Animated Solution for Mathematics - Differentiation: If $P(1)=0$ and $dP(x)/dx>P(x)$ for all $x\ge 1$ then prove that $P(x)>0$ for all $x>1$.

Welcome, fellow traveler, to the elegant world of differential inequalities. We are given a function $P(x)$ with two key pieces of information: $P(1)=0$ and $P^{\prime }(x)>P(x)$ for all $x\ge 1$.

Our mission is to prove that $P(x)>0$ for all $x>1$. Imagine you are standing at the point $x=1$ on a graph; you know the function is currently touching the x-axis, and its rate of change is "pushing" it upwards faster than the function itself.

The Integrating Factor

When we look at the inequality $P^{\prime }(x)>P(x)$, it is helpful to bring everything to one side: $P^{\prime }(x)-P(x)>0$. This structure is a classic signature in calculus that suggests the use of an integrating factor.

To make the left side a perfect derivative, we multiply the entire inequality by $e^{-x}$. This yields:

Since the derivative of $e^{-x}$ is $-e^{-x}$, this expression is exactly the derivative of the product $e^{-x}P(x)$.

Let us define a new, auxiliary function to simplify our analysis: $f(x)=e^{-x}P(x)$. We find its derivative, $f^{\prime }(x)$, using the product rule:

This expands to:

Because $e^{-x}>0$ for all real $x$ and we are given $P^{\prime }(x)-P(x)>0$, it follows that $f^{\prime }(x)>0$ for all $x\ge 1$.

Monotonicity and the Boundary

Because $f^{\prime }(x)>0$, we have proven that $f(x)$ is a strictly increasing function. This means that as $x$ moves to the right, $f(x)$ must increase.

We look back at our boundary condition: $P(1)=0$. Substituting this into our definition of $f(x)$, we get:

Since the function starts at zero at $x=1$ and is strictly increasing, it must be that $f(x)>f(1)$ for all $x>1$. Therefore, $f(x)>0$ for all $x>1$.

We have established that $e^{-x}P(x)>0$. Since the exponential function $e^{-x}$ is always strictly positive, we can divide both sides by it without changing the direction of the inequality.

This leaves us with the final result: $P(x)>0$ for all $x>1$. We have successfully navigated the differential inequality by leveraging the integrating factor and the boundary condition.