Differential Equations and Limits
DAILY_CHALLENGE
Grade None
Question:
Let $f(x)$ be a continuously differentiable function on the interval $(0,\infty)$ such that $f(1)=2$ and
$$\lim_{t \to x} \dfrac{t^{10}f(x)-x^{10}f(t)}{t^9-x^9} = 1$$
for each $x>0$. Then, for all $x>0$, $f(x)$ is equal to
$\dfrac{31}{11x} - \dfrac{9}{11}x^{10}$
$\dfrac{9}{11x} + \dfrac{13}{11}x^{10}$
$\dfrac{-9}{11x} + \dfrac{31}{11}x^{10}$
$\dfrac{13}{11x} + \dfrac{9}{11}x^{10}$
Step-by-Step Solution
Key Concept: L'Hôpital on limit with parameter gives ODE; linear ODE solved with integrating factor
Apply L'Hôpital's rule w.r.t. $t$ (the limit is $0/0$ form as $t\to x$):
$$\lim_{t\to x}\frac{10t^9 f(x)-x^{10}f'(t)}{9t^8} = \frac{10x^9 f(x)-x^{10}f'(x)}{9x^8} = 1$$
$$\Rightarrow 10xf(x)-x^2 f'(x)=9 \Rightarrow f'(x)-\frac{10}{x}f(x)=-\frac{9}{x^2}$$
Integrating factor: $x^{-10}$.
$$\frac{d}{dx}[x^{-10}f(x)]=-9x^{-12}$$
$$x^{-10}f(x)=\frac{9}{11}x^{-11}+C$$
$$f(x)=\frac{9}{11x}+Cx^{10}$$
$f(1)=2$: $\dfrac{9}{11}+C=2\Rightarrow C=\dfrac{13}{11}$.
$$f(x)=\frac{9}{11x}+\frac{13}{11}x^{10}$$
Correct Answer: B