Animated Solution for Mathematics - Differential Equations: For a differentiable function $f:\mathbb{R}\to \mathbb{R}$, suppose $f^{\prime }(x)=3f(x)+\alpha$, where $\alpha \in \mathbb{R},f(0)=1$ and ${\lim }_{x\to -\infty }f(x)=7$. Then $9f(-{\log }_{e}3)$ is equal to_________

Welcome, fellow traveler on the path to JEE excellence. Today, we aren't just solving a differential equation; we are uncovering the hidden architecture of a function.

When you look at the equation $f^{\prime }(x)=3f(x)+\alpha$, don't see a wall of symbols. See a dynamic system—a relationship where the rate of change of a function is tied to its current value. This is the heartbeat of calculus.

Our first step is to bring order to chaos. We rearrange the given equation into the standard linear form:

This is the classic $\frac{dy}{dx}+Py=Q$ structure. Here, our $P$ is $-3$ and our $Q$ is the constant $\alpha$.

By identifying this, we unlock the power of the Integrating Factor, or $IF$. The $IF$ acts as a mathematical 'multiplier' that collapses the derivative of a product into a single, integrable form. We calculate it as:

Now, we multiply our entire differential equation by this $IF$. This transforms the left side into the derivative of the product $f(x)e^{-3x}$.

Integrating both sides with respect to $x$ gives us:

By multiplying through by $e^{3x}$, we isolate our function: $f(x)=-\frac{\alpha }{3}+C{e}^{3x}$. This is the general solution, the family of all possible functions that satisfy our differential equation.

Now, we act like detectives. We have two unknowns, $\alpha$ and $C$, and two clues.

First, the limit: ${\lim }_{x\to -\infty }f(x)=7$. As $x$ plunges toward negative infinity, the term $C{e}^{3x}$ vanishes into the void of zero. We are left with $-\frac{\alpha }{3}=7$, which immediately reveals $\alpha =-21$.

With $\alpha$ in hand, our function becomes $f(x)=7+C{e}^{3x}$. Now, we use our second clue: $f(0)=1$. Substituting $x=0$ gives $1=7+C$, so $C=-6$.

Our function is fully revealed: $f(x)=7-6e^{3x}$.

We are asked to evaluate $9f(-\ln 3)$. Let's substitute $x=-\ln 3$ into our function:

Using the elegance of logarithmic properties, $3(-\ln 3)=\ln (3^{-3})=\ln (1/27)$. Thus, $e^{\ln (1/27)}=1/27$.

The expression simplifies to:

Finally, multiplying by 9, we arrive at our destination: $9\cdot \frac{61}{9}=61$.

Take a moment to appreciate this. We started with a vague differential equation and, through the systematic application of calculus, stripped away the layers to find a precise numerical truth. This is the beauty of physics and mathematics—the ability to predict the specific from the general.