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 \[\operatorname{P}= tan\,x\,\,and\,\,Q={{x}^{m}}\,cos\,x\] Now integrating factor \[(I.F.) =\,\,{{e}^{\int{^{Pdx}}}}={{e}^{\int{^{\tan \,\,dx}}}}\] \[=\,\,\,{{e}^{\log \,\,\sec \,\,x}}=\sec \,\,x\] Thus sol. is given by, \[y.{{e}^{\int{^{Pdx}}}}=\int{\,Q.{{e}^{\int{^{Pdx}}}}\,\,dx+c}\] \[\Rightarrow \,\,\,\operatorname{y}.sec\,x= \int{{{x}^{m}}}.cos\,x.sec\,xdx\,+c\] \[\Rightarrow \,\,\,\operatorname{y}\,sec\,x= \frac{{{x}^{m+1}}}{m+1}+c\] \[\Rightarrow \,\,\,\,(m+1)\,y=\,\,{{x}^{m+1}}\ \cos \,x+c\,(m+1)\,cos\,x\]
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