A) \[y=-x\,{{(\sec x+\tan x)}^{-1}}+\frac{C}{\sec x+\tan x}+1\]
B) \[y=x+\frac{C}{\sec x+\tan x}+\frac{1}{\tan x}\]
C) \[y=\frac{x+1}{\sec x+\tan x}+C\]
D) \[y=x+\sec x-\tan \,x+C\]
Correct Answer: A
Solution :
Given, \[\frac{dy}{dx}+\sec x.y=\tan x\] \[\left( 0\le x\le \frac{\pi }{2} \right)\] \[IF={{e}^{\int{\sec \,x\,\,dx}}}={{e}^{log|\,\,\sec \,x+tan\,\,x+}}\] \[=|\sec \,x+\tan x|\] \[\therefore \] General solution is \[y.(IF)=\int{Q.\,\,(IF)\,dx+C}\] \[\Rightarrow \] \[y.|\sec \,x+\tan x|\] \[=\int{\tan x.\,(\sec \,x+\tan x)\,dx+C}\] \[\Rightarrow \] \[y(\sec \,x+\tan \,x)\] \[=\int{\tan x.\sec \,x\,dx+\int{{{\tan }^{2}}\,x\,dx+C}}\] \[\Rightarrow \] \[y(\sec x+\tan x)=sec\,x+\int{({{\sec }^{2}}\,x-1)\,dx}\] \[+C\] \[\Rightarrow \] \[y\,(\sec x+\tan x)=(\sec \,x+\tan x)-x+C\] \[\Rightarrow \] \[y=1+\frac{(C-x)}{(\sec x+\tan x)}\] \[=-x{{(\sec x+\tan x)}^{-1}}+\frac{C}{(\sec \,x+\tan x)}+1\]
You need to login to perform this action.
You will be redirected in
3 sec
