VC1.5

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.