A) \[-2\]
B) \[2\]
C) \[-1\]
D) \[0\]
Correct Answer: C
Solution :
\[x={{\tan }^{-1}}t\Rightarrow \frac{dx}{dt}=\frac{1}{1+{{t}^{2}}}\] \[\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}=\frac{dy}{dt}(1+{{t}^{2}})\] ? (1) \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dt}\left[ \frac{dy}{dt}(1+{{t}^{2}}) \right]\cdot \frac{dt}{dx}\] \[=\left[ \frac{dy}{dt}2t+(1+{{t}^{2}})\frac{{{d}^{2}}y}{d{{t}^{2}}} \right](1+{{t}^{2}})\] ? (2) Hence the given differential equation \[\frac{{{d}^{2}}y}{d{{x}^{2}}}+xy\frac{dy}{dx}+{{\sec }^{2}}x>0\], becomes \[(1+{{t}^{2}})\left[ 2t\frac{dy}{dt}+(1+{{t}^{2}})\frac{{{d}^{2}}y}{d{{t}^{2}}} \right]\] \[+y{{\tan }^{-1}}t\left[ \frac{dy}{dt}(1+{{t}^{2}}) \right]+(1+{{t}^{2}})=0\] Cancelling \[(1+{{t}^{2}})\] throughout we get \[(1+{{t}^{2}})\frac{{{d}^{2}}y}{d{{t}^{2}}}+(2t+y{{\tan }^{-1}}t)\frac{dy}{dt}=-1\Rightarrow k=-1\]
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