For x ∈ ℝ, let y(x) be a solution of the differential equation (x² - 5) dy/dx - 2xy = -2x(x² - 5)² s

Q: Why do we divide by $(x^2 - 5)$? Isn't that dangerous if $x^2 = 5$?

That is a sharp observation! In the context of solving the differential equation, we are looking for a general solution on intervals where the equation is well-defined. By dividing by $(x^2 - 5)$, we are operating on the domain $x \in \mathbb{R} \setminus \{\sqrt{5}, -\sqrt{5}\}$. Since the initial condition $y(2) = 7$ is given at $x=2$, we are clearly working within an interval where $x^2 - 5 \neq 0$. Thus, the division is perfectly valid in this region.

Q: Why does the Integrating Factor simplify to $\frac{1}{x^2 - 5}$?

Great question! When we compute the integral $\int -\frac{2x}{x^2 - 5} dx$, we use the substitution $u = x^2 - 5$, so $du = 2x dx$. The integral becomes $-\int \frac{1}{u} du = -\ln|u| = -\ln|x^2 - 5|$. When we raise $e$ to this power, we get $e^{-\ln|x^2 - 5|}$, which is $e^{\ln(|x^2 - 5|^{-1})}$. Since $e^{\ln(A)} = A$, this simplifies directly to $(x^2 - 5)^{-1}$, or $\frac{1}{x^2 - 5}$.

Q: Why can't we just find the maximum of $y(x) = -x^4 + 2x^2 + 15$ by taking the derivative directly?

You absolutely can! Taking the derivative $y'(x) = -4x^3 + 4x$ and setting it to zero gives $x(x^2 - 1) = 0$, leading to critical points at $x = 0, 1, -1$. However, using the substitution $t = x^2$ is a powerful 'JEE-style' shortcut. It turns a cubic equation into a linear one, reducing the risk of calculation errors and helping you visualize the function as a simple parabola, which is often faster and more intuitive.

Question

For $x \in \mathbb{R}$, let $y(x)$ be a solution of the differential equation $(x^2 - 5) \frac{dy}{dx} - 2xy = -2x(x^2 - 5)^2$ such that $y(2) = 7$. Then the maximum value of the function $y(x)$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that, at first glance, might seem like a tangled mess of polynomials.

As we peel back the layers, you will see the elegance of the **Linear Differential Equation (LDE)** structure. Let us look at our equation:

$$(x^2 - 5) \frac{dy}{dx} - 2xy = -2x(x^2 - 5)^2$$

The first thing to notice is that the degree of $y$ and $\frac{dy}{dx}$ are both linear. This is our signal to transform this into the standard form: $\frac{dy}{dx} + P(x)y = Q(x)$.

By dividing the entire equation by $(x^2 - 5)$, we obtain:

$$\frac{dy}{dx} - \frac{2x}{x^2 - 5}y = -2x(x^2 - 5)$$

Now, the path forward is illuminated.

### The Multiplier of Destiny: The Integrating Factor

With our equation in standard form, we identify our key players: $P(x) = -\frac{2x}{x^2 - 5}$ and $Q(x) = -2x(x^2 - 5)$.

To solve this, we need the **Integrating Factor (IF)**, defined as $IF = e^{\int P(x) dx}$. This is where the magic happens.

Look at the integral $\int -\frac{2x}{x^2 - 5} dx$. The numerator is the derivative of the denominator! This is a classic $\int \frac{f'(x)}{f(x)} dx$ form, which integrates to $\ln|f(x)|$.

Thus, our integral is $-\ln|x^2 - 5|$. When we exponentiate this, $e^{-\ln|x^2 - 5|}$ becomes $(x^2 - 5)^{-1}$, or simply:

$$IF = \frac{1}{x^2 - 5}$$

This is our Integrating Factor, the key that unlocks the solution.

### Anchoring the Solution

The general solution for an LDE is $y \cdot (IF) = \int Q(x) \cdot (IF) dx + c$. Substituting our values, we have:

$$y \cdot \frac{1}{x^2 - 5} = \int [-2x(x^2 - 5)] \cdot \frac{1}{x^2 - 5} dx + c$$

Notice how the term $(x^2 - 5)$ cancels out perfectly? We are left with $\int -2x dx + c$, which simplifies to $-x^2 + c$.

Multiplying by $(x^2 - 5)$, we get the general solution:

$$y(x) = (x^2 - 5)(c - x^2)$$

Now, we use the initial condition $y(2) = 7$ to anchor our solution. Substituting $x=2$ and $y=7$:

$$7 = (4 - 5)(c - 4) \Rightarrow 7 = -1(c - 4) \Rightarrow c = -3$$

Our specific solution is $y(x) = (x^2 - 5)(-3 - x^2) = -x^4 + 2x^2 + 15$.

### The Final Peak: Maximization

We have arrived at the final stage: maximizing $y(x) = -x^4 + 2x^2 + 15$. Instead of dealing with a fourth-degree polynomial, let us use a substitution that simplifies our life.

Let $t = x^2$. Since $x$ is a real number, $t \ge 0$. Our function becomes:

$$y(t) = -t^2 + 2t + 15$$

This is a downward-opening parabola. The maximum occurs at the vertex, $t = -\frac{b}{2a} = -\frac{2}{2(-1)} = 1$.

Since $t=1$ satisfies our constraint $t \ge 0$, we substitute $t=1$ back into the quadratic:

$$y_{max} = -(1)^2 + 2(1) + 15 = 16$$

We have reached the summit. The **maximum value is 16**. Remember, every complex problem is just a series of simple, logical steps waiting to be connected.

Animated Solution for Mathematics - Differential Equations: For $x \in R$ , let $y (x)$ be a solution of the differential equation $(x^{2} - 5) \frac{d y}{d x} - 2 x y = - 2 x (x^{2} - 5)^{2}$ such that $y (2) = 7$ . Then the maximum value of the function $y (x)$ is

Enter Numerical Value:

Visualized Solution

Analyze the Differential Equation

Convert to Standard Form

Identify $P (x)$ and $Q (x)$

Compute Integrating Factor

Simplify Integrating Factor

General Solution Formula

Integrate the Right Side

Express $y$ Explicitly

Apply Initial Condition

Find the Specific Solution

Setup for Maximization

Find Maximum of the Quadratic

Calculate the Maximum Value

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

The Multiplier of Destiny

Anchoring the Solution

The Final Peak

Similar Questions

If the solution curve $y = f (x)$ of the differential equation $(x^{2} - 4) y^{'} - 2 x y + 2 x (4 - x^{2})^{2} = 0, x > 2$ , passes through the point $(3, 15)$ , then the local maximum value of $f$ is ......... .

If $y = y (x)$ is the solution of the differential equation, $\frac{d y}{d x} + 2 y tan x = sin x, y (\frac{π}{3}) = 0$ , then the maximum value of the function $y (x)$ over $R$ is equal to :

If $\frac{d y}{d x} + 2 y tan x = sin x, 0 < x < \frac{π}{2}$ and $y (\frac{π}{3}) = 0$ , then the maximum value of $y (x)$ is

Let $y = y (x)$ be the solution of the differential equation $\frac{d y}{d x} = 2 (y + 2 sin x - 5) x - 2 cos x$ such that $y (0) = 7$ . Then $y (π)$ is equal to :

If the curve $y = y (x)$ is the solution of the differential equation $2 (x^{2} + x^{5/4}) d y - y (x + x^{1/4}) d x = 2 x^{9/4} d x, x > 0$ which passes through the point $(1, 1 - \frac{4}{3} lo g_{e} 2)$ , then the value of $y (16)$ is equal to :

Let $y = y (x)$ be the solution of the differential equation $x lo g_{e} x \frac{d y}{d x} + y = x^{2} lo g_{e} x, (x > 1)$ . If $y (2) = 2$ , then $y (e)$ 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 _______.

If $y = y (x)$ is the solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2}$ satisfying $y (1) = 1$ , then $y (1/2)$ is equal to :

A solution of the differential equation $(\frac{d y}{d x})^{2} - x \frac{d y}{d x} + y = 0$ is

The solution of the differential equation $x \frac{d y}{d x} + 2 y = x^{2} (x \neq = 0)$ with $y (1) = 1$ , is