Question

eliminate the arbitrary constant and obtain the differential equation satisfied by it y=ccos(pt-a) where p is an arbitrary constant​

Answer 1

Answer:

The answer is $y''+p^2y=0$

Step-by-step explanation:

Given  $y=c\cos (pt-a)$...(1)

Differentiate both sides w.r.t. t

  $\Rightarrow y'=c(-\sin(pt-a))p=-cp\sin(pt-a)$ ...(2)

Again differentiate both sides w.r.t. t

 $\Rightarrow y''=-cp(\cos(pt-a))(p)=-p^2(c \cos(pt-a))=-p^2(y)\\\Rightarrow y''+p^2y=0$

( Since using (1) $c\cos(pt-a)=y$ )