Exercise · ZW-SECOND-ORDER-LINEAR-EQUATION-006 · ★★☆ Intermediate
For \(k>0\), which value makes \(y=\sin(kt)\) solve \(y''+16y=0\)?
Choose one option.
Show answer
- C. \(k=4\)
Exercise · ZW-SECOND-ORDER-LINEAR-EQUATION-006 · ★★☆ Intermediate
Choose one option.
Since \(y''=-k^2\sin(kt)\), we need \(-k^2+16=0\), so \(k=4\).
It is a second-order differential equation, linear in the unknown \(y\) and its derivatives, whose coefficients \(p\), \(q\) and term \(g\) depend only on \(t\).
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