}\right).$$ Put $$x=\dfrac{\pi}{4}$$; recall that $$\tan\dfrac{\pi}{4}=1.$$ Then

$$\tan x - x = 1-\frac{\pi}{4}.$$

Introduce the unknown value $$y_0=y\!\left(\dfrac{\pi}{4}\right).$$ Substituting $$x=\dfrac{\pi}{4},\;y=y_0$$ into the implicit relation gives

$$\Bigl(y_0+1\Bigr)\Bigl(1-\frac{\pi}{4}\Bigr)+y_0\cdot\frac{\pi}{4}=1.$$

Let us now isolate $$y_0$$ step by step:

$$\begin{aligned} \bigl(1-\tfrac{\pi}{4}\bigr)y_0 &+ \bigl(1-\tfrac{\pi}{4}\bigr) + \tfrac{\pi}{4}\,y_0 =1,\\[4pt] \Bigl(\,1-\tfrac{\pi}{4}+\tfrac{\pi}{4}\Bigr)y_0 &+ \bigl(1-\tfrac{\pi}{4}\bigr) =1,\\[4pt] 1\cdot y_0 &+ \bigl(1-\tfrac{\pi}{4}\bigr)=1,\\[4pt] y_0 &= 1-\Bigl(1-\tfrac{\pi}{4}\Bigr)=\frac{\pi}{4}. \end{aligned}$$

Thus

$$y\!\left(\frac{\pi}{4}\right)=\frac{\pi}{4}.$$

Hence, the correct answer is Option C.