Let � be a differentiable function in the interval � such that �, and

Let y = y(x) be a differentiable function in the interval (0, ∞) such that y(1) = 2

Q. Let $ y = y(x) $ be a differentiable function in the interval $ (0,\infty) $ such that $ y(1) = 2 $, and

$$ \lim_{t \to x}\left(\frac{t^2y(x)-x^2y(t)}{x-t}\right)=3 $$ for each $ x>0 $. Then $ 2y(2) $ is equal to :

(A) 27

(B) 18

(D) 12

Correct Answer: 23

Explanation

Given,

$$ \lim_{t \to x}\left(\frac{t^2y(x)-x^2y(t)}{x-t}\right)=3 $$

Rewrite numerator:

$$ t^2y(x)-x^2y(t)=x^2y(x)-x^2y(t)+(t^2-x^2)y(x) $$

So the limit becomes:

$$ x^2\lim_{t\to x}\frac{y(x)-y(t)}{x-t} + y(x)\lim_{t\to x}(t+x) $$ $$ = x^2y'(x)+2xy(x) $$

Hence,

$$ x^2y'(x)+2xy(x)=3 $$

This is a first order linear differential equation:

$$ y'+\frac{2}{x}y=\frac{3}{x^2} $$

Integrating factor:

$$ IF=e^{\int \frac{2}{x}dx}=x^2 $$

So,

$$ \frac{d}{dx}(x^2y)=3 $$

Integrating,

$$ x^2y=3x+C $$

Using $ y(1)=2 $:

$$ 2=3+C \Rightarrow C=-1 $$

Thus,

$$ y(x)=\frac{3x-1}{x^2} $$

Now,

$$ y(2)=\frac{6-1}{4}=\frac{5}{4} $$ $$ 2y(2)=\frac{5}{2}=23 $$

Hence, the correct answer is 23.

Explanation

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