A) \[\tan ({{x}^{2}}+{{y}^{2}})=\frac{{{x}^{2}}}{{{y}^{2}}}+c\]
B) \[cot({{x}^{2}}+{{y}^{2}})=\frac{{{x}^{2}}}{{{y}^{2}}}+c\]
C) \[tan({{x}^{2}}+{{y}^{2}})=\frac{{{y}^{2}}}{{{x}^{2}}}+c\]
D) \[cot({{x}^{2}}+{{y}^{2}})=\frac{{{y}^{2}}}{{{x}^{2}}}+c\]
Correct Answer: A
Solution :
[a] The given equation can be written as \[\frac{x\,dx+y\,dy}{(y\,dx-x\,dy)/{{y}^{2}}}={{y}^{2}}\frac{x}{{{y}^{3}}}{{\cos }^{2}}({{x}^{2}}+{{y}^{2}})\] Or \[\frac{xdx+ydx}{{{\cos }^{2}}({{x}^{2}}+{{y}^{2}})}=\frac{x}{y}\left( \frac{ydx-xdy}{{{y}^{2}}} \right)\] Or \[\frac{1}{2}{{\sec }^{2}}({{x}^{2}}+{{y}^{2}})d({{x}^{2}}+{{y}^{2}})=\frac{x}{y}d\left( \frac{x}{y} \right)\] On integrating, we get \[\frac{1}{2}\tan ({{x}^{2}}+{{y}^{2}})=\frac{1}{2}{{\left( \frac{x}{y} \right)}^{2}}+\frac{c}{2}\] Or \[\tan ({{x}^{2}}+{{y}^{2}})=\frac{{{x}^{2}}}{{{y}^{2}}}+c\]
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