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← PC用は別頁
…… 1階微分方程式の鳥瞰図 ……
○1階の常微分方程式の中で初等的、入門的なものは、有限回の変形と積分計算によって(初等的に,求積法によって)解が求まるものであり、変数分離形やこれに連なる幾つかの形をしたものがよく知られています. ○これ以外の形の1階の常微分方程式の解き方は,よく分かっていません. 他方で,べき級数が使える場合には,1点の近傍でいくらでも精密な解を求めることができ,その関数の性質を調べることができます. また,フーリエ級数を使えば大域的に面積の誤差がほとんどないような解を求めることができます. 特筆すべきこととして,適当な初期条件が与えられた微分方程式に対しては,解の存在と一意性(=ただ1つ存在するということ)を証明することができます.(定理があります) ![]() ※目的や見方によって,各々の長所は短所と裏腹の関係にあります. 有限回の変形と積分計算で解けるものは,入門的・理論的な練習には適していますが,解ける形が限られています. 級数解は,多くに場合に解が求まるかわりに,仮に求まっても初等的にどのような関数に対応しているか分からないというもどかしさがあります. 近似解は厳密解でないので,いい加減なものに見えますが,エネルギー効率のよいスクリューやエンジンを設計しているような場合には,例えば金属表面を1000分の1mmの精度(10−6m)で削ることは至難の業なので,初めの数項(例えば6項)からなる近似解で十分で,それ以上精密な解があっても利用できず,精密過ぎる解は使い道がありません. ○この頁では,有限回の変形と積分計算でよって(初等的に,求積法によって)解が求まるもののうちで変数変換によって「変数分離形」に直せる「同次形」を扱います. ■同次形の微分方程式 ![]() ![]() と呼ばれ, ![]() 解くことができます.
微分方程式を解くとき,変形の途中経過において分母のx, y, uが0になる場合でも,結果的に一般解の1つの場合として表せることがほとんどなので,以下においてはこのような途中経過で分母が0になるときの場合分けは行わず,それらが0でない場合から得られる一般解のみを扱います.
(解説)![]() (右辺)=f(u) また,積の微分法:(fg)'=f 'g+fg'により ![]() ![]() ![]() ![]() となるから
←この形は,以下において何度も登場するので覚えておく方がよい
(左辺)=u+x ![]() したがって,微分方程式は次の形になります. u+x ![]() この式は次のように変形できるので,変数分離形になります. x ![]() ![]() ![]()
【例題1】
(解答)次の微分方程式の一般解を求めてください. x ![]() ![]() ![]() ![]() と変形できるから ![]() ![]() ![]() u+x ![]() x ![]() du= ![]() ∫ du=∫ ![]() u=log|x|+C 元のyに戻すと ![]() ゆえに y=x(log|x|+C)…(答)
【例題2】
(解答)次の微分方程式の一般解を求めてください. x2+y2=2xyy' x2+y2=2xy ![]() ![]() ![]() ![]() ![]() ![]() と変形できるから ![]() ![]() ![]() u+x ![]() ![]() x ![]() ![]() ![]() ![]() ![]() ∫ ![]() ![]() (u2−1)'=2uだから log|u2−1|=−log|x|+A log|u2−1|+log|x|=A log|(u2−1)x|=A=logeA |(u2−1)x|=eA=Bとおく (u2−1)x=±B=Cとおく 元のyに戻すと {( ![]() (y2−x2)x=Cx2 y2−x2=Cx y2=x2+Cx…(答) |
※正しい番号をクリックしてください.
それぞれの問題は暗算では解けませんので,計算用紙が必要です.
y'=2+
![]() ![]() ![]() ![]() u+x ![]() x ![]() du=2 ![]() ∫ du=2∫ ![]() u=2log|x|+C 変数を元のyに戻すと ![]() y=x(2log|x|+C)→4 |
x
![]() ![]() ![]() ![]() ![]() ![]() u+x ![]() x ![]() ![]() ![]() ∫ ![]() ![]() ![]() log|2u−1|=−2log|x|+2C1=−2log|x|+C2とおく log|2u−1|+log(x2)=C2 log|(2u−1)x2|=C2=log eC2=log C3とおく |(2u−1)x2|=C3 (2u−1)x2=±C3=C4とおく 変数を元のyに戻すと (2 ![]() 2yx−x2=C4 2yx=C4+x2 y= ![]() ![]() ![]() ![]() |
![]() ![]() ![]() ![]() ![]() (右辺)= ![]() (左辺)=u+x ![]() u+x ![]() ![]() x ![]() ![]() ![]() ![]() ![]() ∫ ![]() ![]() 左辺の被積分関数は,分子u+1が分母u2+2u−1の微分(÷2)に直せるから ![]() ![]() ![]() ![]() log|u2+2u−1|=−2log|x|+2C1=−2log|x|+C2とおく log|u2+2u−1|+log(x2)=C2=logeC2 |(u2+2u−1)x2|=eC2=C3とおく (u2+2u−1)x2=±C3=C4とおく 変数を元のyに戻すと { ( ![]() ![]() y2+2xy−x2=C4 ここままでもよいが,答の選択肢の形に合わせると,C4=−Cとおいて x2−2xy−y2=C→4 |
![]() ![]() ![]() ![]() ![]() ![]() ![]() (右辺)= ![]() (左辺)=u+x ![]() u+x ![]() ![]() x ![]() ![]() ![]() ![]() ![]() ∫ ![]() ![]() ここで,左辺の被積分関数を部分分数分解する ![]() ![]() ![]() ![]() が恒等式となる定数a, b, cを求めると,a=−1, b=2, c=0 ∫ (− ![]() ![]() ![]() −log|u|+log|u2+1|=−log|x|+C1 −log|u|+log|u2+1|+log|x|=C1=log eC1 log| ![]() | ![]() ![]() (u2+1)x=C3u 変数を元のyに戻すと { ( ![]() ![]() ![]() ![]() y2+x2=C3y y2+x2=Cy→2 |
y'=
![]() ![]() ![]() ![]() ![]() ![]() ![]() (右辺)= ![]() (左辺)=u+x ![]() u+x ![]() ![]() x ![]() ![]() ![]() ![]() ![]() ![]() ∫ ![]() ![]() 左辺の被積分関数は,分子2u−1が分母u2−uの微分になっているから log|u2−u|=−3log|x|+C1 log|u2−u|+3log|x|=C1 log|(u2−u)x3|=C1=log eC1 |(u2−u)x3|=eC1=C2とおく (u2−u)x3=±C2=C3とおく 変数を元のyに戻すと { ( ![]() ![]() xy2−x2y=C3 xy(y−x)=C→1 |
![]() ![]() ![]() ![]() ![]() ![]() (右辺)=−u−2u2 (左辺)=u+x ![]() u+x ![]() x ![]() ![]() ![]() ∫ ![]() ![]() ここで,左辺の被積分関数を部分分数分解すると ![]() ![]() ![]() ∫ ( ![]() ![]() ![]() log|u|−log|u+1|=−2log|x|+C1 log|u|−log|u+1|+log(x2)=C1=log eC1 log| ![]() | ![]() ![]() ux2=C3(u+1) 変数を元のyに戻すと ![]() ![]() xy=C3( ![]() x2y=C3(y+x) x2y=C(x+y)→2 |
同次型微分方程式[例と解]
の形に変形できる微分方程式について,類題と解を示す.
(1)
両辺を(検算) のとき だから,微分方程式を満たす. |
(2)
両辺を(検算) のとき だから,微分方程式 |
(3)
両辺を
この問題で,右辺の
であれば,左辺の部分分数分解は大変! (検算) のとき に等しくなることが分かる(結構長い計算になる)から,微分方程式 |
(参考)
両辺を(3’) であれば,左辺の部分分数分解は大変であるが,途中経過は (検算) のとき 両辺を微分すると,長い計算の結果 が得られる.次に(*1)÷(*2)により が成り立つから,微分方程式の解となってる. |
(4)
両辺を(検算) のとき に等しくなるから,微分方程式を満たす. |
(5)
両辺をのとき,両辺を (*1)÷(*2) だから,微分方程式を満たす. |
(6)
両辺を(検算) のとき だから,微分方程式 |
(7)
両辺を(検算) のとき,両辺を微分すると だから,微分方程式を満たす. |
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■[個別の頁からの質問に対する回答][同次形 微分方程式について/17.1.10]
問題6
右辺 = -u - 2u^2
=>[作者]:連絡ありがとう.解説の中で係数の2が抜けているところがあるというご指摘だと理解しました.訂正しました. |
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