Let f(x) be a continuously differentiable function on the interval (0, ∞)  such that $f(1)=2$ and $\underset{t\to <!--\; \to \; -->x}{lim}\frac{t^{10}f(x)-<!--\; -\; -->x^{10}f(t)}{t^9-<!--\; -\; -->x^9}=1$

for each x > 0. Then, for all x > 0, f(x) is equal to

To find the function $f(x)$, we start by evaluating the given limit expression for a fixed $x>0$:

$\underset{t\to x}{lim}\frac{t^{10}f(x)-x^{10}f(t)}{t^9-x^9}=1$

As $t\to x$, both the numerator and denominator approach 0, giving an indeterminate form of $\frac{0}{0}$. Since $f(x)$ is continuously differentiable with respect to $t$, we can apply L'Hôpital's Rule by differentiating the numerator and the denominator with respect to $t$:

$\underset{t\to x}{lim}\frac{10t^{9}f(x)-x^{10}{f}^{\prime }(t)}{9t^8}=1$

Now, substitute $t=x$ into the limit expression:

$\frac{10x^{9}f(x)-x^{10}{f}^{\prime }(x)}{9x^8}=1$

Multiply both sides by $9x^8$:

$10x^{9}f(x)-x^{10}{f}^{\prime }(x)=9x^8$

Divide the entire equation by $x^{10}$ (since $x>0$):

$\frac{10}{x}f(x)-f^{\prime }(x)=\frac{9}{x^2}$

Rearranging into standard first-order linear differential equation form $f^{\prime }(x)+P(x)f(x)=Q(x)$:

$f^{\prime }(x)-\frac{10}{x}f(x)=-\frac{9}{x^2}$

The integrating factor $I(x)$ is:

$I(x)=e^{\int -\frac{10}{x}dx}=e^{-10\ln x}=x^{-10}$

Multiplying the differential equation by the integrating factor yields:

$\frac{d}{dx}[x^{-10}f(x)]=-9x^{-12}$

Integrating both sides with respect to $x$:

$x^{-10}f(x)=\int -9x^{-12}dx=-9\left(\frac{x^{-11}}{-11}\right)+C$

$x^{-10}f(x)=\frac{9}{11x^{11}}+C$

Multiplying by $x^{10}$:

$f(x)=\frac{9}{11x}+C{x}^{10}$

Using the initial condition $f(1)=2$:

$2=\frac{9}{11}+C\Rightarrow C=2-\frac{9}{11}=\frac{13}{11}$

Substituting $C$ back into the solution gives:

$f(x)=\frac{9}{11x}+\frac{13}{11}{x}^{10}$