第一章　微分方程式

§1 微分方程式と曲線群

問１

Eliminate[{y == c Exp[2  x],  
        y' == D[c Exp[2  x], x]},  c]

Plot[
      Evaluate[Table[
          c Exp[2  x],  {c,  -10,  10,   .5}]],  {x,  -5,  
        5},  PlotStyle -> 
        Table[Hue[c/10], {c,  -10,  10,   .5}]]

SetAttributes[c, Constant]

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

<< Graphics`ImplicitPlot`

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}]]

問2

SetAttributes[ {a, b}, Constant]

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}]

Eliminate[{y - (a Sin[2  x] + b Cos[2  x]) == 0,  
        Dt[y - (a Sin[2  x] + b Cos[2  x]), x] == 0, 
        Dt[y - (a Sin[2  x] + b Cos[2  x]), {x, 2}] == 0},  {a, 
        b}]

問題

１．（１）

Eliminate[{y - Sqrt[x + c] == 0,  
        Dt[y - Sqrt[x + c], x] == 0},  c]

Plot[
      Evaluate[Table[ 
          If[x + c >= 0, Evaluate[ Sqrt[x + c]],  0],  {c,  -10,  
            10,   .5}]],  {x,  -10, 10},  
      PlotStyle -> 
        Table[Hue[c/10], {c,  -10,  10,   .5}]]

１．（２）

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

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。

DSolve[x y[x] Dt[y[x], x] == 1 + y[x]^2,  y[x],  x]

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}]]

１．（３）

Eliminate[{y - c/x - c == 0,  Dt[y - c/x - c, x] == 0},  
      c]

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。

DSolve[y[x] == (-1 - x) x Dt[y[x], x],  y[x],  
      x]

Simplify[%]

Plot[
      Evaluate[Table[ -c/x - c,  {c,  -2, 
            2,   .1}]],  {x,  -5, 5},  
      PlotStyle -> 
        Table[Hue[c/10], {c,  -10,  10,   .5}]]

２，（１）

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}]]

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

２，（２）

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}]]

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

DSolve[y[x] Dt[y[x], x] == 4  x,  y[x], x]

２，（３）

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}]]

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

§2 微分方程式の解

例題１

DSolve[x + y[x] y'[x] == 0,  y[x], x]

<< Graphics`ImplicitPlot`

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}]]

例題２

DSolve[y'[x]^2 == 4  y[x],  y[x], x]

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

演習問題　１−１　[A]

1 (1)

Eliminate[{y == Sin[x + c],  y' == D[Sin[x + c], x]},  
      c]

1 (2)

Eliminate[{y == Tan[x + c],  y' == D[Tan[x + c], x]},  
      c]

1 (3)

SetAttributes[{a,  b},  Constant]

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}]

1 (4)

Eliminate[{y == a x + b/x, Dt[ y - a x - b/x, x] == 0, 
        Dt[ y - a x - b/x, {x, 2}] == 0},  {a, b}]

2 (1)

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

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}]]

Show[g2, g1,  PlotRange -> {0, 12.5}]

SetAttributes[c,  Constant]

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

2 (2)

<< Graphics`ImplicitPlot`

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}]]

SetAttributes[c,  Constant]

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

演習問題　１−１　[B]

3(1)

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

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

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

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

3(2)

SetAttributes[{a,  b},  Constant]

?a

Eliminate[{y == a Exp[x]  + b Exp[2  x],  
        Dt[a Exp[x]  + b Exp[2  x] - y ,  x] == 0,  
        Dt[a Exp[x]  + b Exp[2  x] - y,  {x,  2}] == 0},  {a,  
        b}]

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

２，（２）

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}]]

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

DSolve[y[x] Dt[y[x], x] == 4  x,  y[x], x]

２，（３）

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}]]

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