A) 1
B) -1
C) -5
D) 5
Correct Answer: C
Solution :
Let\[\frac{dy}{dx}+y\tan x=\sin 2x\] \[I.F={{e}^{\int_{{}}^{{}}{\tan xdx}}}={{e}^{-\log \cos x}}=\sec x\] Required solution is \[y(\sec x)=\int_{{}}^{{}}{\sin 2x\sec xdx+c}\] \[y(\sec x)=\int_{{}}^{{}}{\frac{2\sin x\cos x}{\cos x}dx+c}\] \[y(\sec x)=2\int_{{}}^{{}}{\sin xdx+c}\] \[y(\sec x)=-2\cos x+c\] ......(1) Given\[y(0)=1\] \[\therefore \]put \[x=0\]and \[y=1,\]we get \[\Rightarrow \]\[c=1+2\Rightarrow c=3\] \[\therefore \]from eqn (1), we have \[y\sec x=-2\cos x+3\] To find \[y(\pi ),\] put \[x=\pi \] in eqn (2), we get \[y(\sec \pi )=-2\cos \pi +3\] \[y=2(-1)(-1)+3(1)=23=5\]
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