If y = y(x) is the solution curve of the differential equation dy/dx + y tan x = x sec x, 0 ≤ x ≤ π/

Q: Wait, why do we use the Integrating Factor method here? Can't we just separate the variables?

That is a fantastic question! If you look closely at the term $y \tan x$, you will see that $y$ is tied to a function of $x$ in a way that makes simple separation impossible. The Integrating Factor is specifically designed for these 'linear' differential equations where the variable $y$ appears linearly, allowing us to 'force' the left side into a product rule derivative.

Q: How do I know when to use the ILATE rule for integration?

ILATE is your compass in the wilderness of integration! Whenever you see a product of two different types of functions—like an algebraic $x$ and a trigonometric $\sec^2 x$—Integration by Parts is almost always the way to go. ILATE helps you decide which part to differentiate (the $u$) and which to integrate (the $dv$) to make the integral simpler, not harder.

Q: Why does the constant $C$ appear, and how do I find its value?

The constant $C$ represents the family of all possible solution curves. Because we are given a specific condition, $y(0) = 1$, we are essentially 'pinning' our curve to a specific point in the coordinate plane. By substituting $x=0$ and $y=1$, we filter out all other curves and find the unique one that passes through our target.

Q: Is it okay if I leave the answer in terms of $\ln$ instead of $\log_e$?

Absolutely! In the context of JEE Advanced, $\ln$ and $\log_e$ are mathematically identical. Just ensure your final expression matches the form provided in the options. Sometimes, a simple log property like $\ln(a/b) = -\ln(b/a)$ is all that stands between you and the correct option.

Q: I got stuck at the very end with the trigonometry values. Any tips?

Trigonometry is the backbone of these problems! Always keep your standard values for $\frac{\pi}{6}$ and $\frac{\pi}{4}$ handy. Remember that $\sec(\frac{\pi}{6}) = \frac{2}{\sqrt{3}}$ and $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$. Precision in these small steps prevents the 'silly mistake' trap!

Question

If $y = y(x)$ is the solution curve of the differential equation $\frac{dy}{dx} + y 	an x = x \sec x, 0 \le x \le \frac{\pi}{3}, y(0) = 1$, then $y\left(\frac{\pi}{6}ight)$ is equal to

Options:

(A) $\frac{\pi}{12} - \frac{\sqrt{3}}{2} \log_e \left(\frac{2}{\sqrt{3}}ight) + \frac{\sqrt{3}}{2}$

(B) $\frac{\pi}{12} + \frac{\sqrt{3}}{2} \log_e \left(\frac{2\sqrt{3}}{e}ight)$

(C) $\frac{\pi}{12} + \frac{\sqrt{3}}{2} \log_e \left(\frac{2\sqrt{3}}{e}ight)$

(D) $\frac{\pi}{12} + \frac{\sqrt{3}}{2} \log_e \left(\frac{2}{e\sqrt{3}}ight)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The given differential equation is:

$$\frac{dy}{dx} + y 	an x = x \sec x$$

This is a **first-order linear differential equation** of the standard form:

$$\frac{dy}{dx} + P(x)y = Q(x)$$

Here, we identify $P(x) = 	an x$ and $Q(x) = x \sec x$.

### The Magic of the Integrating Factor

Our first mission is to find the Integrating Factor ($I.F.$). We define it as:

$$I.F. = e^{\int P(x) dx}$$

Substituting $P(x) = 	an x$, we calculate the integral:

$$\int 	an x \, dx = \ln|\sec x|$$

Thus, the integrating factor simplifies beautifully:

$$I.F. = e^{\ln|\sec x|} = \sec x$$

### The Dance of Integration by Parts

Multiplying the entire original equation by the $I.F. = \sec x$, the left side becomes the derivative of a product, and the right side becomes $x \sec^2 x$:

$$\frac{d}{dx}(y \sec x) = x \sec^2 x$$

Integrating both sides with respect to $x$:

$$y \sec x = \int x \sec^2 x \, dx + C$$

To solve $\int x \sec^2 x \, dx$, we apply the **ILATE rule** for integration by parts, setting $u = x$ and $dv = \sec^2 x \, dx$:

$$y \sec x = x 	an x - \int 	an x \, dx + C$$

$$y \sec x = x 	an x - \ln|\sec x| + C$$

### Pinning the Curve

We are given the boundary condition $y(0) = 1$. Substituting $x = 0$ and $y = 1$ into our general solution:

$$1 \cdot \sec(0) = 0 \cdot 	an(0) - \ln|\sec(0)| + C$$

Since $\sec(0) = 1$ and $\ln(1) = 0$, the equation simplifies to:

$$1 = 0 - 0 + C \Rightarrow C = 1$$

The unique curve is defined by:

$$y \sec x = x 	an x - \ln|\sec x| + 1$$

### The Final Reveal

To find $y$ at $x = \frac{\pi}{6}$, we substitute the value into our specific solution:

$$y \sec\left(\frac{\pi}{6}ight) = \frac{\pi}{6} 	an\left(\frac{\pi}{6}ight) - \ln\left|\sec\left(\frac{\pi}{6}ight)ight| + 1$$

Using the trigonometric values $\sec\left(\frac{\pi}{6}ight) = \frac{2}{\sqrt{3}}$ and $	an\left(\frac{\pi}{6}ight) = \frac{1}{\sqrt{3}}$:

$$y \left(\frac{2}{\sqrt{3}}ight) = \frac{\pi}{6} \left(\frac{1}{\sqrt{3}}ight) - \ln\left(\frac{2}{\sqrt{3}}ight) + 1$$

Multiplying by $\frac{\sqrt{3}}{2}$ to isolate $y$, we obtain the **final answer**:

$$y = \frac{\pi}{12} - \frac{\sqrt{3}}{2} \ln\left(\frac{2}{\sqrt{3}}ight) + \frac{\sqrt{3}}{2}$$

Animated Solution for Mathematics - Differential Equations: If $y = y (x)$ is the solution curve of the differential equation $\frac{d y}{d x} + y tan x = x sec x, 0 \leq x \leq \frac{π}{3}, y (0) = 1$ , then $y (\frac{π}{6})$ is equal to

Select Answer:

Visualized Solution

Identify the Differential Equation

Calculate the Integrating Factor

Formulate the General Solution

Simplify the Integrand

Integrate using Integration by Parts

Complete the Integration

Apply Initial Condition

Write the Particular Solution

Evaluate at $x = \frac{π}{6}$

Final Answer

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Magic of the Integrating Factor

The Dance of Integration by Parts

Pinning the Curve

The Final Reveal

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