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◆◆◆ 復習 ◆◆◆
(I)同次形の微分方程式
[ポイント]![]() ![]() ![]() zがxに関して変数分離形になります. z= ![]() ![]() ![]() ![]() ![]() ![]() ![]()
【例】
x2y'−xy+2y2=0
y'=![]() ![]() ![]() において,z= ![]()
一般解は
z'x+z=z−2z2y= ![]() になります. z'x=−2z2 ![]() ![]() ![]() ![]() となって変数分離形になります.
(II)ベルヌーイ形の微分方程式
[ポイント]y'+P(x)y=Q(x)yn (n≠0, 1) の形に書けるものは, y−n y'+P(x)y1−n=Q(x) と変形してz=y1−nとおけば, zがxに関して線形微分方程式になります. z=y1−nとおけば z'=(1−n)y−n y' となって ![]() は線形微分方程式になります.
【例】
xy'+y=xy2
y'+![]()
一般解は
y−2 y'+![]() になります. ![]() においてz=y−1とおけば z'=−y−2 y' だから −z'+ ![]() z'− ![]() は線形微分方程式になります.(この結果は同次形と見なしても解けます.)
(III)リッカチ形の微分方程式
[ポイント]y'+P(x)y2+Q(x)y+R(x)=0 の形に書けるものは,その1つの特別解y1がわかれば,y=y1+zとおくことにより,zがxに関してベルヌーイ形の微分方程式になります. y1'+P(x)y12+Q(x)y1+R(x)=0 だから,y1がR(x)を”持って行ってしまう”ので,残りのzについては z'+S(x)z2+T(x)z=0 となってベルヌーイ形になります.
【例】
xy'−y+y2=x2
y1=xが1つの特別解になるから,y=x+uとおくと,uは次のベルヌーイ形の微分方程式を満たします.u'+(2− ![]() ![]() u−2 u'+(2− ![]() ![]() においてw=u−1とおけば w'+( ![]() ![]() は線形微分方程式になります.
一般解は,順に
になります. |
◆◆◆この頁で扱う内容◆◆◆ …上記以外で,変数変換によって既知の解き方に持ち込めるもの
(IV)y'=f(ax+by+c)
[ポイント]の形に書けるものは,z=ax+by+cとおくことにより,zがxに関して変数分離形の微分方程式になります. z'=a+by' ⇔ y'= ![]()
←yが全部取れたらOK
![]() z'=bf(z)+a は変数分離形になります.
【例IV.1】
(解答)y'= ![]() z=2x+y−1とおくと z'=2+y' ⇔ y'=z'−2になるから z'−2= ![]()
一般解は
z'=y=Ce ![]() ![]() は変数分離形になります.
【例IV.2】
(解答)y'= ![]()
※この問題のように,右辺の分子と分母のx,yの係数について,a'x+b'y+c' / ax+by+cとしたときに
(解答)a'=ka, b'=kbを満たすとき すなわち分母=0が表す直線と分子=0が表す直線が平行になっているとき(定数項cは同じ比でない) a'x+b'y+c'=k(ax+by+c)+...の形に書くことができます. 4x−2y+3=2(2x−y+1)+1だから z=2x−y+1とおくと (右辺)= ![]() また,z'=2−y'だから (左辺)=2−z' したがって方程式は
一般解は
2−z'=(2x−y+1)2+2x=C になります. ![]() z'=− ![]() となって,変数分離形になります. |
(V)y'=f(
[ポイント]![]() の形に書けるものは, ![]() ax+by+c=0 の連立方程式の解をx=p, y=qとするとき(a:a'≠b:b'だから2直線は交わる)
とする変換により,定数項が消えて
Y'=f(
z=![]() ![]() ![]() ap+bq+c=0 が成り立つ場合は ![]() a(X+p)+b(Y+q)+c=aX+bY+(ap+bq+c) において,定数項が消えます.
【例V.1】
(解答)y'=− ![]() ![]() 3x−4y+5=0 の解はx=1, y=2だから, y'=− ![]() ![]() ![]() ![]() ![]() ![]() ![]() ![]()
一般解は
(x+2y−5)(2x−y)=C になります. |
z=x−2y+1とおくと
z'=1−2y' ⇔ y'= ![]() ![]() ![]() ![]() ![]() ∫ ![]() log|z−1|=−x+C1 |z−1|=e−x+C1=eC1e−x z−1=±eC1e−x=C2e−x x−2y+1−1=C2e−x x−2y=C2e−x y= ![]() ![]() |
【問題2】
![]() ![]() 1 ![]() ![]() 3 ![]() ![]()
z=x−yとおくと
z'=1−y' ⇔ y'=1−z' 1−z'= ![]() ![]() ![]() ![]() は変数分離形 ∫ ![]() これを計算して,最後にx,yに戻す ![]() |
【問題3】
![]() ![]() 1(2x+y+3)2+2x=C 2(2x+y+3)2−2x=C 3(2x+y+3)2+2y=C 4(2x+y+3)2−2y=C 解説
z=2x+y+3とおくと
z'=2+y' ⇔ y'=z'−2 z'−2=− ![]() z'= ![]() は変数分離形になる. これを解くと z2=2x+C (2x+y+3)z2−2x=C →2 |
【問題4】
![]() ![]() 1(2y+x+7)(y+3x)=C 2(2y+x+7)(y−3x)=C 3(2y−x+7)(y+3x)=C 4(2y−x−7)(y+3x)=C 解説
連立方程式6x−5y+21=0 , 5x+4y−7=0の解はx=−1 , y=3
![]() ![]() X=x+1 , Y=y−3とおくと,dX=dx , dY=dy ![]() ![]() 右辺の分母分子をXで割って,z= ![]() ![]() z'X=− ![]() ![]() ![]() は変数分離形になる. これを解くと (2y−x−7)(y+3x)=C→4 |
(VI)
[ポイント]![]() ![]() の形に書けるものは,z=xyとおくと変数分離形になります. z=xyのときz'=y+xy' ⇔ y'= ![]() ![]() ![]() z'−y=yf(z) z'− ![]() ![]() z'= ![]() ![]() ![]() となって変数分離形になります.
【例VI.1】
(解答)![]() ![]() z=xyとおくとz'=y+xy' ⇔ y'= ![]() ![]() ![]() z'−y=y(z2−1) z'=y(z2−1+1)=yz2= ![]() ![]() ![]() は変数分離形になるので,これを解いて,最後にx,yに戻します. y2= ![]() になります. |
![]() ![]() ![]() と変形できるので,z=xyとおくと変数分離形になります. …(途中経過略)… ![]() ![]() になり,これを解くと ![]() ![]() 変数をx,yに戻すと ![]() ![]() ![]() ![]() ![]() と変形できるので,z=xyとおくと変数分離形になります. …(途中経過略)… ![]() ![]() になり,これを解くと ![]() 変数をx,yに戻すと y(xy+2)=Cx→3 |
![]() ![]() |
■[個別の頁からの質問に対する回答][微分方程式:変数変換による解き方 について/17.4.4]
例Ⅳ2の解答にz&aposと表示されていますが、z'の誤りではないかと思われます。
=>[作者]:連絡ありがとう.ブラウザ用の文字符号でセミコロンが1つ抜けていましたので訂正しました(&apos → ') |
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