Question ID: #899
Let $y=y(x)$ be a differentiable function in the interval $(0, \infty)$ such that $y(1)=2$ and $\lim_{t\rightarrow x}\left(\frac{t^{2}y(x)-x^{2}y(t)}{x-t}\right)=3$ for each $x > 0$. Then $2y(2)$ is equal to
- (1) 18
- (2) 23
- (3) 27
- (4) 12
Solution:
Given Limit:
$$\lim_{t \to x} \frac{t^2y(x) – x^2y(t)}{x-t} = 3$$
Using L’Hospital’s Rule (differentiating with respect to $t$):
$$\lim_{t \to x} \frac{2t y(x) – x^2 y'(t)}{-1} = 3$$
$$x^2 y'(x) – 2x y(x) = 3$$
Divide by $x^2$:
$$\frac{dy}{dx} – \frac{2}{x}y = \frac{3}{x^2}$$
Integrating Factor (I.F.):
$$I.F. = e^{\int -\frac{2}{x} dx} = e^{-2\ln x} = \frac{1}{x^2}$$
Solution of LDE:
$$y \cdot \frac{1}{x^2} = \int \frac{3}{x^2} \cdot \frac{1}{x^2} dx = \int 3x^{-4} dx$$
$$\frac{y}{x^2} = 3\frac{x^{-3}}{-3} + C = -\frac{1}{x^3} + C$$
$$y(x) = -\frac{1}{x} + Cx^2$$
Given $y(1) = 2$:
$$2 = -1 + C \Rightarrow C = 3$$
$$y(x) = 3x^2 – \frac{1}{x}$$
We need $2y(2)$:
$$y(2) = 3(2^2) – \frac{1}{2} = 12 – 0.5 = 11.5$$
$$2y(2) = 2(11.5) = 23$$
Ans. (2)
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