Animated Solution for Mathematics - Differentiation: If $f:\mathbb{R}\to \mathbb{R}$ is a differentiable function such that $f^{\prime }(x)>2f(x)$ for all $x\in \mathbb{R}$, and $f(0)=1$, then

Welcome, future engineers! Today, we are diving into a problem that might look like a standard differential equation at first glance, but it hides a beautiful, deeper truth about how functions grow.

We are given a differentiable function $f:\mathbb{R}\to \mathbb{R}$ with the condition $f^{\prime }(x)>2f(x)$ and an initial anchor point $f(0)=1$. Our goal is to understand the behavior of this function for $x>0$.

Let us start by rearranging our given inequality: $f^{\prime }(x)-2f(x)>0$.

Does this structure look familiar? It should! It is the hallmark of a first-order linear differential equation.

If we had an equality, we would immediately reach for the Integrating Factor method. Even with an inequality, the same logic applies. We need to find a function that, when multiplied by our expression, turns the left-hand side into a perfect derivative.

The coefficient of $f(x)$ is $-2$. Therefore, our Integrating Factor is $e^{\int -2dx}=e^{-2x}$.

This exponential term is the magic key. When we multiply the entire inequality by $e^{-2x}$, we get:

Because $e^{-2x}$ is always strictly positive, the inequality sign remains unchanged. Now, look at the left-hand side. It is the exact expansion of the product rule:

Let us define an auxiliary function $g(x)=f(x)e^{-2x}$. Our inequality tells us that $g^{\prime }(x)>0$ for all $x$.

This is a powerful statement! It means that $g(x)$ is a strictly increasing function. For any $x>0$, it must be true that $g(x)>g(0)$.

We know $f(0)=1$, so:

Thus, $g(x)>1$. Substituting back, we find $f(x)e^{-2x}>1$, which simplifies to $f(x)>e^{2x}$. This confirms that our function $f(x)$ grows faster than the exponential $e^{2x}$.

Since $f(x)>e^{2x}$ and $e^{2x}>0$, it follows that $f(x)>0$. Returning to our original inequality $f^{\prime }(x)>2f(x)$, since $f(x)$ is positive, $2f(x)$ is also positive, which forces $f^{\prime }(x)>0$.

A positive derivative means $f(x)$ is strictly increasing for $x>0$.

We have successfully navigated the trap and proven that $f(x)$ is increasing and $f(x)>e^{2x}$. Keep this technique in your toolkit—it is a favorite of JEE examiners!