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Question: Answered & Verified by Expert
If $\mathrm{a}$ and $\mathrm{b}$ are the arbitrary constants, then the differential equation corresponding to the family of curves given by $y=x[a \cos (\log x)+b \sin (\log x)]$ is
MathematicsDifferential EquationsTS EAMCETTS EAMCET 2023 (14 May Shift 2)
Options:
  • A $x \frac{d^2 y}{d x^2}+x \frac{d y}{d x}-2 y=0$
  • B $x^2 \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+2 y=0$
  • C $x^2 \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-2 y=0$
  • D $x^2 \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+y=0$
Solution:
1211 Upvotes Verified Answer
The correct answer is: $x^2 \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+2 y=0$
$y=x[a \cos (\log x)+b \sin (\log x)]$
$\begin{aligned} & y=a x \cos \log x+b x \sin \log x \\ & \frac{d y}{d x}=a\left[-x \sin \log x \cdot \frac{1}{x}+\cos \log x\right] \\ & +b\left[x(\cos \log x) \cdot \frac{1}{x}+\sin (\log x)\right] \\ & =-a \sin \log x+a \cos \log x+b \cos \log x+b \sin \log x \\ & =(b-a) \sin \log x+(a+b) \cos \log x \\ & \frac{d^2 y}{d x^2}=\frac{(b-a)}{x} \cdot \cos \log x-\frac{(a+b)}{x} \cdot \sin (\log x) \\ & \Rightarrow \quad x \frac{d^2 y}{d x^2}=(b-a) \cos \log x-(a+b) \sin \log x \\ & \end{aligned}$
$\Rightarrow \quad x^2 \frac{d^2 y}{d x^2}=x(b-a) \cos \log x-x(a+b) \sin \log x$ ...(i)
$\Rightarrow \quad x \frac{d y}{d x}=x(b-a) \sin \log x+x(a+b) \cos \log x$ ...(ii)
(i)-(ii)
$\begin{aligned} & x^2 \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=-2 a x \cos \log x-2 b x \sin \log x \\ & \Rightarrow x^2 \frac{d^2 y}{d x^2}-x \frac{d y}{d x}+2 y=0 .\end{aligned}$

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