VC1.5

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Given,r=e^{{nt}}a+e^{{-nt}}b\,

From the above, {\frac  {dr}{dt}}={\frac  {d}{dt}}(e^{{nt}})a+e^{{nt}}{\frac  {da}{dt}}+{\frac  {d}{dt}}(e^{{-nt}})b+e^{{-nt}}{\frac  {db}{dt}}\,

Since a,b are constant vectors, {\frac  {da}{dt}}={\frac  {db}{dt}}=0\,

Therefore, {\frac  {dr}{dt}}=ne^{{nt}}a-ne^{{-nt}}b=n(e^{{nt}}a-e^{{-nt}}b)\,

Again differentiating this w.r.t 't', we have

{\frac  {d^{2}r}{dt^{2}}}=n[ne^{{nt}}a+ne^{{-nt}}b]=n^{2}[e^{{nt}}a+e^{{-nt}}b]\,

From the given equation, the above can be written as

{\frac  {d^{2}r}{dt^{2}}}=n^{2}r\,

So, {\frac  {d^{2}r}{dt^{2}}}-n^{2}r=0\, ,which is to be proved.

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