Subjective Type

if $$y = A{e^{ - kt}}\cos (pt + c)$$ , then prove that $${{{d^2}y} \over {d{t^2}}} + 2k{{dy} \over {dt}} + {n^2}y = 0$$ , where $${n^2} = {p^2} + {k^2}$$

Solution

$$\cfrac{dy}{dt}=Ae^{-kt}(-\sin(pt+c))p+\cos (pt+c)\times Ax-e^{-kt}(k)$$
$$\cfrac{dy}{dt}=Ae^{-kt}(-p\sin(pt+c)-k\cos(pt+c))$$
$$\cfrac{d^{2}y}{dt^{2}}=-Ae^{-kt}p^{2}\cos(pt+c)-Ap\sin(pt+c)e^{-kt}\times (-k)+pAk\sin(pt+c)e^{-kt}+k^{2}A\cos (pt+c)e^{-kt}$$
$$\cfrac{d^{y}}{dt^{2}}$$=$$2pAk\sin(pt+c)\,e^{-kt}$$+$$Ae^{-kt}\cos(pt+c)(p^{2}-k^{2})$$
Now, by substituting values of $$\cfrac{d^{2}y}{dt^{2}},\cfrac{dy}{dt}$$ and $$y$$,
we get such that $$\cfrac{d^{2}y}{dt^{2}}+2k\cfrac{dy}{dt}+x^{2}y=0$$
and $$n^{2}=p^{2}+k^{2}$$
$$LHS=0=RHS$$.
Hence, proved

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