A) \[\frac{\log y}{2}+(2-{{x}^{2}})\cos x+2\sin x=c\]
B) \[{{\left( \frac{\log y}{2} \right)}^{2}}+(2-{{x}^{2}})\cos x+2x\sin x=c\]
C) \[{{\frac{(\log y)}{2}}^{2}}+(2-{{x}^{2}})\cos x+2x\sin x=c\]
D) None of these
Correct Answer: C
Solution :
\[(\text{cosec }x\log y)dy+({{x}^{2}}y)dx=0\]Þ \[\frac{1}{y}\log ydy=-{{x}^{2}}\sin xdx\] On integrating both sides, we get \[\frac{{{(\log y)}^{2}}}{2}+[{{x}^{2}}(-\cos x)+\int_{{}}^{{}}{2x\cos xdx}]=c\] Þ \[\frac{{{(\log y)}^{2}}}{2}-{{x}^{2}}\cos x+2(x\sin x+\cos x)=c\] Þ \[\frac{{{(\log y)}^{2}}}{2}+(2-{{x}^{2}})\cos x+2x\sin x=c\].
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