Chapter 1 - First-Order Differential Equations - 1.1 Differential Equations Everywhere - Problems - Page 12: 30

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Given: $y(t)=A\cos(\omega t-\phi)\\ y'(t)=-A\omega \sin(\omega t-\phi)\\ y''(t)=-A\omega^2 \cos(\omega t-\phi)$ Therefore, $y+\omega^2 y=-A\omega^2 \cos(\omega t-\phi)+A\omega^2(\omega -\phi)=0$ Find $\phi$: $-A\omega \sin(\omega(0)-\phi)=0\\ A\omega \sin (\phi)=0\\ \sin(\phi)=0\\ \phi=0$ and $a=A\cos(\omega(0)-(0))\\ a=A\cos(0)\\ a=A$