SATHEE: Differential Equations 2 Question 11

Differential Equations 2 Question 11

11. Let $f$ be a differentiable function such that $f^{'} (x) = 7 - \frac{3}{4} \frac{f (x)}{x}, (x > 0)$ and $f (1) \neq 4$ . Then, $lim_{x \to 0^{+}} x f \frac{1}{x}$

(2019 Main, 10 Jan II)

(a) does not exist

(b) exists and equals $\frac{4}{7}$

(d) exists and equals 4

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Answer:

Correct Answer: 11. (d)

Solution:

Given, $f^{'} (x) = 7 - \frac{3}{4} \frac{f (x)}{x}, (x > 0)$

On putting $f (x) = y$ and $f^{'} (x) = \frac{d y}{d x}$ , then we get

$\begin{aligned} \frac{d y}{d x} & = 7 - \frac{3}{4} \frac{y}{x} \\ \Rightarrow \frac{d y}{d x} + \frac{3}{4 x} y & = 7 \end{aligned}$

which is a linear differential equation of the form $\frac{d y}{d x} + P y = Q$ , where $P = \frac{3}{4 x}$ and $Q = 7$ .

Now, integrating factor (IF) $= e^{\int \frac{3}{4 x} d x}$

$= e^{\frac{3}{4} \log x} = e^{\log x^{3 / 4}} = x^{3 / 4}$

and solution of differential Eq. (i) is given by

$\begin{aligned} y (I F) = \int (Q \cdot (I F)) d x + C \\ y x^{3 / 4} = \int 7 x^{3 / 4} d x + C \\ \Rightarrow y x^{3 / 4} = 7 \frac{x^{\frac{3}{4} + 1}}{\frac{3}{4} + 1} + C \\ \Rightarrow y x^{3 / 4} = 4 x^{\frac{7}{4}} + C \\ \Rightarrow y = 4 x + C x^{- 3 / 4} \\ So, y = f (x) = 4 x + C \cdot x^{- 3 / 4} \\ Now, f \frac{1}{x} = \frac{4}{x} + C \cdot x^{3 / 4} \\ ∴ lim_{x \to 0^{+}} x f \frac{1}{x} = lim_{x \to 0^{+}} x \frac{4}{x} + C x^{3 / 4} = lim_{x \to 0^{+}} (4 + C x^{7 / 4}) = 4 \end{aligned}$

Differential Equations 2 Question 11

11. Let $f$ be a differentiable function such that $f^{'} (x) = 7 - \frac{3}{4} \frac{f (x)}{x}, (x > 0)$ and $f (1) \neq 4$ . Then, $lim_{x \to 0^{+}} x f \frac{1}{x}$

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Differential Equations 2 Question 11

11. Let f be a differentiable function such that f′(x)=7−34f(x)x,(x>0) and f(1)≠4. Then, limx→0+xf1x

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11. Let $f$ be a differentiable function such that $f^{'} (x) = 7 - \frac{3}{4} \frac{f (x)}{x}, (x > 0)$ and $f (1) \neq 4$ . Then, $lim_{x \to 0^{+}} x f \frac{1}{x}$