A) 2
B) 1
C) ½
D) 3
Correct Answer: A
Solution :
\[\frac{{{d}^{2}}y}{d{{x}^{2}}}-\sqrt{\frac{dy}{dx}-3}=x\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}-x=\sqrt{\frac{dy}{dx}-3}\] Squaring both sides, we get \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}}-x \right)}^{2}}=\left( \frac{dy}{dx}-3 \right)\] Þ\[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{2}}+{{x}^{2}}-2x\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{dy}{dx}-3\]. Clearly, degree = 2.
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