a. Nếu \(y = A\sin \left( {\omega t + \varphi } \right) + B\cos \left( {\omega t + \varphi } \right),\) trong đó A, B, ω và φ là những hằng số, thì \(y” + {\omega ^2}y = 0.\)
b. Nếu \(y = \sqrt {2x – {x^2}} \) thì \({y^3}y” + 1 = 0.\)
a.
\(\begin{array}{l}
y = A\sin \left( {\omega t + \varphi } \right) + B\cos \left( {\omega t + \varphi } \right)\,\text{ nên }\\
y’ = A\omega \cos \left( {\omega t + \varphi } \right) – B\omega \sin \left( {\omega t + \varphi } \right)\\
y” = – A{\omega ^2}\sin \left( {\omega t + \varphi } \right) – B{\omega ^2}\cos \left( {\omega t + \varphi } \right)\\
Suy\,ra\,\\\,y” + {\omega ^2}y = – \left[ {A{\omega ^2}\sin \left( {\omega t + \varphi } \right)+B{\omega ^2}\cos \left( {\omega t + \varphi } \right)} \right]\\
+ {\omega ^2}\left[ {A\sin \left( {\omega t + \varphi } \right) + B\cos \left( {\omega t + \varphi } \right)} \right] = 0
\end{array}\)
b. Ta có:
\(\begin{array}{l}
y’ = \frac{{2 – 2x}}{{2\sqrt {2x – {x^2}} }} = \frac{{1 – x}}{{\sqrt {2x – {x^2}} }}\\
y” = \frac{{ – \sqrt {2x – {x^2}} – \left( {1 – x} \right).\frac{{1 – x}}{{\sqrt {2x – {x^2}} }}}}{{\left( {2x – {x^2}} \right)}}\\
= \frac{{ – 2x + {x^2} – 1 + 2x – {x^2}}}{{\sqrt {{{\left( {2x – {x^2}} \right)}^3}} }} = \frac{{ – 1}}{{\sqrt {{{\left( {2x – {x^2}} \right)}^3}} }}\\
Suy\,ra\,\\{y^3}.y” + 1 = \sqrt {{{\left( {2x – {x^2}} \right)}^3}} .\frac{{ – 1}}{{\sqrt {{{\left( {2x – {x^2}} \right)}^3}} }} + 1 = 0
\end{array}\)