Animated Solution for Mathematics - Differential Equations: Let $f:(0,\infty )\to R$ be a function which is differentiable at all points of its domain and satisfies the condition $x^2{f}^{\prime }(x)=2xf(x)+3$ with $f(1)=4.$ Then $2f(2)$ is equal to :

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to peel back the layers of a differential equation that might look intimidating at first glance, but hides a beautiful, elegant symmetry.

The problem asks us to solve $x^2{f}^{\prime }(x)=2xf(x)+3$ given the anchor point $f(1)=4$. Our mission is to find $2f(2)$. Let us embark on this journey.

The Art of Rearrangement

When you encounter a differential equation, your first task is to organize the chaos. We have $f^{\prime }(x)$ and $f(x)$ on opposite sides.

Let us bring them together:

Now, pause and look at this expression. Does it stir a memory? In calculus, we are often looking for patterns that match the derivative rules we learned in our early days.

Specifically, look at the quotient rule:

If we set $v=x^2$, then $v^{\prime }=2x$. The numerator of our quotient rule would be $x^2{f}^{\prime }(x)-f(x)(2x)$. This is exactly what we have on our left side!

The Strategic Division

We have the numerator of the quotient rule, but we are missing the denominator, $v^2$. Since $v=x^2$, $v^2$ must be $x^4$.

This is the 'Aha!' moment. By dividing the entire equation by $x^4$, we transform the left side into a perfect derivative:

Now, the left side is simply the derivative of a quotient:

We have successfully condensed a complex relationship into a single, manageable derivative.

The Integration

With the equation in the form $\frac{d}{dx}\left(\frac{f(x)}{x^2}\right)=3x^{-4}$, the path forward is clear. We integrate both sides with respect to $x$.

The integral of a derivative is the function itself, so the left side becomes $\frac{f(x)}{x^2}$. On the right, we apply the power rule for integration:

Thus, we arrive at the general solution:

Multiplying through by $x^2$, we isolate our function:

Finding the Anchor

We are almost there. We have a general solution, but we need the specific one. We use our anchor, $f(1)=4$.

Substituting $x=1$ and $f(1)=4$ into our equation, we get:

Solving for $C$, we find $C=5$. Our specific function is revealed:

The Final Calculation

Finally, we calculate $f(2)$:

The question asks for $2f(2)$, so we multiply by $2$:

And there it is—a clean, satisfying integer. Remember, JEE problems aren't just about calculation; they are about recognizing the hidden structures of mathematics. Keep practicing, keep observing, and the patterns will reveal themselves to you. The final answer is 39.