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第一章　微分方程式

T､1 微分方程式と曲線群

T問１

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[Graphics:HTMLFiles/MathAChap1_4.gif]

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$Eliminate[{x^2 + y^2 - 2 c x == 0, Dt[x^2 + y^2 - 2 c x, x] == 0}, c]$

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$ImplicitPlot[Evaluate[Table[x^2 + y^2 - 2 c x == 0, {c, -10, 10, .5}]], {x, -20, 20}, PlotStyle -> Table[Hue[c/10], {c, -10, 10, .5}]]$

[Graphics:HTMLFiles/MathAChap1_11.gif]

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T問2

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$Eliminate[{a y^2 - 4 (x + b) == 0, Dt[a y^2 - 4 (x + b), x] == 0, Dt[a y^2 - 4 (x + b), {x, 2}] == 0}, {a, b}]$

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$y Dt[y, {x, 2}] == -Dt[y, x]^2$

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T問題

T１．（１）

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[Graphics:HTMLFiles/MathAChap1_21.gif]

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T１．（２）

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$Eliminate[{c x^2 - y^2 == 1, Dt[c x^2 - y^2, x] == 0}, c]$

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Mathematicaで逆に微分方程式を解いてみよう。以下の解が元の
式と等しいことに注意せよ。

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${{y[x] -> -(-1 + e^(2 (C[1] + Log[x])))^(1/2)}, {y[x] -> (-1 + e^(2 (C[1] + Log[x])))^(1/2)}}$

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$ImplicitPlot[Evaluate[Table[ c x^2 - y^2 == 1, {c, -2, 2, .1}]], {x, -5, 5}, PlotStyle -> Table[Hue[c/10], {c, -10, 10, .5}]]$

[Graphics:HTMLFiles/MathAChap1_28.gif]

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T１．（３）

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Mathematicaで逆に微分方程式を解いてみよう。以下の解が元の
式と等しいことに注意せよ。

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${{y[x] -> e^(-Log[x] + Log[1 + x]) C[1]}}$

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[Graphics:HTMLFiles/MathAChap1_37.gif]

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T２，（１）

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$ImplicitPlot[Evaluate[Table[ y^2 + (x - c)^2 == 4, {c, -2, 2, .1}]], {x, -5, 5}, PlotStyle -> Table[Hue[c/10], {c, -10, 10, .5}]]$

[Graphics:HTMLFiles/MathAChap1_40.gif]

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$Eliminate[{y^2 + (x - c)^2 == 4, Dt[y^2 + (x - c)^2, x] == 0}, c]$

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T２，（２）

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$ImplicitPlot[Evaluate[Table[ y^2 - 4 c (x + c) == 0, {c, -2, 2, .1}]], {x, -5, 5}, PlotStyle -> Table[Hue[c/10], {c, -10, 10, .5}]]$

[Graphics:HTMLFiles/MathAChap1_45.gif]

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$Eliminate[{y^2 - 4 c (x + c) == 0, Dt[y^2 - 4 c (x + c), x] == 0}, c]$

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${{y[x] -> -2^(1/2) (2 x^2 + C[1])^(1/2)}, {y[x] -> 2^(1/2) (2 x^2 + C[1])^(1/2)}}$

T２，（３）

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$ImplicitPlot[Evaluate[Table[ y^2 - (2 x)^2 == c^2, {c, -2, 2, .1}]], {x, -5, 5}, PlotStyle -> Table[Hue[c/10], {c, -10, 10, .5}]]$

[Graphics:HTMLFiles/MathAChap1_52.gif]

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$Eliminate[{y^2 - (2 x)^2 == c^2, Dt[y^2 - (2 x)^2, x] == 0}, c]$

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T､2 微分方程式の解

T例題１

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${{y[x] -> -2^(1/2) (-x^2/2 + C[1])^(1/2)}, {y[x] -> 2^(1/2) (-x^2/2 + C[1])^(1/2)}}$

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$ImplicitPlot[Evaluate[Table[x^2 + y^2 == c^2, {c, 0, 10, .5}]], {x, -10, 10}, PlotStyle -> Table[Hue[c/10], {c, 0, 10, 0.5}]]$

[Graphics:HTMLFiles/MathAChap1_60.gif]

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T例題２

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${{y[x] -> 1/4 (4 x^2 - 4 x C[1] + C[1]^2)}, {y[x] -> 1/4 (4 x^2 + 4 x C[1] + C[1]^2)}}$

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$Plot[Evaluate[Table[(x - c)^2, {c, 0, 10, .5}]], {x, -5, 10}, PlotStyle -> Table[Hue[c/10], {c, 0, 10, 0.5}]]$

[Graphics:HTMLFiles/MathAChap1_65.gif]

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T演習問題　１−１　[A]

T1 (1)

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T1 (2)

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T1 (3)

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$Eliminate[{a x^2 + b y^2 == 1, Dt[ a x^2 + b y^2, x] == 0, Dt[ a x^2 + b y^2, {x, 2}] == 0}, {a, b}]$

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T1 (4)

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T2 (1)

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$g1 = Plot[x^2/2, {x, -5, 5}]$

[Graphics:HTMLFiles/MathAChap1_79.gif]

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$g2 = Plot[Evaluate[Table[c (x - c) + c^2/2, {c, -5, 5, .5}]], {x, -5, 5}, PlotStyle -> Table[Hue[c/10], {c, -5, 5, .5}]]$

[Graphics:HTMLFiles/MathAChap1_82.gif]

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[Graphics:HTMLFiles/MathAChap1_84.gif]

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[Graphics:HTMLFiles/MathAChap1_87.gif]

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$Eliminate[{y == c (x - c) + c^2/2, Dt[y - c (x - c) - c^2/2, x] == 0}, c]$

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T2 (2)

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$ImplicitPlot[Evaluate[Table[(y - c)^2 + x^2 == c^2, {c, 0, 5, .2}]], {x, -5, 5}, PlotStyle -> Table[Hue[c/5], {c, 0, 5, .2}]]$

[Graphics:HTMLFiles/MathAChap1_94.gif]

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$Eliminate[{(y - c)^2 + x^2 == c^2, Dt[(y - c)^2 + x^2, x] == 0}, c]$

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T演習問題　１−１　[B]

T3(1)

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$Solve[{(1 - a)^2 + b^2 == c^2, (-1 - a)^2 + b^2 == c^2}, {a, b}]$

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${{b -> -(-1 + c^2)^(1/2), a -> 0}, {b -> (-1 + c^2)^(1/2), a -> 0}}$

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$ImplicitPlot[Evaluate[Flatten[Table[{x^2 + (y - (-1 + c^2)^(1/2))^2 == c^2, x^2 + (y + (-1 + c ... 2))^2 == c^2}, {c, 1, 10, .5}]]], {x, -10, 10}, PlotStyle -> Table[Hue[c/10], {c, 1, 10, .5} ]]$

[Graphics:HTMLFiles/MathAChap1_103.gif]

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$Eliminate[{x^2 + (y - (-1 + c^2)^(1/2))^2 == c^2, Dt[x^2 + (y - (-1 + c^2)^(1/2))^2 , x] == 0}, c]$

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$Eliminate[{x^2 + (y + (-1 + c^2)^(1/2))^2 == c^2, Dt[x^2 + (y + (-1 + c^2)^(1/2))^2 , x] == 0}, c]$

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T3(2)

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Global`a

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Converted by Mathematica ﾊ(April 23, 2003)

第一章 微分方程式