Animated Solution for Mathematics - Differential Equations: If $y(x)$ is the solution of the differential equation $\frac{dy}{dx}+(\frac{2x+1}{x})y=e^{-2x},x>0$, where $y(1)=\frac{1}{2}{e}^{-2}$, then :

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through the language of change. Differential equations are the heartbeat of physics and engineering, and mastering them is like learning to read the underlying code of the universe.

Let us look at the equation before us:

At first glance, it might seem like a daunting collection of symbols, but look closer. It follows the elegant, standard structure of a First-Order Linear Differential Equation:

Recognizing this form is your first victory. It tells us exactly which path to take.

To solve this, we need a key. That key is the Integrating Factor, or $I.F.$, defined by the formula:

Here, our $P(x)$ is $\frac{2x+1}{x}$. A seasoned mathematician knows that before integrating, we simplify the expression:

Now, the integration becomes trivial:

When we raise $e$ to this power, we get $e^{2x+\ln x}$. Using the laws of exponents, this becomes $e^{2x}\cdot e^{\ln x}$. Since $e^{\ln x}=x$, our Integrating Factor simplifies beautifully to:

This is the magic of mathematics—complexity collapsing into simplicity.

Now, we apply the general solution formula:

Substituting our values, we get:

Look at the integrand on the right. We have $e^{-2x}\cdot e^{2x}$, which is $e^0$, or simply $1$. The entire expression collapses into:

With the initial condition $y(1)=\frac{1}{2}{e}^{-2}$, we substitute to find $C$:

Our particular solution is:

Finally, we must determine where this function is decreasing. We need the derivative $y^{\prime }(x)$. Using the product rule on $y(x)=\frac{1}{2}x{e}^{-2x}$, we get:

Factoring out the common term, we arrive at:

For the function to be decreasing, we need $y^{\prime }(x)<0$. Since the exponential term is always positive, the sign depends entirely on $(1-2x)$.

Solving $1-2x<0$ gives $x>\frac{1}{2}$. Thus, the function is decreasing for all $x>\frac{1}{2}$. This confirms that in the interval $(\frac{1}{2},1)$, the function is indeed decreasing.