A) \[y=c{{e}^{\int{p\,dx}}}\]
B) \[y=c{{e}^{-\int{p\,dx}}}\]
C) \[x=c{{e}^{-\int{p\,dy}}}\]
D) \[x=c{{e}^{\int{p\,dy}}}\]
E) none of these
Correct Answer: B
Solution :
\[\because \]In the given linear differential equation coefficient of Q is 0. \[\therefore \]Solution is \[y.{{e}^{\int{P}\,dx}}=\int{0}\,dx+c\] \[\Rightarrow \] \[y=c{{e}^{-\int{P\,}dx}}\]
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