A) \[{{e}^{x}}\tan y=C\]
B) \[C{{e}^{x}}={{(1-\tan \,\,y)}^{3}}\]
C) \[C\tan \,\,y={{(1-{{e}^{x}})}^{2}}\]
D) \[\tan \,\,y=C{{(1-{{e}^{x}})}^{3}}\]
Correct Answer: D
Solution :
[d] \[3{{e}^{x}}\tan \,\,y\,\,dx+(1-{{e}^{x}}){{\sec }^{2}}y\,\,dy=0\] \[\Rightarrow \frac{3{{e}^{x}}}{1-{{e}^{x}}}dx+\frac{{{\sec }^{2}}y}{\tan y}dy=0\] Integrating we get \[\int{\frac{3{{e}^{x}}}{1-{{e}^{x}}}dx+\int{\frac{{{\sec }^{2}}y}{\tan \,y}dy=D}}\], \[\Rightarrow -3\ell n(1-{{e}^{x}})+\ell n\,tan\,y=D\] \[\Rightarrow -\ell n{{(1-{{e}^{x}})}^{3}}+\ell n\,tan\,y=D\] \[\Rightarrow \ell n\frac{\tan y}{{{(1-{{e}^{x}})}^{3}}}=\ell nC\Rightarrow \tan y=C{{(1-{{e}^{x}})}^{3}}\]
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