Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x)$ and $Q(x)$ are functions of $x$.
• Integrating Factor (IF): For a first-order linear differential equation, the integrating factor is $e^{\int P(x) dx}$. Multiplying the differential equation by the IF allows us to integrate both sides.
• Limit Definition: Understanding how limits can define constants of integration in differential equation solutions.

Step-by-Step Solution

Step 1: Rewrite the given differential equation.

We are given the differential equation: $(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0$ We want to rearrange it into a standard form that allows us to identify $P(x)$ and $Q(x)$. First, we rewrite the equation: $\tan x \, dy = -[(y+1)\tan^2 x + y] \, dx$ $\frac{dy}{dx} = -\frac{(y+1)\tan^2 x + y}{\tan x} = -\frac{y\tan^2 x + \tan^2 x + y}{\tan x} = -y\tan x - \tan x - \frac{y}{\tan x}$ $\frac{dy}{dx} + y\tan x + y\cot x = -\tan x$ $\frac{dy}{dx} + y(\tan x + \cot x) = -\tan x$

Step 2: Simplify the coefficient of y.

Notice that $\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x} = \frac{2}{2\sin x \cos x} = \frac{2}{\sin 2x} = 2\csc 2x$. Therefore, our differential equation becomes: $\frac{dy}{dx} + 2y\csc 2x = -\tan x$

Step 3: Calculate the Integrating Factor (IF).

The integrating factor is given by $IF = e^{\int P(x) dx}$, where $P(x) = 2\csc 2x$. $IF = e^{\int 2\csc 2x \, dx} = e^{\int \frac{2}{\sin 2x} \, dx}$ Let $u = 2x$, then $du = 2dx$, so $IF = e^{\int \csc u \, du} = e^{\ln |\tan \frac{u}{2}|} = e^{\ln |\tan x|} = |\tan x|$ Since $x \in (0, \frac{\pi}{2})$, $\tan x > 0$, so $IF = \tan x$.

Step 4: Multiply the differential equation by the Integrating Factor.

Multiplying the differential equation by $\tan x$, we get: $\tan x \frac{dy}{dx} + 2y\csc 2x \tan x = -\tan^2 x$ $\tan x \frac{dy}{dx} + 2y \frac{1}{\sin 2x} \frac{\sin x}{\cos x} = -\tan^2 x$ $\tan x \frac{dy}{dx} + 2y \frac{1}{2\sin x \cos x} \frac{\sin x}{\cos x} = -\tan^2 x$ $\tan x \frac{dy}{dx} + y \frac{1}{\cos^2 x} = -\tan^2 x$ $\tan x \frac{dy}{dx} + y \sec^2 x = -\tan^2 x$

Step 5: Recognize the left-hand side as a derivative.

Notice that $\frac{d}{dx}(y \tan x) = \tan x \frac{dy}{dx} + y \sec^2 x$. Therefore, we can rewrite the equation as: $\frac{d}{dx}(y \tan x) = -\tan^2 x$

Step 6: Integrate both sides with respect to x.

Integrating both sides with respect to $x$, we get: $\int \frac{d}{dx}(y \tan x) \, dx = \int -\tan^2 x \, dx$ $y \tan x = -\int \tan^2 x \, dx = -\int (\sec^2 x - 1) \, dx = -\int \sec^2 x \, dx + \int 1 \, dx = -\tan x + x + C$ $y \tan x = -\tan x + x + C$

Step 7: Solve for y(x).

Dividing by $\tan x$, we get: $y(x) = -1 + x \cot x + C \cot x$

Step 8: Apply the limit condition to find C.

We are given that $\lim_{x \to 0^+} xy(x) = 1$. Substituting our expression for $y(x)$: $\lim_{x \to 0^+} x(-1 + x \cot x + C \cot x) = 1$ $\lim_{x \to 0^+} (-x + x^2 \cot x + Cx \cot x) = 1$ We know that $\lim_{x \to 0} x \cot x = \lim_{x \to 0} x \frac{\cos x}{\sin x} = \lim_{x \to 0} \cos x \cdot \frac{x}{\sin x} = 1 \cdot 1 = 1$. Also, $\lim_{x \to 0} x^2 \cot x = \lim_{x \to 0} x \cdot x \cot x = 0 \cdot 1 = 0$. Therefore, $\lim_{x \to 0^+} (-x + x^2 \cot x + Cx \cot x) = -0 + 0 + C(1) = C$ So, $C = 1$.

Step 9: Write the particular solution.

The particular solution is: $y(x) = -1 + x \cot x + \cot x$

Step 10: Evaluate y(π/4).

We want to find $y(\frac{\pi}{4})$. $y\left(\frac{\pi}{4}\right) = -1 + \frac{\pi}{4} \cot\left(\frac{\pi}{4}\right) + \cot\left(\frac{\pi}{4}\right) = -1 + \frac{\pi}{4}(1) + 1 = \frac{\pi}{4} - 1 + 1 = \frac{\pi}{4} -1$

Common Mistakes & Tips

• Trigonometric Identities: Make sure to correctly apply trigonometric identities, especially when simplifying the coefficient of $y$ and when integrating.
• Integrating Factor: Double-check the calculation of the integrating factor, as errors here propagate through the rest of the solution.
• Limit Evaluation: Be careful when evaluating limits involving trigonometric functions. Recall that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ and $\lim_{x \to 0} x \cot x = 1$.

Summary

We solved the given differential equation by first rewriting it in the standard form of a first-order linear differential equation. We then calculated the integrating factor, multiplied the equation by it, and integrated both sides. Using the given limit condition, we found the value of the integration constant. Finally, we evaluated the particular solution at $x = \frac{\pi}{4}$ to get the answer $\frac{\pi}{4}-1$.

Final Answer The final answer is $\boxed{\frac{\pi}{4} - 1}$, which corresponds to option (B).