A) \[x\sec \frac{y}{x}=\sqrt{2}\]
B) \[x\tan \frac{y}{x}=1\]
C) \[x{{\tan }^{2}}\frac{y}{x}=1\]
D) \[x{{\sec }^{2}}\frac{y}{x}=2\]
Correct Answer: B
Solution :
[b] The given differential equation is \[\left( \tan \left( \frac{y}{x} \right)-\frac{y}{x}{{\sec }^{2}}\left( \frac{y}{x} \right) \right)+{{\sec }^{2}}\left( \frac{y}{x} \right)\frac{dy}{dx}=0\] Let \[y=vx.\] Then \[\frac{dy}{dx}=v+x\frac{dv}{dx}.\] \[\therefore \,\,\,\,\left( \tan v-v{{\sec }^{2}}v \right)+{{\sec }^{2}}v\left( v+x\frac{dv}{dx} \right)=0\] \[\Rightarrow \,\,\,\,\,\int{\frac{{{\sec }^{2}}v}{\tan v}}dv=-\int{\frac{dx}{x}}\] \[\Rightarrow \,\,\,\,\,{{\log }_{e}}\tan v=-{{\log }_{e}}x+{{\log }_{e}}c\] \[\Rightarrow \,\,\,\,\,x\tan v=c\] \[\Rightarrow \,\,\,\,\,x\tan \frac{y}{x}=c\] Using \[y(1)=\frac{\pi }{4},\] we get \[c=1.\] \[\therefore \,\,\,\,\,\,\,x\tan \frac{y}{x}=1\]
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