Q105 / 160 · TS EAMCET 2024 (10 May Shift 2) · Mathematics
If $y=\log \left(x-\sqrt{x^2-1}\right)$, then $\left(x^2-1\right) y^{\prime \prime}+x y^{\prime}+e^y+\sqrt{x^2-1}=$
- A0
- B1
- C$\sqrt{x^2-1}$
- Correct answerD$x$
Solution
$y=\log \left(x-\sqrt{x^2-1}\right) \Rightarrow e^y=x-\sqrt{x^2-1}$
Differentiate w.r.t x , we get $\begin{aligned} & \Rightarrow e^y y^{\prime}=1-\frac{x}{\sqrt{x^2-1}} \Rightarrow e^y y^{\prime}=-\frac{e^y}{\sqrt{x^2-1}} \\ & \Rightarrow y^{\prime}=\frac{-1}{\sqrt{x^2-1}} \Rightarrow y^{\prime \prime}=\frac{x}{\left(x^2-1\right)^{\frac{3}{2}}}\end{aligned}$
Then $\left(x^2-1\right) y^{\prime \prime}+y^{\prime}+e^y+\sqrt{x^2-1}=x$
Differentiate w.r.t x , we get $\begin{aligned} & \Rightarrow e^y y^{\prime}=1-\frac{x}{\sqrt{x^2-1}} \Rightarrow e^y y^{\prime}=-\frac{e^y}{\sqrt{x^2-1}} \\ & \Rightarrow y^{\prime}=\frac{-1}{\sqrt{x^2-1}} \Rightarrow y^{\prime \prime}=\frac{x}{\left(x^2-1\right)^{\frac{3}{2}}}\end{aligned}$
Then $\left(x^2-1\right) y^{\prime \prime}+y^{\prime}+e^y+\sqrt{x^2-1}=x$
Hence, the correct answer is option: 'D'.