- LG a
- LG b
LG 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 \[y" + {\omega ^2}y = 0.\]
Lời giải chi tiết:
\[\begin{array}{l}
y = A\sin \left[ {\omega t + \varphi } \right] + B\cos \left[ {\omega t + \varphi } \right]\,\text{ nên }\\
y' = A\omega \cos \left[ {\omega t + \varphi } \right] - B\omega \sin \left[ {\omega t + \varphi } \right]\\
y" = - A{\omega ^2}\sin \left[ {\omega t + \varphi } \right] - B{\omega ^2}\cos \left[ {\omega t + \varphi } \right]\\
Suy\,ra\,:\\\,y" + {\omega ^2}y = - \left[ {A{\omega ^2}\sin \left[ {\omega t + \varphi } \right]+B{\omega ^2}\cos \left[ {\omega t + \varphi } \right]} \right]\\
+ {\omega ^2}\left[ {A\sin \left[ {\omega t + \varphi } \right] + B\cos \left[ {\omega t + \varphi } \right]} \right] = 0
\end{array}\]
LG b
Nếu \[y = \sqrt {2x - {x^2}} \]thì \[{y^3}y" + 1 = 0.\]
Lời giải chi tiết:
Ta có:
\[\begin{array}{l}
y' = \frac{{2 - 2x}}{{2\sqrt {2x - {x^2}} }} = \frac{{1 - x}}{{\sqrt {2x - {x^2}} }}\\
y''= \frac{{\left[ {1 - x} \right]'\sqrt {2x - {x^2}} - \left[ {1 - x} \right]\left[ {\sqrt {2x - {x^2}} } \right]'}}{{2x - {x^2}}}\\ = \frac{{ - \sqrt {2x - {x^2}} - \left[ {1 - x} \right].\frac{{\left[ {2x - {x^2}} \right]'}}{{2\sqrt {2x - {x^2}} }}}}{{2x - {x^2}}} \\= \frac{{ - \sqrt {2x - {x^2}} - \left[ {1 - x} \right].\frac{{2 - 2x}}{{2\sqrt {2x - {x^2}} }}}}{{2x - {x^2}}}\\= \frac{{ - \sqrt {2x - {x^2}} - \left[ {1 - x} \right].\frac{{1 - x}}{{\sqrt {2x - {x^2}} }}}}{{\left[ {2x - {x^2}} \right]}}\\
= \frac{{ - 2x + {x^2} - 1 + 2x - {x^2}}}{{\sqrt {{{\left[ {2x - {x^2}} \right]}^3}} }}= \frac{{ - 1}}{{\sqrt {{{\left[ {2x - {x^2}} \right]}^3}} }}\\
Suy\,ra\,\\{y^3}.y" + 1 \\= \sqrt {{{\left[ {2x - {x^2}} \right]}^3}} .\frac{{ - 1}}{{\sqrt {{{\left[ {2x - {x^2}} \right]}^3}} }} + 1 \\= -1+1=0
\end{array}\]