Let f : R → R be a thrice differentiable odd function satisfying f'(x) ≥ 0, f'(x) = f(x), f(0) = 0,

Q: Wait, why do we multiply by $f'(x)$? Is that a standard trick?

It is indeed a classic and powerful technique in differential equations! When you have an equation of the form $f''(x) = g(f(x))$, multiplying by $f'(x)$ allows you to transform the equation into a form where you can integrate with respect to $x$. It essentially converts a second-order equation into a first-order one by utilizing the chain rule in reverse: $\frac{d}{dx}(\frac{1}{2}(f'(x))^2) = f'(x)f''(x)$. It is a beautiful way to reduce complexity.

Q: Why did we reject the negative root when taking the square root of $(f'(x))^2 = (f(x))^2 + 9$?

Great observation! The problem explicitly states that $f'(x) \ge 0$. This condition is our guiding light. If we had chosen the negative root, we would be violating the given constraint that the slope of the function is non-negative. Always pay close attention to these constraints; they are often the key to selecting the correct branch of a solution.

Q: Could I have just used the general solution $f(x) = A e^x + B e^{-x}$?

Yes, you absolutely could! Since $f''(x) = f(x)$, the characteristic equation is $r^2 - 1 = 0$, giving $r = \pm 1$. So $f(x) = A e^x + B e^{-x}$ is a valid approach. However, the integration method we used is more robust and works even when the coefficients aren't constant. Both methods are elegant, but understanding the integration trick gives you a deeper intuition for how these functions behave.

Question

Let $f: \mathbb{R} 	o \mathbb{R}$ be a thrice differentiable odd function satisfying $f'(x) \ge 0, f'(x) = f(x), f(0) = 0, f'(0) = 3$. Then $9f(\log_e 3)$ is equal to _______.

Sigma Percentile · Accepted Answer

### The Symphony of Derivatives: Unlocking the Mystery of $f(x)$

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a problem; we are peeling back the layers of a mathematical mystery.

We are given a thrice differentiable odd function $f(x)$ that obeys a beautiful, self-referential law: $f''(x) = f(x)$. This is not just any equation; it is the hallmark of exponential growth and decay, a fundamental pattern in the universe.

We are also armed with two critical anchors: $f(0) = 0$ and $f'(0) = 3$. Our mission is to find the value of $9f(\ln 3)$.

### Phase 1: The Art of Reduction

When we look at $f''(x) = f(x)$, our instinct might be to jump straight to the general solution $f(x) = Ae^x + Be^{-x}$. While that is a valid path, let us explore a more profound technique—one that reveals the underlying structure of the function.

We want to reduce the order of this differential equation by introducing a catalyst. We multiply both sides of our equation by $f'(x)$.

Because $f'(x)f''(x)$ is the derivative of $\frac{1}{2}(f'(x))^2$, we transform our equation into:

$$f'(x) \cdot f''(x) = f'(x) \cdot f(x)$$

This is the moment of clarity. Both sides are now exact derivatives. We are essentially saying that the rate of change of the square of the slope is proportional to the rate of change of the square of the function itself.

### Phase 2: The Integration

Now, we integrate both sides with respect to $x$. The left side becomes $\frac{1}{2}(f'(x))^2$, and the right side becomes $\frac{1}{2}(f(x))^2$, plus our constant of integration, $C$.

Multiplying by $2$ to clean up the fractions, we get:

$$(f'(x))^2 = (f(x))^2 + C'$$

Here, $C'$ is simply $2C$. Now, we invoke our initial conditions. We know $f(0) = 0$ and $f'(0) = 3$.

Substituting these into our equation, we find $3^2 = 0^2 + C'$, which gives us $C' = 9$. Our equation is now locked in:

$$(f'(x))^2 = (f(x))^2 + 9$$

### Phase 3: The First-Order Transformation

We are now looking at $f'(x) = \sqrt{(f(x))^2 + 9}$. We take the positive root because the problem implies the growth of the function.

This is a first-order separable differential equation. Let $y = f(x)$. Then, $\frac{dy}{dx} = \sqrt{y^2 + 9}$. Separating the variables, we get:

$$\frac{dy}{\sqrt{y^2 + 9}} = dx$$

Integrating both sides, we recall the standard integral $\int \frac{dy}{\sqrt{y^2 + a^2}} = \ln|y + \sqrt{y^2 + a^2}|$. Applying this, we obtain:

$$\ln|y + \sqrt{y^2 + 9}| = x + C_2$$

Using $f(0) = 0$, we find $C_2 = \ln 3$. Thus, $\ln|y + \sqrt{y^2 + 9}| = x + \ln 3$.

### Phase 4: The Grand Finale

Exponentiating both sides, we get $y + \sqrt{y^2 + 9} = 3e^x$. Now, we evaluate at $x = \ln 3$.

The right side becomes $3e^{\ln 3} = 3(3) = 9$. So, $y + \sqrt{y^2 + 9} = 9$.

Rearranging, $\sqrt{y^2 + 9} = 9 - y$. Squaring both sides, we obtain:

$$y^2 + 9 = 81 - 18y + y^2$$

The $y^2$ terms cancel, leaving $18y = 72$, so $y = 4$. Finally, the question asks for $9f(\ln 3)$, which is $9 	imes 4 = \mathbf{36}$.

We have arrived at the summit. The beauty of this problem lies not just in the answer, but in the elegance of the transformation. Keep practicing, keep questioning, and never lose your wonder for the mathematics that govern our world.

Animated Solution for Mathematics - Differential Equations: Let $f : R \to R$ be a thrice differentiable odd function satisfying $f^{'} (x) \geq 0, f^{'} (x) = f (x), f (0) = 0, f^{'} (0) = 3$ . Then $9 f (lo g_{e} 3)$ is equal to _______.

Enter Numerical Value:

Visualized Solution

Analyzing the Core Equation

The Integration Trick

Integrating Both Sides

Determining the Constant $C^{'}$

Simplifying to First-Order

Variable Separable Method

Finding the Second Constant

Converting to Exponential Form

Evaluating at $x = ln 3$

Solving for $y$

Final Calculation

The Sigma Insight: Variable Separable Method

The Symphony of Derivatives

Phase 1

Phase 2

Phase 3

Phase 4

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