Chứng minh các hệ thức sau: a. \({x^2}y'' - 2\left( {{x^2} + {y^2}} \right)\left( {1 + y} \right) = 0\) với
Chứng minh các hệ thức sau:
a. \({x^2}y'' - 2\left( {{x^2} + {y^2}} \right)\left( {1 + y} \right) = 0\) với \(y = x\tan x\)
b. \({y^3}y'' + 1 = 0\) với \(y = \sqrt {2x - {x^2}} \)
c. \(4\left( {{x^2} + 1} \right)y'' + 4xy' - y = 0\) với \(y = \sqrt {x + \sqrt {1 + {x^2}} } \)
d. \(\left( {1 + {x^2}} \right)y'' + xy' - {k^2}y = 0\) với \(y = {\left( {x + \sqrt {1 + {x^2}} } \right)^k}\)
a) \(y = x\tan x\)
\(\begin{array}{l}y' = \tan x + x\left( {1 + {{\tan }^2}x} \right)\\y'' = 1 + {\tan ^2}x + 1 + {\tan ^2}x + x.2\tan x\left( {1 + {{\tan }^2}x} \right)\\\,\,\,\,\,\,\, = 2\left( {1 + {{\tan }^2}x} \right) + 2x\tan x\left( {1 + {{\tan }^2}x} \right)\\\,\,\,\,\,\,\, = 2\left( {1 + {{\tan }^2}x} \right)\left( {1 + x\tan x} \right)\\ \Rightarrow {x^2}y'' - 2\left( {{x^2} + {y^2}} \right)\left( {1 + y} \right)\\ = 2{x^2}\left( {1 + x\tan x} \right)\left( {1 + {{\tan }^2}x} \right) - 2\left( {{x^2} + {x^2}{{\tan }^2}x} \right)\left( {1 + x\tan x} \right)\\ = 2{x^2}\left( {1 + x\tan x} \right)\left( {1 + {{\tan }^2}x} \right) - 2{x^2}\left( {1 + {{\tan }^2}x} \right)\left( {1 + x\tan x} \right)\\ = 0\,\,\left( {dpcm} \right)\end{array}\)
b) \(y = \sqrt {2x - {x^2}} \)
\(\begin{array}{l}y' = \dfrac{{2 - 2x}}{{2\sqrt {2x - {x^2}} }} = \dfrac{{1 - x}}{{\sqrt {2x - {x^2}} }}\\y'' = \dfrac{{ - \sqrt {2x - {x^2}} - \left( {1 - x} \right)\dfrac{{1 - x}}{{\sqrt {2x - {x^2}} }}}}{{2x - {x^2}}}\\\,\,\,\,\,\, = \dfrac{{ - \left( {2x - {x^2}} \right) - {{\left( {1 - x} \right)}^2}}}{{\sqrt {2x - {x^2}} \left( {2x - {x^2}} \right)}}\\\,\,\,\,\,\, = \dfrac{{ - 2x + {x^2} - 1 + 2x - {x^2}}}{{\sqrt {2x - {x^2}} \left( {2x - {x^2}} \right)}}\\\,\,\,\,\,\, = \dfrac{{ - 1}}{{\sqrt {2x - {x^2}} \left( {2x - {x^2}} \right)}} = - \dfrac{1}{{y.{y^2}}} = - \dfrac{1}{{{y^3}}}\\ \Rightarrow {y^3}y'' + 1 = {y^3}.\dfrac{{ - 1}}{{{y^3}}} + 1 = - 1 + 1 = 0\,\,\,\left( {dpcm} \right)\end{array}\)
c) \(y = \sqrt {x + \sqrt {1 + {x^2}} } \)
\(\begin{array}{l}y' = \dfrac{{1 + \dfrac{x}{{\sqrt {1 + {x^2}} }}}}{{2\sqrt {x + \sqrt {1 + {x^2}} } }} = \dfrac{{x + \sqrt {1 + {x^2}} }}{{2\sqrt {1 + {x^2}} \sqrt {x + \sqrt {1 + {x^2}} } }} = \dfrac{{\sqrt {x + \sqrt {1 + {x^2}} } }}{{2\sqrt {1 + {x^2}} }} = \dfrac{y}{{2\sqrt {1 + {x^2}} }}\\y'' = \dfrac{1}{2}\left( {\dfrac{{y'\sqrt {1 + {x^2}} - y.\dfrac{x}{{\sqrt {1 + {x^2}} }}}}{{1 + {x^2}}}} \right) = \dfrac{1}{2}.\dfrac{{y'\left( {1 + {x^2}} \right) - xy}}{{\left( {1 + {x^2}} \right)\sqrt {1 + {x^2}} }}\\ \Rightarrow 4\left( {{x^2} + 1} \right)y'' + 4xy' - y\\ = 4\left( {{x^2} + 1} \right).\dfrac{1}{2}.\dfrac{{y'\left( {1 + {x^2}} \right) - xy}}{{\left( {1 + {x^2}} \right)\sqrt {1 + {x^2}} }} + 4x.\dfrac{y}{{2\sqrt {1 + {x^2}} }} - y\\ = 2.\dfrac{{y'\left( {1 + {x^2}} \right) - xy}}{{\sqrt {1 + {x^2}} }} + 4x.y' - y\\ = 2.\dfrac{y}{{2\sqrt {1 + {x^2}} }}.\sqrt {1 + {x^2}} - \dfrac{{2xy}}{{\sqrt {1 + {x^2}} }} + 4x.\dfrac{y}{{2\sqrt {1 + {x^2}} }} - y\\ = y - \dfrac{{2xy}}{{\sqrt {1 + {x^2}} }} + \dfrac{{2xy}}{{\sqrt {1 + {x^2}} }} - y = 0\,\,\,\left( {dpcm} \right)\end{array}\)
d) \(y = {\left( {x + \sqrt {1 + {x^2}} } \right)^k}\)
\(\begin{array}{l}y' = k{y^{\dfrac{1}{k}\left( {k - 1} \right)}}.\left( {1 + \dfrac{x}{{\sqrt {1 + {x^2}} }}} \right) = \dfrac{{k{y^{1 - \dfrac{1}{k}}}}}{{\sqrt {1 + {x^2}} }}{y^{\dfrac{1}{k}}} = \dfrac{{ky}}{{\sqrt {1 + {x^2}} }}\\y'' = \dfrac{{ky'\sqrt {1 + {x^2}} - ky\dfrac{x}{{\sqrt {1 + {x^2}} }}}}{{1 + {x^2}}}\\\,\,\,\,\,\, = \dfrac{{ky'\left( {1 + {x^2}} \right) - kxy}}{{\left( {1 + {x^2}} \right)\sqrt {1 + {x^2}} }}\\ \Rightarrow \left( {1 + {x^2}} \right)y'' + xy' - {k^2}y\\ = \left( {1 + {x^2}} \right)\dfrac{{ky'\left( {1 + {x^2}} \right) - kxy}}{{\left( {1 + {x^2}} \right)\sqrt {1 + {x^2}} }} + x.\dfrac{{ky}}{{\sqrt {1 + {x^2}} }} - {k^2}y\\ = \dfrac{{ky'\left( {1 + {x^2}} \right) - kxy}}{{\sqrt {1 + {x^2}} }} + x.\dfrac{{ky}}{{\sqrt {1 + {x^2}} }} - {k^2}y\\ = \dfrac{{ky'\left( {1 + {x^2}} \right) - kxy + kxy}}{{\sqrt {1 + {x^2}} }} - {k^2}y\\ = k.\dfrac{{ky}}{{\sqrt {1 + {x^2}} }}\sqrt {1 + {x^2}} - {k^2}y\\ = {k^2}y - {k^2}y = 0\,\,\,\left( {dpcm} \right)\end{array}\)
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