Animated Solution for Mathematics - Differential Equations: Let $f$ be a differentiable function in the interval $(0,\infty )$ such that $f(1)=1$ and ${\lim }_{t\to x}\frac{t^{2}f(x)-x^{2}f(t)}{t-x}=1$ for each $x>0$. Then $2f(2)+3f(3)$ is equal to _______

Imagine you are standing at the edge of a mathematical precipice. You are presented with the limit:

At first glance, it looks like a standard limit problem, but it is actually a hidden door to a differential equation. When you test the limit by substituting $t=x$, you immediately see the $0/0$ indeterminate form.

This is the universe telling you that there is more beneath the surface. By applying L'Hopital's Rule, differentiating with respect to $t$, and treating $x$ as a constant, we transform this limit into the elegant equation:

This is the moment where the problem shifts from a limit exercise to a differential equation challenge.

Unlocking the Function

Now that we have $2xf(x)-x^2{f}^{\prime }(x)=1$, our mission is to isolate $f(x)$. We must bring this into the standard linear form $f^{\prime }(x)+P(x)f(x)=Q(x)$.

By dividing the entire equation by $-x^2$, we get:

Here, $P(x)=-\frac{2}{x}$ and $Q(x)=-\frac{1}{x^2}$. The integrating factor, $I.F.=e^{\int P(x)dx}$, is the key that unlocks the solution.

Calculating the integrating factor:

This is a beautiful step where the logarithms and exponentials cancel out perfectly.

Solving for $f(x)$

With the integrating factor in hand, we multiply our standard form equation by $\frac{1}{x^2}$ to get:

Integrating both sides, we find:

Multiplying by $x^2$, we arrive at the general solution:

Using the initial condition $f(1)=1$, we find $1=\frac{1}{3}+C$, which gives $C=\frac{2}{3}$. Thus, our function is:

Finally, calculating $2f(2)+3f(3)$ is a matter of simple substitution. We find:

Plugging these into $2f(2)+3f(3)$ gives:

You have successfully navigated the bridge between limits and differential equations. The final result is 24.