A) \[{{x}^{2}}{{y}_{2}}+y=0\]
B) \[{{x}^{4}}{{y}_{2}}+y=0\]
C) \[x{{y}_{2}}-y=0\]
D) \[{{x}^{4}}{{y}_{2}}-y=0\]
Correct Answer: B
Solution :
\[y=ax\cos \left( \frac{1}{x}+b \right)\] .?. (i) Differentiate (i), we get \[{{y}_{1}}=a\,\left[ \cos \left( \frac{1}{x}+b \right)-x\sin \left( \frac{1}{x}+b \right)\text{ }\left( \frac{-1}{{{x}^{2}}} \right) \right]\] \[=a\left[ \cos \,\left( \frac{1}{x}+b \right)+\frac{1}{x}\sin \left( \frac{1}{x}+b \right) \right]\] .....(ii) Again, differentiate (ii), we get \[{{y}_{2}}=\frac{-a}{{{x}^{3}}}\cos \left( \frac{1}{x}+b \right)\] \[=\frac{-ax}{{{x}^{4}}}\cos \left( \frac{1}{x}+b \right)\]\[=\frac{-y}{{{x}^{4}}}\] Þ \[\left( \frac{1}{{{y}^{2}}}-\frac{1}{y} \right)dy+\left( \frac{1}{{{x}^{2}}}+\frac{1}{x} \right)dx=0\].
You need to login to perform this action.
You will be redirected in
3 sec
