question_answer

If \[y=1-x+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{4}}}{4!}-....to\,\,\infty ,\] then prove that \[\frac{{{d}^{2}}y}{d{{x}^{2}}}-y=0\]

Answer:

We have, \[y=1-x+\frac{{{x}^{2}}}{2!}-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{4}}}{4!}-...\infty \] \[\Rightarrow \]   \[y={{e}^{-x}}\] \[\therefore \]      \[\frac{dy}{dx}=-\,{{e}^{-x}}\] \[\Rightarrow \]   \[\frac{{{d}^{2}}y}{d{{x}^{2}}}={{e}^{-x}}\] \[\Rightarrow \]   \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=y\] \[\Rightarrow \]   \[\frac{{{d}^{2}}y}{d{{x}^{2}}}-y=0\]                     Hence proved.