Proving the double differential of z = -z implies z=sin⁡xd2zdx2=−zimplies z is of the form asin⁡x+bcos⁡x. 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 sin⁡xorcos⁡x?

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Proving the double differential of z = -z implies z=sin⁡xd2zdx2=−zimplies z is of the form asin⁡x+bcos⁡x. Is there a proof for the same. I was trying to arrive at the desired function but couldn&#039;t understand how to get these trigonometric functions in the equations by integration. Does it require the use of taylor polynomial expansion of sin⁡xorcos⁡x?

Luke Kane · Accepted Answer

We assume the solution is of the formz=∑n=0∞anxn=a0+a1x+a2x2+⋯+anxn+⋯thenz&amp;nbsp;″=∑n=2∞n(n−1)anxn−2=1⋅2a2+2⋅3a3x+3⋅4a4x2+⋯from z&#039;&#039;=-z then with arbitrary&amp;nbsp;a0&amp;nbsp;and&amp;nbsp;a1&amp;nbsp;we havea2=-11×2a0&amp;nbsp;&amp;nbsp;a4=-13×4a2=14!a0,&amp;nbsp;&amp;nbsp;a6=-15×6a4=-16!a0,⋯a3=-12×3a1&amp;nbsp;&amp;nbsp;a5=-14×5a3=15!a1&amp;nbsp;&amp;nbsp;a7=-16×7a5=-17!a0&amp;nbsp;&amp;nbsp;⋯thenz=a0(1−x22!+x44!−⋯)+a1(x−x33!+x55!−⋯)=a0cos⁡x+a1sin⁡x

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

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