siutulysr5

2022-01-26

Given $y=A\mathrm{cos}\left(kt\right)+B\mathrm{sin}\left(kt\right)$ where A,B,k are constants, prove that ${y}^{\left(2\right)}+\left({k}^{2}\right)y=0$
There was a silly typo, ${y}^{n}$ is actually y′′

Amina Hall

Expert

${y}^{}=\frac{{d}^{2}}{{dt}^{2}}y=\frac{d}{dt}\left(\left\{\frac{d}{dt}y\right\}\right)=\frac{d}{dt}\left(-Ak\mathrm{sin}\left(kt\right)+Bk\mathrm{cos}\left(kt\right)\right)=k\frac{d}{dt}\left(-A\mathrm{sin}\left(kt\right)+B\mathrm{cos}\left(kt\right)\right)=$
${k}^{2}\left(-A\mathrm{cos}\left(kt\right)-B\mathrm{sin}\left(kt\right)\right)=-{k}^{2}y$
may be the derivatives of $\mathrm{sin}\left(x\right)$ and $\mathrm{cos}\left(x\right)$

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