Question

Let $y=y(x)$ be a differentiable function in the interval $(0,\infty)$ such that $y(1)=2$, and \[ \lim_{t\to x}\left(\frac{t^2y(x)-x^2y(t)}{x-t}\right)=3 \text{ for each } x>0. \] Then $2y(2)$ is equal to

• A. 23
• B. 12
• C. 18
• D. 27

Hint:

Limits involving functions at $x$ and $t$ often reduce to derivatives—try rewriting them in derivative form.

Answer 1

The Correct Option is A

Step 1: Simplify the given limit.
Rewrite the expression as \[ \lim_{t\to x}\frac{t^2y(x)-x^2y(t)}{x-t} =\lim_{t\to x}\frac{x^2y(t)-t^2y(x)}{t-x} \] Step 2: Apply differentiation.
This limit is equal to \[ \frac{d}{dt}\left[x^2y(t)-t^2y(x)\right]_{t=x} \] \[ = x^2y'(x)-2xy(x) \] Given that this equals $3$, \[ x^2y'(x)-2xy(x)=3 \] Step 3: Solve the differential equation.
\[ y'(x)-\frac{2}{x}y(x)=\frac{3}{x^2} \] Integrating factor: \[ \text{IF}=e^{\int -\frac{2}{x}dx}=\frac{1}{x^2} \] \[ \frac{d}{dx}\left(\frac{y}{x^2}\right)=\frac{3}{x^4} \] Integrating, \[ \frac{y}{x^2}=-\frac{1}{x^3}+C \] \[ y=-\frac{1}{x}+Cx^2 \] Step 4: Use the given condition $y(1)=2$.
\[ 2=-1+C \Rightarrow C=3 \] \[ y(x)=3x^2-\frac{1}{x} \] Step 5: Find $2y(2)$.
\[ y(2)=12-\frac12=\frac{23}{2} \] \[ 2y(2)=23 \]