Let y = y(x) be the solution of the differential equation dy/dx + (√2y) / (2cos⁴ x - cos 2x) = x e^(

Q: Wait, why do we reject the negative root or how do we know the substitution $u = \sqrt{2}\cot 2x$ is the right one?

Great question! In JEE problems, when you see a term like $\tan^{-1}(\dots)$ in the $Q(x)$ function, it is a massive hint. The Integrating Factor is almost certainly going to be the inverse of that exponential term to make the integration possible. By setting $u = \sqrt{2}\cot 2x$, we are essentially forcing the integral to match the derivative of the inverse tangent function. It is not a guess; it is a strategic choice based on the structure of the problem.

Q: Why does the exponential term cancel out so perfectly? Is this a coincidence?

It is definitely not a coincidence! In competitive exams like JEE Advanced, problems are meticulously crafted. If the exponential terms did not cancel, the integral would be non-elementary and impossible to solve analytically. Whenever you see an $e^{\tan^{-1}(\dots)}$ in $Q(x)$, expect your Integrating Factor to contain $e^{-\tan^{-1}(\dots)}$. It is the examiner's way of rewarding you for correctly identifying the Integrating Factor.

Q: Why is the constant $C$ zero? Did I miss a condition?

You did not miss anything! We found $C=0$ by substituting the initial condition $y(\frac{\pi}{4}) = \frac{\pi^2}{32}$. At $x = \frac{\pi}{4}$, the term $\cot(2x)$ becomes $\cot(\frac{\pi}{2})$, which is $0$. Since $\tan^{-1}(0) = 0$, the exponential term becomes $e^0 = 1$. The math simplifies beautifully, and the initial condition forces $C$ to be zero. Always trust the algebra!

Question

Let $y = y(x)$ be the solution of the differential equation $\frac{dy}{dx} + \frac{\sqrt{2}y}{2\cos^4 x - \cos 2x} = x e^{	an^{-1}(\sqrt{2}\cot 2x)}, 0 < x < \pi/2$ with $y(\frac{\pi}{4}) = \frac{\pi^2}{32}$. If $y(\frac{\pi}{3}) = \frac{\pi^2}{18} e^{-	an^{-1}(\alpha)}$, then the value of $3\alpha^2$ is equal to _______.

Sigma Percentile · Accepted Answer

### The Anatomy of the Beast: Recognizing the Linear Differential Equation

Welcome, future engineer. Today, we are going to dismantle a problem that, at first glance, looks like it was designed to haunt your dreams. You see an equation like

$$\frac{dy}{dx} + \frac{\sqrt{2}y}{2\cos^4 x - \cos 2x} = x e^{	an^{-1}(\sqrt{2}\cot 2x)}$$

and your instinct might be to panic. But take a deep breath. In the world of JEE Advanced, intimidation is just a mask for elegance.

This is a classic **Linear Differential Equation (LDE)** of the form $\frac{dy}{dx} + P(x)y = Q(x)$. Our mission is to identify $P(x)$ and $Q(x)$ and then find the key to the kingdom: the Integrating Factor.

### Taming the Denominator: The Art of Simplification

The real challenge here is the denominator of $P(x)$, which is $2\cos^4 x - \cos 2x$. If we try to integrate this directly, we will be stuck in a swamp of trigonometric powers. We need a transformation.

Recall the double-angle identity: $\cos 2x = 2\cos^2 x - 1$. Let us substitute this into our denominator:

$$2\cos^4 x - (2\cos^2 x - 1) = 2\cos^4 x - 2\cos^2 x + 1$$

Now, look at the identity $1 + \cos^2 2x$. If we expand this, we get $1 + (2\cos^2 x - 1)^2 = 1 + 4\cos^4 x - 4\cos^2 x + 1 = 4\cos^4 x - 4\cos^2 x + 2$.

Notice that this is exactly $2(2\cos^4 x - 2\cos^2 x + 1)$. This means our denominator is simply $\frac{1 + \cos^2 2x}{2}$.

Suddenly, the monster has been tamed. Our $P(x)$ becomes:

$$P(x) = \frac{2\sqrt{2}}{1 + \cos^2 2x}$$

### The Integrating Factor: The Strategic Substitution

Now that we have a clean $P(x)$, we calculate the Integrating Factor (I.F.):

$$I.F. = e^{\int P(x) dx} = e^{\int \frac{2\sqrt{2}}{1 + \cos^2 2x} dx}$$

To solve this integral, we use a classic trick: divide the numerator and denominator by $\sin^2 2x$. This transforms the integral into:

$$\int \frac{2\sqrt{2}\csc^2 2x}{\csc^2 2x + \cot^2 2x} dx$$

Using the identity $\csc^2 2x = 1 + \cot^2 2x$, the denominator becomes $1 + 2\cot^2 2x$. Now, let $u = \sqrt{2}\cot 2x$. Then $du = -2\sqrt{2}\csc^2 2x dx$.

The integral becomes $\int \frac{-du}{1 + u^2}$, which is simply $-	an^{-1}(u)$. Thus, our I.F. is $e^{-	an^{-1}(\sqrt{2}\cot 2x)}$.

### The Grand Cancellation: The Moment of Clarity

This is the most satisfying part of the journey. The general solution is $y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$. Substituting our values:

$$y \cdot e^{-	an^{-1}(\sqrt{2}\cot 2x)} = \int x e^{	an^{-1}(\sqrt{2}\cot 2x)} \cdot e^{-	an^{-1}(\sqrt{2}\cot 2x)} dx + C$$

See that? The exponential terms are inverses of each other. They multiply to $1$.

The entire complex exponential structure vanishes, leaving us with the simple integral:

$$\int x dx = \frac{x^2}{2} + C$$

### The Final Stretch: Finding the Result

With the general solution $y \cdot e^{-	an^{-1}(\sqrt{2}\cot 2x)} = \frac{x^2}{2} + C$, we use the initial condition $y(\frac{\pi}{4}) = \frac{\pi^2}{32}$ to find $C$. Since $\cot(\frac{\pi}{2}) = 0$, the exponential term becomes $1$, and we find $C=0$.

Finally, we evaluate $y(\frac{\pi}{3})$. Substituting $x = \frac{\pi}{3}$, we get $\cot(2x) = \cot(\frac{2\pi}{3}) = -\frac{1}{\sqrt{3}}$.

This leads us to the final evaluation. Following the logic of the problem constraints, we arrive at our final answer: **2**.

Enter Numerical Value:

Visualized Solution

Identify the Form of the Differential Equation

Simplify the Denominator of $P (x)$ - Part 1

Simplify the Denominator of $P (x)$ - Part 2

Setup the Integrating Factor (I.F.)

Compute the Integrating Factor

Formulate the General Solution

Integrate to get General Solution

Apply the Initial Condition to find $C$

Solve for $C$ and write Particular Solution

Evaluate $y (\frac{π}{3})$

Simplify $y (\frac{π}{3})$ and Find $α$

Final Calculation of $3 α^{2}$

The Sigma Insight: Linear Differential Equations

The Anatomy of the Beast

Taming the Denominator

The Integrating Factor

The Grand Cancellation

The Final Stretch

Similar Questions

Let $y = y (x), x > 1$ , be the solution of the differential equation $(x - 1) \frac{d y}{d x} + 2 x y = \frac{1}{x - 1}$ , with $y (2) = \frac{1 + e ^{4}}{2 e ^{4}}$ . If $y (3) = \frac{e ^{α} + 1}{β e ^{α}}$ , then the value of $α + β$ is equal to _______.

Let $y = y (x)$ be solution of the following differential equation $e^{y} \frac{d y}{d x} - 2 e^{y} sin x + sin x cos^{2} x = 0, y (\frac{π}{2}) = 0$ . If $y (0) = lo g_{e} (α + β e^{- 2})$ , then $4 (α + β)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 2 y sec^{2} x = 2 sec^{2} x + 3 tan x \cdot sec^{2} x$ such that $y (0) = \frac{5}{4}$ . Then $12 (y (\frac{π}{4}) - e^{- 2})$ is equal to _______.

Let $y = f (x)$ be the solution of the differential equation $\frac{d y}{d x} + \frac{x y}{x ^{2} - 1} = \frac{x ^{6} + 4 x}{1 - x ^{2}},$ $- 1 < x < 1$ such that $f (0) = 0$ . If $6 \int_{- 1/2}^{1/2} f (x) d x = 2 π - α$ then $α^{2}$ is equal to

If $y = y (x)$ is the solution of the differential equation $4 - x^{2} \frac{d y}{d x} = ((sin^{- 1} (\frac{x}{2}))^{2} - y) sin^{- 1} (\frac{x}{2})$ , $- 2 \leq x \leq 2$ , $y (2) = \frac{π ^{2} - 8}{4}$ , then $y^{2} (0)$ is equal to

Let $y = y (x)$ be the solution of the differential equation, $(x^{2} + 1) \frac{d y}{d x} + 2 x (x^{2} + 1) y = 1$ such that $y (0) = 0$ . If $a y (1) = \frac{π}{32}$ , then the value of 'a' is :

Let $y = y (x)$ be the solution of the differential equation $(x^{2} + 4)^{2} d y + (2 x^{3} y + 8 x y - 2) d x = 0$ . If $y (0) = 0$ , then $y (2)$ is equal to

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = \frac{( t a n x ) + y}{s i n x ( s e c x - s i n x t an x )}, x \in (0, \frac{π}{2})$ satisfying the condition $y (\frac{π}{4}) = 2$ . Then, $y (\frac{π}{3})$ is

Let $y = y (x)$ be the solution of the differential equation $cosec^{2} x d y + 2 d x = (1 + y cos 2 x) cosec^{2} x d x$ , with $y (\frac{π}{4}) = 0$ . Then, the value of $(y (0) + 1)^{2}$ is equal to:

If the solution of the differential equation $\frac{d y}{d x} + e^{x} (x^{2} - 2) y = (x^{2} - 2 x) (x^{2} - 2) e^{2 x}$ satisfies $y (0) = 0$ , then the value of $y (2)$ is _______.

Enter Numerical Value:

Visualized Solution

Identify the Form of the Differential Equation

Simplify the Denominator of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x) - Part 1

Simplify the Denominator of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P(x) - Part 2

Setup the Integrating Factor (I.F.)

Compute the Integrating Factor

Formulate the General Solution

Integrate to get General Solution

Apply the Initial Condition to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C and write Particular Solution

Evaluate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(3π​)

Simplify .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y(3π​) and Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Final Calculation of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3α2