A) \[p>q\]
B) \[\frac{p}{q}=\frac{1}{2}\]
C) \[p=q\]
D) \[p<q\]
Correct Answer: D
Solution :
\[{{y}^{2}}{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{2}}+{{x}^{2}}{{y}^{2}}-\sin x=-3x{{\left( \frac{dy}{dx} \right)}^{1/3}}\] |
\[{{\left( {{y}^{2}}{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{2}}+{{x}^{2}}{{y}^{2}}-\sin x \right)}^{3}}=-9{{x}^{3}}\left( \frac{dy}{dx} \right)\] |
Here order = 2 = p |
Degree = 6 = q \[\therefore \] \[p<q\] |
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