A) \[{{y}^{2}}=1-\frac{x}{\sec x\,+\tan x}\]
B) \[{{y}^{2}}=1+\frac{x}{\sec x\,+\tan x}\]
C) \[y=1+\frac{x}{\sec x\,+\tan x}\]
D) \[y=1-\frac{x}{\sec x\,+\tan x}\]
Correct Answer: D
Solution :
\[2y\frac{dy}{dx}+{{y}^{2}}\sec x=\tan x\] put \[{{y}^{2}}=t\] \[\Rightarrow \] \[2y=\frac{dy}{dx}=\frac{dt}{dx}\] \[\frac{dt}{dx}+t\,\sec =\tan x\] \[I.F.\,\,{{e}^{\int{\sec \,xdx}}}={{e}^{\ell n(\sec x+\tan x)}}=\sec x+\tan x\] \[t.(\sec x+\tan x)=\int{(\sec x+\tan x)\tan x\,\,dx}\] \[=\int{\sec x\ \tan x+}\int{{{\tan }^{2}}xdx}\] \[{{y}^{2}}(\sec x+\tan x)=\sec x+\tan x-x+c\] \[y(0)=1\] \[\Rightarrow \] \[c=0\] \[\Rightarrow \] \[{{y}^{2}}=1-\frac{x}{\sec x\,+\tan x}\]You need to login to perform this action.
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