A) \[y+\log \,\,y+{{e}^{x}}{{\cos }^{2}}x=2\]
B) \[\log \,\,(y+1)+{{e}^{x}}{{\cos }^{2}}x=1\]
C) \[y+\log \,y={{e}^{x}}{{\cos }^{2}}x\]
D) \[(y+1)+{{e}^{x}}{{\cos }^{2}}x=2\]
Correct Answer: A
Solution :
We have \[{{e}^{-x}}(y+1)dy+(co{{s}^{2}}x-\sin 2x)y\,\,dx=0\] \[\Rightarrow \]\[\frac{(y+1)}{y}dy=-{{e}^{x}}\,({{\cos }^{2}}x-\sin 2x)dx\] \[\Rightarrow \]\[\left( 1+\frac{1}{y} \right)dy=-{{e}^{x}}({{\cos }^{2}}x-\sin 2x)dx\] On integrating both sides, we get \[y+\log y=-{{e}^{x}}{{\cos }^{2}}x+\int{{{e}^{x}}\sin 2xdx}\] \[-\int{{{e}^{x}}\sin 2x\,\,dx}+c\] \[\Rightarrow \] \[y+\log =-{{e}^{x}}\,{{\cos }^{2}}x+c\]
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