Let y = y(x) satisfies the equation dy/dx - |A| = 0, for all x 0, where A = [[y, sin x, 1], [0, -1

Q: Wait, why do we treat $y$ as a constant when expanding the determinant? Isn't $y$ a function of $x$?

That is a brilliant observation! When we calculate the determinant $|A|$, we are evaluating the value of the matrix at a specific, arbitrary point $x$. At that instantaneous moment, $y$ has a specific value $y(x)$. Therefore, for the purpose of the determinant expansion, $y$ acts as a scalar value. We only treat it as a variable when we plug it back into the differential equation $\frac{dy}{dx} - |A| = 0$.

Q: I got stuck on the integration $\int x \sin x dx$. Is there a shortcut?

Integration by parts is your best friend here! Remember the LIATE rule: $u=x$ (Algebraic) and $dv=\sin x dx$ (Trigonometric). It is a classic JEE step. If you find yourself struggling, just remember: $\int u dv = uv - \int v du$. It is a beautiful dance of functions—don't rush the signs!

Q: Why does the constant $c$ turn out to be zero? Did I make a calculation error?

Not at all! In many JEE problems, the initial conditions are specifically chosen to simplify the final expression. When you substituted $x = \pi$ and $y = \pi + 2$, the terms on both sides balanced perfectly, leaving $c = 0$. It is a sign that your derivation is elegant and correct.

Q: Can I use a different method instead of the Integrating Factor?

For a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, the Integrating Factor is the most robust and standard method. While you could technically look for an exact differential form, the I.F. method is the 'gold standard' that will never fail you in an exam setting.

Q: Is it possible to solve this without expanding the determinant first?

Mathematically, no. The differential equation is defined by the value of the determinant. You must reduce the matrix to a scalar function of $x$ and $y$ before you can apply the rules of calculus. Think of the determinant as the 'engine' of the problem—you have to build it before you can drive the car!

Question

Let $y = y(x)$ satisfies the equation $\frac{dy}{dx} - |A| = 0$, for all $x > 0$, where $A = \begin{bmatrix} y & \sin x & 1 \ 0 & -1 & 1 \ 2 & 0 & \frac{1}{x} \end{bmatrix}$. If $y(\pi) = \pi + 2$, then the value of $y\left(\frac{\pi}{2}ight)$ is :

Options:

(A) $\frac{\pi}{2} + \frac{4}{\pi}$

(B) $\frac{\pi}{2} - \frac{1}{\pi}$

(C) $\frac{3\pi}{2} - \frac{1}{\pi}$

(D) $\frac{\pi}{2} - \frac{4}{\pi}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the intersection of two vast mathematical landscapes: Linear Algebra and Calculus. Often, these subjects feel like separate islands, but today, we are building a bridge between them.

The problem before us is a masterclass in how different branches of mathematics speak the same language. We are given a differential equation involving the determinant of a matrix $A$, and our goal is to find the specific path of the function $y(x)$.

### Unmasking the Determinant

At first glance, the matrix $A$ is defined as:

$$A = \begin{bmatrix} y & \sin x & 1 \ 0 & -1 & 1 \ 2 & 0 & \frac{1}{x} \end{bmatrix}$$

A determinant is a single scalar value that summarizes the 'volume' or 'scaling factor' of the matrix. To find it, we expand along the first row.

We take $y$ and multiply it by the minor $\left( -\frac{1}{x} - 0 ight)$, subtract $\sin x$ times its minor $(0 - 2)$, and add $1$ times its minor $(0 - (-2))$. When we simplify this, we obtain:

$$|A| = -\frac{y}{x} + 2\sin x + 2$$

Suddenly, the matrix has vanished, leaving behind a beautiful, manageable expression. This is the heart of the problem: transforming a complex structure into a simple, solvable differential equation.

### The Dance of the Differential Equation

Now, we substitute our determinant back into the given equation: $\frac{dy}{dx} - |A| = 0$. This yields:

$$\frac{dy}{dx} - \left( -\frac{y}{x} + 2\sin x + 2 ight) = 0$$

Rearranging this into the standard linear form, we get:

$$\frac{dy}{dx} + \frac{1}{x}y = 2\sin x + 2$$

This is a classic first-order linear differential equation where $P(x) = \frac{1}{x}$ and $Q(x) = 2\sin x + 2$. To solve this, we calculate the Integrating Factor (I.F.), defined as $e^{\int P(x) dx}$.

Since $\int \frac{1}{x} dx = \ln x$, our I.F. is simply $e^{\ln x} = x$. This is a moment of pure elegance—the complexity of the exponential and logarithmic functions collapses into a simple variable $x$.

### The Integration Journey

Multiplying our entire equation by $x$, we obtain:

$$\frac{d}{dx}(xy) = x(2\sin x + 2)$$

Now, we integrate both sides. The right side requires us to evaluate $\int (2x\sin x + 2x) dx$. We split this into two parts.

The integral $\int 2x dx$ is trivial, giving us $x^2$. The integral $\int 2x \sin x dx$ requires Integration by Parts. By setting $u = x$ and $dv = \sin x dx$, we find that:

$$\int x \sin x dx = -x \cos x + \sin x$$

Combining these, we arrive at the general solution:

$$xy = x^2 - 2x\cos x + 2\sin x + c$$

### The Final Reveal

We use the initial condition $y(\pi) = \pi + 2$. Substituting $x = \pi$ and $y = \pi + 2$ into our equation, we find:

$$\pi(\pi + 2) = \pi^2 - 2\pi \cos \pi + 2 \sin \pi + c$$

Since $\cos \pi = -1$ and $\sin \pi = 0$, the equation simplifies to $\pi^2 + 2\pi = \pi^2 + 2\pi + c$, which forces $c = 0$. With $c$ gone, our specific solution is $xy = x^2 - 2x\cos x + 2\sin x$.

Finally, to find $y\left(\frac{\pi}{2}ight)$, we substitute $x = \frac{\pi}{2}$. The term $\cos\left(\frac{\pi}{2}ight)$ becomes $0$, and $\sin\left(\frac{\pi}{2}ight)$ becomes $1$. We are left with:

$$\frac{\pi}{2} y = \frac{\pi^2}{4} + 2$$

Dividing by $\frac{\pi}{2}$, we reach our destination:

**$$y = \frac{\pi}{2} + \frac{4}{\pi}$$**

Animated Solution for Mathematics - Differential Equations: Let $y = y (x)$ satisfies the equation $\frac{d y}{d x} - ∣ A ∣ = 0$ , for all $x > 0$ , where $A = y 02 sin x - 1 0 11 \frac{1}{x}$ . If $y (π) = π + 2$ , then the value of $y (\frac{π}{2})$ is :

Select Answer:

Visualized Solution

Understanding the Problem

Calculating the Determinant $∣ A ∣$

Forming the Differential Equation

Standard Linear Form

Finding the Integrating Factor (I.F.)

Setting up the General Solution

Integration by Parts

Final General Solution

Applying the Initial Condition

Solving for Constant $c$

Finding $y (π /2)$

Final Calculation

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Unmasking the Determinant

The Dance of the Differential Equation

The Integration Journey

The Final Reveal

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