A) \[(1-{{x}^{2}}){{y}_{2}}+x{{y}_{1}}+{{p}^{2}}y=0\]
B) \[(1-{{x}^{2}}){{y}_{2}}+x{{y}_{1}}-{{p}^{2}}y=0\]
C) \[(1+{{x}^{2}}){{y}_{2}}-x{{y}_{1}}+{{p}^{2}}y=0\]
D) \[(1-{{x}^{2}}){{y}_{2}}-x{{y}_{1}}+{{p}^{2}}y=0\]
Correct Answer: D
Solution :
Given mat \[x=\sin \,t,\,y=\cos \,pt\] \[\frac{dx}{dt}=\cos \,t,\,\frac{dy}{dt}=-p\,\sin \,pt\] \[\frac{dx}{dt}=-\,\frac{p\,\sin \,pt}{\cos \,t}\] \[\Rightarrow \] \[{{y}_{1}}=\frac{-p\,\sqrt{1-{{y}^{2}}}}{\sqrt{1-{{x}^{2}}}}\] \[\Rightarrow \,\,\,{{y}_{1}}\sqrt{1-{{x}^{2}}}=-p\,\,\sqrt{1-y}\] On squaring both sides, we get \[y_{1}^{2}\,(1-{{x}^{2}})={{p}^{2}}(1-{{y}^{2}})\] Again differentiating \[2{{y}_{1}}{{y}_{2}}\,(1-{{x}^{2}})-2x\,y_{1}^{2}=-2\,y\,{{y}_{1}}\,{{p}^{2}}\] or \[(1-{{x}^{2}}){{y}_{2}}-x\,{{y}_{1}}+{{p}^{2}}y=0\]
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