Let \(y = y(x)\) be a solution curve of the differential equation \((y + 1) \tan^2x dx + \tan x dy + y dx = 0, x \in \left(0, \frac{\pi}{2}\right)\). If \(\lim_{x \to 0^+} xy(x) = 1\), then the value of \(y\left(\frac{\pi}{4}\right)\) is :
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Always check if a differential equation can be written as \(\frac{d}{dx}(y \cdot g(x)) = h(x)\). Here, noticing that \(\frac{d}{dx}(\tan x) = \sec^2 x\) allowed for an easy exact derivative identification.
Step 1: Understanding the Concept:
We rearrange the differential equation into a linear form. The equation is solvable using an integrating factor or by direct identification of an exact derivative. We then use the initial condition to find the constant. Step 2: Key Formula or Approach:
1. \(\frac{dy}{dx} + P(x)y = Q(x)\).
2. \(y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) dx\). Step 3: Detailed Explanation:
Rearrange the equation:
\[ \tan x \frac{dy}{dx} + y(\tan^2x + 1) = -\tan^2x \implies \tan x \frac{dy}{dx} + y \sec^2x = -\tan^2x \]
Dividing by \(\tan x\):
\[ \frac{dy}{dx} + y\frac{\sec^2x}{\tan x} = -\tan x \]
This is a linear differential equation. \(\text{I.F.} = e^{\int \frac{\sec^2x}{\tan x} dx} = e^{\ln(\tan x)} = \tan x\).
Multiplying through:
\[ \frac{d}{dx}(y \tan x) = -\tan^2x = 1 - \sec^2x \]
Integrating:
\[ y \tan x = \int (1 - \sec^2x) dx = x - \tan x + C \]
\[ y = \frac{x}{\tan x} - 1 + \frac{C}{\tan x} \]
Given \(\lim_{x \to 0^+} xy = 1\):
\[ \lim_{x \to 0^+} \left( \frac{x^2}{\tan x} - x + \frac{Cx}{\tan x} \right) = 0 - 0 + C = 1 \implies C = 1 \]
Thus, \(y = \frac{x+1}{\tan x} - 1\).
At \(x = \pi/4\):
\[ y(\pi/4) = \frac{\pi/4 + 1}{1} - 1 = \frac{\pi}{4} \] Step 4: Final Answer:
The value of \(y(\pi/4)\) is \(\frac{\pi}{4}\).
