Let f be a differentiable function such that 2(x+2)²f(x)-3(x+2)²=10∫_0^x (t+2)f(t)dt, x≥0 Then f(2)

Q: Wait, why do we differentiate? Can't we just integrate the right side directly?

That is a common trap! We cannot integrate the right side because $f(t)$ is an unknown function. We don't know its form yet. Differentiation is the only way to turn this integral equation into a differential equation that we can actually solve.

Q: How do I know what value to plug in to find the constant $k$?

Look for the 'zero-point'. In the original equation, the integral is from $0$ to $x$. If we set $x=0$, the integral becomes the integral from $0$ to $0$, which is always zero. This allows us to isolate the constant term and find the initial condition $f(0)$.

Q: Is the Leibniz rule always this simple?

In this specific case, yes! Because the upper limit is just $x$ and the lower limit is a constant, the derivative is simply the integrand evaluated at $x$. If the limits were more complex, like $x^2$ or $\sin(x)$, we would need to use the chain rule as well. Always check your limits!

Question

Let f be a differentiable function such that $2(x+2)^{2}f(x)-3(x+2)^{2}=10\int_{0}^{x}(t+2)f(t)dt,x\ge0$ Then $f(2)$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a complex mathematical puzzle. You see an equation:

$$2(x+2)^{2}f(x)-3(x+2)^{2}=10\int_{0}^{x}(t+2)f(t)dt$$

At first glance, it feels like a fortress. The function $f(x)$ is locked away inside an integral, and you are tasked with finding its value at $x=2$.

In the world of JEE Advanced, an integral equation is not a dead end; it is a challenge waiting for the right tool. That tool is the **Leibniz Rule**.

### Phase 1: The Leibniz Transformation

To free $f(x)$, we must perform a surgical strike. We differentiate both sides with respect to $x$.

On the right-hand side, the Leibniz Rule acts like a key. Since our integral is defined from $0$ to $x$, the derivative simplifies beautifully:

$$\frac{d}{dx} \left[ 10 \int_0^x (t+2) f(t) dt \right] = 10(x+2)f(x)$$

On the left-hand side, we apply the product rule to $2(x+2)^2 f(x)$, yielding $4(x+2)f(x) + 2(x+2)^2 f'(x)$. The derivative of the second term, $-3(x+2)^2$, is $-6(x+2)$.

Equating these, we get:

$$4(x+2)f(x) + 2(x+2)^2 f'(x) - 6(x+2) = 10(x+2)f(x)$$

### Phase 2: Simplifying the Landscape

Now, look at the equation. Every term shares a common factor of $2(x+2)$.

Since $x \ge 0$, we know $x+2$ is never zero, allowing us to divide through safely. This leaves us with a much friendlier linear differential equation:

$$2f(x) + (x+2)f'(x) - 3 = 5f(x)$$

Rearranging this, we find $(x+2)f'(x) = 3f(x) + 3$, or more elegantly:

$$(x+2)f'(x) = 3(f(x) + 1)$$

### Phase 3: The Hunt for the Constant

This is a classic variable-separable differential equation. We isolate the terms:

$$\frac{f'(x)}{f(x)+1} = \frac{3}{x+2}$$

Integrating both sides gives us $\ln|f(x)+1| = 3\ln|x+2| + C$, which transforms into $f(x) + 1 = k(x+2)^3$.

To find the constant $k$, we return to the original integral equation. By setting $x=0$, the integral vanishes:

$$2(0+2)^2 f(0) - 3(0+2)^2 = 0$$

Solving this, we find $8f(0) - 12 = 0$, so $f(0) = \frac{3}{2}$. Substituting this back into our general solution, we find $k = \frac{5}{16}$.

### The Final Strike

With $f(x) = \frac{5}{16}(x+2)^3 - 1$, finding $f(2)$ is now a simple matter of substitution.

Plugging in $x=2$:

$$f(2) = \frac{5}{16}(4)^3 - 1 = \frac{5}{16}(64) - 1 = 20 - 1 = 19$$

You have successfully navigated the trap and unlocked the function. The final answer is **19**.

Animated Solution for Mathematics - Differential Equations: Let f be a differentiable function such that $2 (x + 2)^{2} f (x) - 3 (x + 2)^{2} = 10 \int_{0}^{x} (t + 2) f (t) d t, x \geq 0$ Then $f (2)$ is equal to

Enter Numerical Value:

Visualized Solution

The Integral Equation

Applying Leibniz Rule

Executing the Differentiation

Simplifying the Equation

Forming the Differential Equation

Separating the Variables

Solving the Integrals

Finding the Initial Condition

Calculating $f (0)$

Solving for the Constant $k$

The Final Function $f (x)$

Calculating $f (2)$

Summary and Key Takeaways

The Sigma Insight: Linear Differential Equations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Final Strike

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