A) \[{{x}^{2}}{{y}_{2}}+x{{y}_{1}}-{{n}^{2}}y=0\]
B) \[{{x}^{2}}{{y}_{2}}-x{{y}_{1}}-{{n}^{2}}y=0\]
C) \[{{x}^{2}}{{y}_{2}}+x{{y}_{1}}+{{n}^{2}}y=0\]
D) \[{{x}^{2}}{{y}_{2}}-x{{y}_{1}}+{{n}^{2}}y=0\]
Correct Answer: C
Solution :
\[y=b\cos \{n\log (x/n)\}\] On differentiating w.r.t.\[x,\]we get \[{{y}_{1}}=\frac{dy}{dx}=-b\sin \left[ n\log \frac{x}{n} \right].n.\frac{1}{x/n}.\frac{1}{n}\] \[{{y}_{1}}=-b\sin \left\{ n\log \frac{x}{n} \right\}.\frac{n}{x}\] Again, differentiating w.r.t.\[x,\]we get \[{{y}_{2}}=\frac{{{d}^{2}}y}{d{{x}^{2}}}=-b\cos \left( n\log \frac{x}{n} \right)\frac{n}{x}.\frac{n}{x/n}.\frac{1}{n}\] \[-b\sin \left\{ n\log \frac{x}{n} \right\}\left( -\frac{n}{{{x}^{2}}} \right)\] \[=-b\cos \left( n\log \frac{x}{n} \right).\frac{{{n}^{2}}}{{{x}^{2}}}+b\sin \left( n\log \frac{x}{n} \right).\frac{n}{{{x}^{2}}}\] \[\Rightarrow \] \[{{y}_{2}}=-\frac{{{n}^{2}}}{{{x}^{2}}}+(-{{y}_{1}}).\frac{1}{x}\] \[\Rightarrow \] \[{{y}_{2}}=\frac{-y{{n}^{2}}}{{{x}^{2}}}-\frac{{{y}_{1}}}{x}\] \[\Rightarrow \] \[{{x}^{2}}{{y}_{2}}=-y{{n}^{2}}-x{{y}_{1}}\] \[\Rightarrow \] \[{{x}^{2}}{{y}_{2}}+x{{y}_{1}}+{{n}^{2}}y=0\]
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