Proving the double differential of z = -z implies z= \sin x \frac{d^2z}{dx^2}=-z

predorsomwy

predorsomwy

Answered question

2022-05-02

Proving the double differential of z = -z implies z=sinx
d2zdx2=z
implies z is of the form asinx+bcosx. Is there a proof for the same. I was trying to arrive at the desired function but couldn't understand how to get these trigonometric functions in the equations by integration. Does it require the use of taylor polynomial expansion of sinxorcosx?

Answer & Explanation

Luke Kane

Luke Kane

Beginner2022-05-03Added 14 answers

We assume the solution is of the form
z=n=0anxn=a0+a1x+a2x2++anxn+
then
z =n=2n(n1)anxn2=12a2+23a3x+34a4x2+
from z''=-z then with arbitrary a0 and a1 we have
a2=-11×2a0  a4=-13×4a2=14!a0,  a6=-15×6a4=-16!a0,
a3=-12×3a1  a5=-14×5a3=15!a1  a7=-16×7a5=-17!a0  
then
z=a0(1x22!+x44!)+a1(xx33!+x55!)=a0cosx+a1sinx

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