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 :
Given that,\[y=ax\cos \left( \frac{1}{x}+b \right)\] On differentiating w.r.t.\[x\], we get \[{{y}_{1}}=a\left[ \cos \left( \frac{1}{x}+b \right)-x\sin \left( \frac{1}{x}+b \right)\left( -\frac{1}{{{x}^{2}}} \right) \right]\] \[\Rightarrow \] \[{{y}_{1}}=a\left[ \cos \left( \frac{1}{x}+b \right)+\frac{1}{x}\sin \left( \frac{1}{x}+b \right) \right]\] Again differentiating, we get \[{{y}_{2}}=a\left[ -\sin \left( \frac{1}{x}+b \right)\left( -\frac{1}{{{x}^{2}}} \right)-\frac{1}{{{x}^{2}}}\sin \left( \frac{1}{x}+b \right) \right.\] \[\left. -\frac{1}{{{x}^{3}}}\cos \left( \frac{1}{x}+b \right) \right]\] \[\Rightarrow \] \[{{y}_{2}}=-\frac{ax}{{{x}^{4}}}\cos \left( \frac{1}{x}+b \right)\] \[\Rightarrow \] \[{{x}^{4}}{{y}_{2}}+y=0\]
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