蜷梧ｬ｡蠖｢縺ｫ蟶ｰ逹縺ｧ縺阪ｋ譁ｹ遞句ｼ/A> 隨ｬ吝屓隨ｬ大屓隧ｦ鬨/A> 隨ｬ10蝗荳髫守ｷ壼ｽ｢蠕ｮ蛻婿遞句ｼ/A> 隨ｬ11蝗繝吶Ν繝後繧､縺ｮ譁ｹ遞句ｼ/A> 隨ｬ12蝗螳悟蠕ｮ蛻婿遞句ｼ/A> 隨ｬ13蝗遨榊蝗蟄/A> 隨ｬ14蝗繧ｯ繝ｬ繝ｼ繝ｭ繝ｼ縺ｮ譁ｹ遞句ｼ/A> 隨ｬ15蝗鬮俶ｬ｡蠕ｮ蛻婿遞句ｼ/A> 隨ｬ16蝗隨ｬ貞屓隧ｦ鬨

(1)

8 同次形に帰着できる微分方程式

y′= a1x + b1y + c1 a2x + b2y + c2 の形の微分方程式は，適当な置換によって変数分離形または同次形に帰着できる。

8.1 a1

: b

1

= a

2

: b

2

の場合

u = a2x + b2y + c2 とおけば変数分離形になる（第 6 回 y′= f (a2x + b2y + c2)の特別の場合）。 例. y′= 2x− 4y − 1 3x− 6y − 2 y′ = 6x− 12y − 3 3(3x− 6y − 2) = 2(3x− 6y − 2) + 1 3(3x− 6y − 2) u = 3x− 6y − 2 とおくと u′ = 3− 6y′ = 3− 6 · 2u + 1 3u = 3u− 4u − 2 u = −u − 2 u ∴ − 1 = u u + 2u ′ = ( 1− 2 u + 2 ) u′ ∫ −1 dx = ∫ ( 1− 2 u + 2 ) du −x + C1 = u− 2 log |u + 2| = 3x− 6y − 2 − 2 log |3x − 6y| ∴ C1+ 2 = 4x− 6y − 2 log |3x − 6y| C = 2x− 3y − log |3x − 6y|

8.2 a

1

: b

1

̸= a2

: b

2

の場合

{ a1x + b1y + c1= a1(x− α) + b1(y− β) a2x + b2y + c2= a2(x− α) + b2(y− β) となる α, β を見つけて， { X = x− α Y = y− β とおくと Y′= dY dX = dY dx dX dx = y ′ 1 = y ′_{= f} ( a1X + b1Y a2X + b2Y ) となり，同次形 Y′ = a1X + b1Y a2X + b2Y が得られる。 なお，α, β は，連立方程式 { a1x + b1y + c1= 0 a2x + b2y + c2= 0 を解いて得られる。

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例. y′= 5x + 9y + 17 3x− y − 9 { 5x + 9y + 17 = 0 3x− y − 9 = 0 ∴ { x = 2 y =−3 ∴ y′= 5(x− 2) + 9(y + 3) 3(x− 2) − (y + 3) { X = x− 2 Y = y + 3 とおく Y′ = 5X + 9Y 3X− Y U = Y X とおくと U′ = Y ′_{− U} X 1 X = 1 Y′− U U ′ = 1 5 + 9U 3− U − U U′ = 3− U 5 + 9U − 3U + U2U′ = 3− U U2_{+ 6U + 5}U′ = 3− U (U + 1)(U + 5)U ′ = ( 1 U + 1 − 2 U + 5 ) U′ ∫ 1 X dX = ∫ ( 1 U + 1 − 2 U + 5 ) dU

log|X| + C1 = log|U + 1| − 2 log |U + 5|

C1 = log|U + 1| − 2 log |U + 5| − log |X|

= log U + 1 (U + 5)2_X = log Y + X (Y + 5X)2 = log y + 3 + x− 2 (y + 3 + 5x− 10)2 = log x + y + 1 (5x + y− 7)2 x + y + 1 (5x + y− 7)2 =±e C1 _{= C}

問題

次の微分方程式を解きなさい。 (1) y′= 10x + 7y− 1 x− y + 5 (2) y′+ 6x + 3y + 2 4x + 2y + 1 = 0 (3) (5x− y − 9)y′+ 3x− 7y + 1 = 0 (4) y′= √ 2x− y + 1 x− 2 （√がついているが同様にしてできる）

(3)

解答 (1) y′= 10x + 7y− 1 x− y + 5 { 10x + 7y− 1 = 0 x− y + 5 = 0 ∴ x =−2, y = 3 { X = x + 2 Y = y− 3 とおく Y′ = 10X + 7Y X− Y = 10 + 7Y /X 1− Y/X U = Y X とおく U′= Y ′_{− U} X 1 X = 1 Y′− U U ′ = 1 10 + 7U 1− U − U U′ = 1− U 10 + 7U− U + U2U′ = 1− U U2_{+ 6U + 10}U′ = −(U + 3) + 4 (U + 3)2_{+ 1} U′ ∫ 1 X dX = ∫ _{−(U + 3) + 4} (U + 3)2_{+ 1} dU log|X| + C1 = − 1 2 log(U + 3) 2_{+ 1} +4 tan−1(U + 3) −2C1 = logU2+ 6U + 10+ 2 log|X| −8 tan−1_{(U + 3)} C = log(U2_{+ 6U + 10)X}2 −8 tan−1_{(U + 3)} = logY2_{+ 6XY + 10X}2 −8 tan−1 ( Y X + 3 ) = log((y− 3)2_{+ 6(x + 2)(y}_{− 3)} +10(x + 2)2) −8 tan−1 ( y− 3 x + 2 + 3 ) = log((y− 3)2_{+ 6(x + 2)(y}_{− 3)} +10(x + 2)2) −8 tan−1 ( y + 3x + 3 x + 2 ) (2) y′+ 6x + 3y + 2 4x + 2y + 1 = 0 y′ = − 12x + 6y + 4 2(4x + 2y + 1) = −3(4x + 2y + 1) + 1 2(4x + 2y + 1) u = 4x + 2y + 1とおく u′ = 4 + 2y′ = 4− 3u + 1 u = u− 1 u 1 = u u− 1u ′ = ( 1 + 1 u− 1 ) u′ x + C1 = u + log|u − 1| = 4x + 2y + 1 + log|4x + 2y| C1− 1 = 3x + 2y + log |4x + 2y| ∴ C = 3x + 2y + log |4x + 2y|

(4)

(3) (5x− y − 9)y′+ 3x− 7y + 1 = 0 y′ = −3x− 7y + 1 5x− y − 9 = −3(x− 2) − 7(y − 1) 5(x− 1) − (y − 1) X = x− 2, Y = y − 1 とおく Y′ = −3X− 7Y 5X− Y = − 3− 7Y X 5− Y X U = Y X とおく U′ = Y ′_{− U} X 1 X = 1 Y′− U U ′ = 1 −3− 7U 5− U − U U′ = 5− U −3 + 7U − 5U + U2U′ = 5− U U2_{+ 2U}− 3U′ = 5− U (U + 3)(U− 2)U ′ =    − 8 5 U + 3 + 3 5 U− 2    U′ log|X| + C1 = − 8 5 log|U + 3| +3 5 log|U − 2| 5C1 = −8 log |U + 3| + 3 log |U − 2| −5 log |X| = log (U− 2) 3 (U + 3)8_X5 C = (U− 2) 3_X3 (U + 3)8_X8 = (Y − 2X) 3 (Y + 3X)8 = ((y− 1) − 2(x − 2)) 3 ((y− 1) + 3(x − 2))8 = (y− 2x + 3) 3 (y + 3x− 7)8 (y− 2x + 3)3 ₌ _{C(y + 3x}_{− 7)}8

(5)

(4) y′= √ 2x− y + 1 x− 2 y′ = √ 2(x− 2) − (y − 5) x− 2 X = x− 2, Y = y − 5 とおく Y′ = √ 2X− Y X = √ 2− Y X U = Y X とおく 1 X = 1 Y′− U U ′ = √ 1 2− U − U U ′ T =√2− U とおく U = 2− T2 ∴ U′=−2T T′ 1 X = 1 T− (2 − T2₎(−2T )T′ = −2T T2_{+ T}_{− 2}T′ = −2T (T + 2)(T− 1)T ′ =    − 4 3 T + 2 + −2 3 T− 1    T′ log|X| + C1 = − 4 3 log|T + 2| −2 3 log|T − 1| −3C1 = 4 log|T + 2| + 2 log |T − 1| +3 log|X| = log(T + 2)4_(T_{− 1)}2_X3 C = (T + 2)4(T− 1)2X3 = (√2− U + 2 )4(√ 2− U − 1 )2 X3 = (√ 2− Y X + 2 )4 (√ 2− Y X − 1 )2 X3 = (√2X− Y + 2√X )4 (√2X− Y −√X )2 = (√4x− y − 3 + 2√x− 2 )4 (√4x− y − 3 −√x− 2 )2