A) \[(m+1)y={{x}^{m+1}}\cos x+c(m+1)\cos x\]
B) \[my=({{x}^{m}}+c)\cos x\]
C) \[y=({{x}^{m+1}}+c)\cos x\]
D) None of these
Correct Answer: A
Solution :
This is the linear equation of the form \[\frac{dy}{dx}+Py=Q\], where \[P=\tan x\] and \[Q={{x}^{m}}\cos x\] Now integrating factor (I.F.)\[={{e}^{\int{Pdx}}}={{e}^{\int{\tan dx}}}\] \[={{e}^{\log \sec x}}=\sec x\] Thus solution is given by, \[y.{{e}^{\int{Pdx}}}=\int{Q}.\,{{e}^{\int{Pdx}}}dx+c\] Þ \[y.\sec x=\int{{{x}^{m}}}.\cos x.\sec xdx+c\]Þ\[y\sec x=\frac{{{x}^{m+1}}}{m+1}+c\] Þ \[(m+1)y={{x}^{m+1}}\cos x+c(m+1)\cos x\].
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