Let y=y(x) be a differentiable function on (0,) such that y(1)=2, and [_t x(t^2y(x)-x^2y(t…

Question

Let $y=y(x)$ be a differentiable function on $(0,\infty)$ such that $y(1)=2$, and
$$\lim_{t\to x}\frac{t^2y(x)-x^2y(t)}{x-t}=3\quad\text{for each }x>0$$
Then $2y(2)$ is equal to:

Accepted Answer

Write $t=x+h$, $h\to0$:
$$\lim_{h\to0}\frac{(x+h)^2y(x)-x^2y(x+h)}{-h}=3$$

Expanding $(x+h)^2=x^2+2xh+h^2$:
$$\lim_{h\to0}\left[\frac{x^2[y(x+h)-y(x)]}{h}\right]-\lim_{h\to0}y(x)(2x+h)=3$$
$x^2y'(x)-2xy(x)=3$

This is a linear ODE: $\dfrac{dy}{dx}-\dfrac2x y=\dfrac{3}{x^2}$. Integrating factor $=e^{\int-2/x\,dx}=\dfrac{1}{x^2}$.

$$y\cdot\frac1{x^2}=\int\frac{3}{x^4}\,dx=-\frac{1}{x^3}+C \implies y=-\frac1x+Cx^2$$

Using $y(1)=2$: $2=-1+C\implies C=3$, so $y(x)=-\dfrac1x+3x^2$.

$$y(2)=-\frac12+12=\frac{23}{2}\implies 2y(2)=23$$

Solutions · 1