Differential Equations 2 Question 11
11. Let $f$ be a differentiable function such that $f^{\prime}(x)=7-\frac{3}{4} \frac{f(x)}{x},(x>0)$ and $f(1) \neq 4$. Then, $\lim _{x \rightarrow 0^{+}} x f \frac{1}{x}$
(2019 Main, 10 Jan II)
(a) does not exist
(c) exists and equals 0
(b) exists and equals $\frac{4}{7}$
(d) exists and equals 4
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Answer:
Correct Answer: 11. (d)
Solution:
- Given, $f^{\prime}(x)=7-\frac{3}{4} \frac{f(x)}{x},(x>0)$
On putting $f(x)=y$ and $f^{\prime}(x)=\frac{d y}{d x}$, then we get
$$ \begin{aligned} \frac{d y}{d x} & =7-\frac{3}{4} \frac{y}{x} \\ \Rightarrow \quad \frac{d y}{d x}+\frac{3}{4 x} y & =7 \end{aligned} $$
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
$$ \begin{aligned} & y(IF)=\int(Q \cdot(IF)) d x+C \\ & y x^{3 / 4}=\int 7 x^{3 / 4} d x+C \\ & \Rightarrow \quad y x^{3 / 4}=7 \frac{x^{\frac{3}{4}+1}}{\frac{3}{4}+1}+C \\ & \Rightarrow \quad y x^{3 / 4}=4 x^{\frac{7}{4}}+C \\ & \Rightarrow \quad y=4 x+C x^{-3 / 4} \\ & \text { So, } \quad y=f(x)=4 x+C \cdot x^{-3 / 4} \\ & \text { Now, } \quad f \frac{1}{x}=\frac{4}{x}+C \cdot x^{3 / 4} \\ & \therefore \lim _{x \rightarrow 0^{+}} x f \frac{1}{x}=\lim _{x \rightarrow 0^{+}} x \frac{4}{x}+C x^{3 / 4}=\lim _{x \rightarrow 0^{+}}\left(4+C x^{7 / 4}\right)=4 \end{aligned} $$
