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.
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\(\begin{array}{l}
y = A\sin \left( {\omega t + \varphi } \right) + B\cos \left( {\omega t + \varphi } \right)\,\text{ nên }\\
y’ = A\omega \cos \left( {\omega t + \varphi } \right) - B\omega \sin \left( {\omega t + \varphi } \right)\\
y” = - A{\omega ^2}\sin \left( {\omega t + \varphi } \right) - B{\omega ^2}\cos \left( {\omega t + \varphi } \right)\\
Suy\,ra\,\\\,y” + {\omega ^2}y = - \left[ {A{\omega ^2}\sin \left( {\omega t + \varphi } \right)+B{\omega ^2}\cos \left( {\omega t + \varphi } \right)} \right]\\
+ {\omega ^2}\left[ {A\sin \left( {\omega t + \varphi } \right) + B\cos \left( {\omega t + \varphi } \right)} \right] = 0
\end{array}\)
b. Ta có:
\(\begin{array}{l}
y’ = \frac{{2 - 2x}}{{2\sqrt {2x - {x^2}} }} = \frac{{1 - x}}{{\sqrt {2x - {x^2}} }}\\
y” = \frac{{ - \sqrt {2x - {x^2}} - \left( {1 - x} \right).\frac{{1 - x}}{{\sqrt {2x - {x^2}} }}}}{{\left( {2x - {x^2}} \right)}}\\
= \frac{{ - 2x + {x^2} - 1 + 2x - {x^2}}}{{\sqrt {{{\left( {2x - {x^2}} \right)}^3}} }} = \frac{{ - 1}}{{\sqrt {{{\left( {2x - {x^2}} \right)}^3}} }}\\
Suy\,ra\,\\{y^3}.y” + 1 = \sqrt {{{\left( {2x - {x^2}} \right)}^3}} .\frac{{ - 1}}{{\sqrt {{{\left( {2x - {x^2}} \right)}^3}} }} + 1 = 0
\end{array}\)