A) \[{{m}^{2}}(a{{e}^{mx}}-b{{e}^{-mx}})\]
B) 1
C) 0
D) none of these
Correct Answer: C
Solution :
[c] \[y=a{{e}^{mx}}+b{{e}^{-mx}}\] \[\therefore \frac{dy}{dx}=am{{e}^{mx}}=mb{{e}^{-mx}}\] Again \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=a{{m}^{2}}{{e}^{mx}}+{{m}^{2}}b{{e}^{-mx}}\] \[={{m}^{2}}(a{{e}^{mx}}+b{{e}^{-mx}})={{m}^{2}}y\] Or \[\frac{{{d}^{2}}y}{d{{x}^{2}}}-{{m}^{2}}y=0\]You need to login to perform this action.
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