A) \[x={{e}^{1+\tan \left( \frac{y}{x} \right)}}\]
B) \[x={{e}^{1-\tan \left( \frac{y}{x} \right)}}\]
C) \[x={{e}^{1+\tan \left( \frac{x}{y} \right)}}\]
D) \[x={{e}^{1-\tan \left( \frac{x}{y} \right)}}\]
Correct Answer: B
Solution :
Given,\[\frac{dy}{dx}=\frac{y}{x}-{{\cos }^{2}}\left( \frac{y}{x} \right)\] \[\Rightarrow \] \[\frac{x\,\,dy-y\,\,dx}{x}=-\left( {{\cos }^{2}}\frac{y}{x} \right)dx\] \[\Rightarrow \] \[{{\sec }^{2}}\left( \frac{y}{x} \right)\left( \frac{x\,\,dy-y\,\,dx}{{{x}^{2}}} \right)=-\frac{dy}{x}\] \[\Rightarrow \] \[{{\sec }^{2}}\frac{y}{x}\cdot d\left( \frac{y}{x} \right)=-\frac{dx}{x}\] \[\Rightarrow \] \[\tan \frac{y}{x}=-\log x+c\] When \[x=1,\,\,y=\frac{\pi }{4}\] \[c=1\] \[\therefore \] \[\tan \left( \frac{y}{x} \right)=1-\log x\] \[\Rightarrow \] \[\log x=1-\tan \left( \frac{y}{x} \right)\] \[\Rightarrow \] \[x={{e}^{1-\tan \left( \frac{y}{x} \right)}}\]
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