Limit Continuity And Differentiability Ques 27
- If $(a+b x) e^{j / x}=x$, then prove that
$ x^3 \frac{d^2 y}{d x^2}=\left(x \frac{d y}{d x}-y\right)^2 \text {. } $
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Answer:
Correct Answer: 27.$(11)$
Solution: Given, $(a+b x) e^{y / x}=x \quad \Rightarrow \quad y=x \log \left(\frac{x}{a+b x}\right)$
$\Rightarrow \quad y=x[\log (x)-\log (a+b x)] \quad ……..(i) $
On differentiating both sides, we get
$ \begin{aligned} \frac{d y}{d x} & =x\left(\frac{1}{x}-\frac{b}{a+b x}\right)+1[\log (x)-\log (a+b x)] \\ \Rightarrow \quad x \frac{d y}{d x} & =x^2\left(\frac{a}{x(a+b x)}\right)+y \\ \Rightarrow \quad x y_1 & =\frac{a x}{a+b x}+y \quad ……..(ii) \end{aligned} $
Again, differentiating both sides, we get
$ \begin{aligned} & x y_2+y_1 =a\left\{\frac{(a+b x) \cdot 1-x \cdot b}{(a+b x)^2}\right\}+y_1 \\ \Rightarrow & \quad x^3 y_2 =\frac{a^2 x^2}{(a+b x)^2} \\ \Rightarrow & \quad x^3 y_2 =\left(\frac{a x}{(a+b x)}\right)^2 \quad [\text{from Eq. (ii)}] \\ \Rightarrow & \quad x^3 y_2 =\left(x y_1-y\right)^2 \\ \Rightarrow & \quad x^3 \frac{d^2 y}{d x^2} =\left(x \frac{d y}{d x}-y\right)^2 \end{aligned} $
BETA
