A) \[y+log\text{ }y+{{e}^{x}}co{{s}^{2}}x=2\]
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 \[{{e}^{-x}}(y+1)dy+({{\cos }^{2}}x-\sin 2x)y\,dx=0\] \[\Rightarrow \]\[\frac{(y+1)}{y}dy=-{{e}^{x}}({{\cos }^{2}}x-\sin 2x)dx\] \[\Rightarrow \]\[\left( 1+\frac{1}{y} \right)dy=-{{e}^{x}}({{\cos }^{2}}x-\sin 2x)dx\] On integrating both sides, we get \[y+\log y=-{{e}^{x}}{{\cos }^{2}}x+\int{{{e}^{x}}\sin 2x\,dx}\] \[-\int{{{e}^{x}}\sin 2xdx+c}\] \[\Rightarrow \] \[y+\log y=-{{e}^{x}}{{\cos }^{2}}x+c\] At \[x=0,y=1\] \[\Rightarrow \] \[1+0=-{{e}^{0}}\cos 0+c\] \[\Rightarrow \] \[c=2\] \[\therefore \]Required solution is \[y+log\text{ }y=-{{e}^{x}}co{{s}^{2}}x+2\]
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