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