A) \[y=\sin x\]
B) \[y=\cos x\]
C) \[y=\sin x+\cos x\]
D) None of these
Correct Answer: A
Solution :
I.F. = \[I.F.={{e}^{\int{-Idx}}}={{e}^{-x}}\] |
\[\therefore \,\,\,y{{e}^{-x}}=\int{{{e}^{-x}}(\cos x-\sin x)dx}\] |
\[=\int{{{e}^{-x}}((-1)\sin x+\cos x)dx}\] |
\[={{e}^{-x}}\sin x+c\] |
\[\therefore \,\,\,y=\sin x+c\,\,{{e}^{x}}\] |
Now \[\underset{x\to \infty }{\mathop{\lim }}\,\,{{e}^{x}}=\infty \]but since y is bounded |
\[\therefore \,\,\,c=0\] |
\[\therefore \,\,\,y=\sin x\]is the solution |
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