利用者:Knotopologynn/sandbox

$\displaystyle {\frac {\displaystyle \;{\frac {dy}{\;dt\;}}\;}{\displaystyle {\frac {dx}{\;dt\;}}}}=\displaystyle {\frac {\;\displaystyle {\frac {1}{\;m\;}}\;}{\displaystyle {\frac {1}{\;\ell +m\psi (t)\;}}}}-{\frac {\;\displaystyle {\frac {\ell }{\;m\;}}\;}{1}}=\displaystyle {\frac {\;\ell +m\psi (t)\;}{\;m\;}}-\displaystyle {\frac {\ell }{\;m\;}}$

$\displaystyle {\frac {\displaystyle \;{\frac {dy}{\;dt\;}}\;}{\displaystyle {\frac {dx}{\;dt\;}}}}=\displaystyle {\frac {\;\ell +m\psi (t)\;}{\;m\;}}-\displaystyle {\frac {\ell }{\;m\;}}$ ・・B・・

$\displaystyle {\frac {\displaystyle \;{\frac {dy}{\;dt\;}}\;}{\displaystyle {\frac {dx}{\;dt\;}}}}=\displaystyle {\frac {\;\displaystyle {\frac {1}{\;m\;}}\;}{\displaystyle {\frac {1}{\;\ell +m\psi (t)\;}}}}-{\frac {\;\displaystyle {\frac {\ell }{\;m\;}}\;}{\displaystyle 1}},,$ ・・A・・

$\displaystyle {\frac {\displaystyle \;{\frac {dy}{\;dt\;}}\;}{\displaystyle {\frac {dx}{\;dt\;}}}}=\displaystyle {\frac {\;\displaystyle {\frac {1}{\;m\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\;}{\displaystyle {\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}}}-{\frac {\;\displaystyle {\frac {\ell }{\;m\;}}\cdot \!{\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\;}{\displaystyle {\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}}}$ *****

$\displaystyle {\frac {\displaystyle \;{\frac {dy}{\;dt\;}}\;}{\displaystyle {\frac {dx}{\;dt\;}}}}=\displaystyle {\frac {\;\displaystyle {\frac {1}{\;m\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\;}{\displaystyle {\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}}}$ **・*

$\displaystyle {\frac {\displaystyle \;{\frac {dy}{\;dt\;}}\;}{\displaystyle {\frac {dx}{\;dt\;}}}}=\displaystyle -{\frac {\;\displaystyle {\frac {\ell }{\;m\;}}\cdot \!{\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\;}{\displaystyle {\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}}}$ **・**

$\displaystyle {\frac {\displaystyle \;{\frac {dy}{\;dt\;}}\;}{\displaystyle {\frac {dx}{\;dt\;}}}}=\displaystyle {\frac {\;\displaystyle {\frac {1}{\;m\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\;}{\displaystyle {\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}}}$ ***

-

$\displaystyle {\frac {\displaystyle \;{\frac {dy}{\;dt\;}}\;}{\displaystyle {\frac {dx}{\;dt\;}}}}=\displaystyle -{\frac {\;\displaystyle {\frac {\ell }{\;m\;}}\cdot \!{\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\;}{\displaystyle {\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}}}$ ****

$\displaystyle {\frac {\displaystyle \;{\frac {dy}{\;dt\;}}\;}{\displaystyle {\frac {dx}{\;dt\;}}}}=\displaystyle {\frac {\;\displaystyle -{\frac {\ell }{\;m\;}}\cdot \!{\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\;}{\displaystyle {\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}}}$ ***

$\displaystyle {\frac {\displaystyle \;{\frac {dy}{\;dt\;}}\;}{\displaystyle {\frac {dx}{\;dt\;}}}}=\displaystyle {\frac {\;\displaystyle -{\frac {\ell }{\;m\;}}\cdot \!{\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\;}{\displaystyle {\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}}}$

$\displaystyle {\frac {\displaystyle \;{\frac {dy}{\;dt\;}}\;}{\displaystyle {\frac {dx}{\;dt\;}}}}=\displaystyle {\frac {\;\displaystyle {\frac {1}{\;m\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\;}{\displaystyle {\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}}}$

$\displaystyle {\frac {\displaystyle \;{\frac {dy}{\;dt\;}}\;}{\displaystyle {\frac {dx}{\;dt\;}}}}$

次に、${\frac {\;{d}x\;}{dt}}$ と ${\frac {\;{d}y\;}{dt}}$ から ${\frac {\;{d}y\;}{dx}}=\psi (t)$ に至る計算の詳細を下に示しておく。

（▲）式を $t$ で微分した式は前掲の、 $\displaystyle {\frac {\;{d}x\;}{dt}}={\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}$ である。一方、（▲▲）式を $t$ で微分すると、

$\displaystyle {\frac {\;{d}y\;}{dt}}={\frac {1}{\;m\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}-{\frac {\ell }{\;m\;}}\cdot \!{\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}$

となる。

$\displaystyle {\frac {\;\displaystyle {\frac {1}{\;m\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}-{\frac {\ell }{\;m\;}}\cdot \!{\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\;}{\displaystyle {\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}}}$

$\displaystyle {\frac {\;\displaystyle {\frac {\;{d}y\;}{dt}}={\frac {1}{\;m\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}-{\frac {\ell }{\;m\;}}\cdot \!{\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\;}{\displaystyle {\frac {\;{d}x\;}{dt}}={\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}}}$

$\displaystyle {\frac {\;{d}y\;}{dt}}={\frac {1}{\;m\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}-{\frac {\ell }{\;m\;}}\cdot \!{\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}$

$\displaystyle {\frac {\;{d}y\;}{dt}}={\frac {1}{\;m\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}-{\frac {\ell }{\;m\;}}\left\{\!{\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\,\right\}.$

$\displaystyle {\frac {\;{d}y\;}{dt}}={\frac {1}{\;m\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}-{\frac {\ell }{\;m\;}}\left\{\!{\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\,\right\}.$

$\displaystyle {\frac {\;{d}x\;}{dt}}={\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}$

上記の（▲）と（▲▲）を用いて、

$\displaystyle {\frac {dy}{\;dx\;}}={\frac {\displaystyle {\frac {dy}{\;dt\;}}}{\displaystyle {\frac {dx}{\;dt\;}}}}$

を計算すると、

$\displaystyle {\frac {dy}{\;dx\;}}=\psi (t)$

を得る。この $\displaystyle {\frac {dy}{\;dx\;}}=\psi (t)$ は、 $\displaystyle F{\Bigl (}k+\ell x+my,\;\;{\frac {dy}{dx}}\;{\Bigr )}=0$ を解く時、 始めに仮定した式と同一である。

$\displaystyle {\frac {dy}{\;dx\;}}={\frac {\displaystyle \;{\frac {dy}{\;dt\;}}\;}{\displaystyle {\frac {dx}{\;dt\;}}}}$

$x=\!\int \!\left\{{\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\,\right\}dt+C,$ ・・・・・（▲）

$\displaystyle y={\frac {1}{\;m\;}}\left[\phi (t)-k-\ell \left\{\!\int \left(\!{\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\,\right)dt+C\right\}\right].$ ・・・・・（▲▲）

$\displaystyle y={\frac {1}{\;m\;}}\left[\phi (t)-\left\{k+\ell \left({\frac {1}{\;m\;}}\cdot \ln {\bigl (}\,\ell +m\phi (t)\,{\bigr )}+C\right)\right\}\right]$

$\displaystyle y={\frac {1}{\;m\;}}\left[\phi (t)-k-\ell \left\{\int \left(\!{\frac {1}{\;\ell +m\psi (t)\;}}\cdot {\frac {\;{d}\phi (t)\;}{dt}}\,\right)dt+C\right\}\right].$

次に、例題2 の（★）式を展開すると、

$\displaystyle y=C\exp(mx)-{\frac {\;\ell x\;}{m}}-{\frac {k}{\;m\;}}-{\frac {\ell }{\;m^{2}\;}},\;$・・・・・（★★）

となる。

$x=X(\phi (t),\;C),\;$ ・・・・・（#）(•)

$y=Y(\phi (t),\;C).\;$・・・・・（##）(••), (• •),

いま，$\left({\frac {dy}{\;dt\;}}\right)/\left({\frac {dx}{\;dt\;}}\right)$ であるから，上記の解、（#）式と（##）式を用いて， ${\frac {dx}{\;dt\;}}\;$ と ${\frac {dy}{\;dt\;}}\;$ を計算してみる．

テスト： ok 下記の三行を投稿に使う． 2021/05/25//12:22/

$u=v,\;$ ・・・・・（＊）

$x=y.\;$・・・・・（＊＊）

いま、（＊）式と（＊＊）式を

$\left({\frac {dy}{\;dt\;}}\right){\big /}\left({\frac {dx}{\;dt\;}}\right)$

$\displaystyle {\frac {dy}{\;dx\;}}={\frac {\displaystyle \;{\frac {dy}{\;dt\;}}}{\displaystyle {\frac {dx}{\;dt\;}}}}$

1, $\displaystyle {\frac {dy}{\;dx\;}}={\frac {\displaystyle \;{\frac {dy}{\;dt\;}}}{\displaystyle {\frac {dx}{\;dt\;}}}}={\frac {\displaystyle {\frac {\phi (t)}{\;\ell +m\phi (t)\;}}\cdot {\frac {\;d\phi (t)\;}{dt}}}{\displaystyle \;{\frac {1}{\;\ell +m\phi (t)\;}}\cdot {\frac {\;d\phi (t)\;}{dt}}\;}}$

${\frac {dx}{\;dt\;}}={\frac {1}{\;\ell +m\phi (t)\;}}\cdot {\frac {\;d\phi (t)\;}{dt}}$

${\frac {dy}{\;dt\;}}={\frac {\phi (t)}{\;\ell +m\phi (t)\;}}\cdot {\frac {\;d\phi (t)\;}{dt}}$

$\displaystyle {\frac {dy}{\;dx\;}}={\frac {\displaystyle \;{\frac {dy}{\;dt\;}}}{\displaystyle {\frac {dx}{\;dt\;}}}}={\frac {\displaystyle {\frac {\phi (t)}{\;\ell +m\phi (t)\;}}\cdot {\frac {\;d\phi (t)\;}{dt}}}{\displaystyle \;{\frac {1}{\;\ell +m\phi (t)\;}}\cdot {\frac {\;d\phi (t)\;}{dt}}\;}}$

$\displaystyle {\frac {dy}{\;dx\;}}={\frac {\displaystyle \;{\frac {dy}{\;dt\;}}}{\displaystyle {\frac {dx}{\;dt\;}}}}={\frac {\displaystyle {\frac {\phi (t)}{\;\ell +m\phi (t)\;}}\cdot {\frac {\;d\phi (t)\;}{dt}}}{\displaystyle \;{\frac {1}{\;\ell +m\phi (t)\;}}\cdot {\frac {\;d\phi (t)\;}{dt}}\;}}=\phi (t)$

が得られた。 例題1 は、$\psi (t)=\phi (t)$ の場合であるから、 上式 $\displaystyle {\frac {dy}{\;dx\;}}=\phi (t)$ は、常微分方程式 $\displaystyle F{\Bigl (}k+\ell x+my,\;\;{\frac {dy}{dx}}\;{\Bigr )}=0$ を解くとき、 始めに仮定した式 $\displaystyle {\frac {dy}{\;dx\;}}=\psi (t)$ と同一である。

$\displaystyle y={\frac {1}{\;m\;}}[{\frac {1}{\;m\;}}{(\exp(mx-mC)-\ell )-k-\ell x}]$

$\displaystyle y={\frac {1}{\;m\;}}\left[{\frac {1}{\;m\;}}\{(\exp(mx-mC)-\ell )-k-\ell x\}\right]$

$\displaystyle y={\frac {1}{\;m\;}}\left[{\frac {1}{\;m\;}}\{\exp(mx-mC)-\ell \}-k-\ell x\right]$

$\displaystyle y={\frac {1}{\;m\;}}\left[{\frac {1}{\;m\;}}\{\exp(mx-mC\;)-\ell \;\}-k-\ell x\;\right].$

$\displaystyle y={\frac {1}{\;m\;}}\left[{\frac {1}{\;m\;}}\{\exp(mx-mC\;)-\ell \;\}-k-\ell x\;\right].$ ・・・・・（•••）

次に、（•）式と（••）式から、 $\phi (t)$ を消去した式を示しておく。

$\displaystyle y={\frac {1}{\;m\;}}\left[{\frac {1}{\;m\;}}\{\exp(mx-mC\;)-\ell \;\}-k-\ell x\;\right].$ ・・・・・（•••）

上式（•••）は、$k+\ell x+my={\frac {dy}{dx}}$ の一般解である。

上記（★）式の一般解は、公式を用いて解くと、

$\displaystyle y=C\exp(mx)-{\frac {1}{\;m\;}}\left\{k+\ell \left(x+{\frac {1}{\;m\;}}\right)\right\}.$・・・・・（★★）

となる。 この（★★）式は（★）式、$k+\ell x+my={\frac {dy}{dx}}$ を満たすので（★）式の一般解である。

以下で、例題1 の（•••）式と （★★）を比較する。 まず、例題1 の（•••）式を展開すると、

$\displaystyle y={\frac {\;\exp(-mC)\;}{m^{2}}}\cdot \exp(mx)-{\frac {\;\ell x\;}{m\;}}-{\frac {\ell }{\;m^{2}\;}}-{\frac {k}{\;m\;}},$