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Chapter 6: Q30E (page 327)

To prove Abel’s identity (26) for n = 3, proceed as follows:

(a) Let\({\bf{W}}\left( {\bf{x}} \right){\bf{ = W}}\left[ {{{\bf{y}}_{\bf{1}}}{\bf{,}}\,{{\bf{y}}_{\bf{2}}}{\bf{,}}\,{{\bf{y}}_{\bf{3}}}} \right]\left( {\bf{x}} \right)\). Use the product rule for differentiation to show

\({\bf{W'}}\left( {\bf{x}} \right){\bf{ = }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{''}}}\end{array}} \right|{\bf{ + }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}}&{{{\bf{y}}_{\bf{2}}}}&{{{\bf{y}}_{\bf{3}}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{''}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{''}}}\end{array}} \right|{\bf{ + }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}}&{{{\bf{y}}_{\bf{2}}}}&{{{\bf{y}}_{\bf{3}}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'''}}}\end{array}} \right|\)

(b) Show that the above expression reduces to

(32) \({\bf{W'}}\left( {\bf{x}} \right){\bf{ = }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}}&{{{\bf{y}}_{\bf{2}}}}&{{{\bf{y}}_{\bf{3}}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'''}}}\end{array}} \right|\)

(c) Since each \({{\bf{y}}_{\bf{i}}}\) satisfies (17), show that

(33) \({{\bf{y}}_{\bf{i}}}^{\left( {\bf{3}} \right)}\left( {\bf{x}} \right){\bf{ = - }}\sum\limits_{{\bf{k = 1}}}^{\bf{3}} {{{\bf{p}}_{\bf{k}}}\left( {\bf{x}} \right){{\bf{y}}_{\bf{i}}}^{\left( {{\bf{3 - k}}} \right)}\left( {\bf{x}} \right),\,\,\,\,\,\,\,\,\,\,\left( {{\bf{i = 1,2,3}}} \right)} \)

(d) Substituting the expressions in (33) into (32), show that(34) \({\bf{W'}}\left( {\bf{x}} \right){\bf{ = - }}{{\bf{p}}_{\bf{1}}}\left( {\bf{x}} \right){\bf{W}}\left( {\bf{x}} \right)\)

(e) Deduce Abel’s identity by solving the first-order differential equation (34).

Short Answer

Expert verified
  1. Proved.
  2. Proved.
  3. Proved.
  4. \({\bf{W'}}\left( {\bf{x}} \right){\bf{ = - }}{{\bf{p}}_{\bf{1}}}\left( {\bf{x}} \right){\bf{W}}\left( {\bf{x}} \right)\)
  5. \({\bf{W = W}}\left( {\bf{0}} \right){{\bf{e}}^{{\bf{ - }}{{\bf{p}}_{\bf{1}}}{\bf{x}}}}\)

Step by step solution

01

(a) Step 1: Use the given information

Given that,

Let \({\bf{W}}\left( {\bf{x}} \right){\bf{ = W}}\left[ {{{\bf{y}}_{\bf{1}}}{\bf{,}}\,{{\bf{y}}_{\bf{2}}}{\bf{,}}\,{{\bf{y}}_{\bf{3}}}} \right]\left( {\bf{x}} \right)\)

Then, express the Wronskian,

\({\bf{W}}\left( {\bf{x}} \right){\bf{ = }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}}&{{{\bf{y}}_{\bf{2}}}}&{{{\bf{y}}_{\bf{3}}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{''}}}\end{array}} \right|\)

Use the product rule for differentiation,

\({\bf{W'}}\left( {\bf{x}} \right){\bf{ = }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{''}}}\end{array}} \right|{\bf{ + }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}}&{{{\bf{y}}_{\bf{2}}}}&{{{\bf{y}}_{\bf{3}}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{''}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{''}}}\end{array}} \right|{\bf{ + }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}}&{{{\bf{y}}_{\bf{2}}}}&{{{\bf{y}}_{\bf{3}}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'''}}}\end{array}} \right|\)

02

(b) Step 2: Find the determinant of the expression of part (a), 

From part (a),

\({\bf{W'}}\left( {\bf{x}} \right){\bf{ = }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{''}}}\end{array}} \right|{\bf{ + }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}}&{{{\bf{y}}_{\bf{2}}}}&{{{\bf{y}}_{\bf{3}}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{''}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{''}}}\end{array}} \right|{\bf{ + }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}}&{{{\bf{y}}_{\bf{2}}}}&{{{\bf{y}}_{\bf{3}}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'''}}}\end{array}} \right|\)

Find the determinant of the above expression,

Here two rows are identical,

Therefore,

\(\begin{array}{c}{\bf{W'}}\left( {\bf{x}} \right){\bf{ = 0 + 0 + }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}}&{{{\bf{y}}_{\bf{2}}}}&{{{\bf{y}}_{\bf{3}}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'''}}}\end{array}} \right|\\{\bf{W'}}\left( {\bf{x}} \right){\bf{ = }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}}&{{{\bf{y}}_{\bf{2}}}}&{{{\bf{y}}_{\bf{3}}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'''}}}\end{array}} \right|\end{array}\)

03

(c) Step 3: Use the given information each \({{\bf{y}}_{\bf{i}}}\) satisfies (17),

Equation (17); Let \({{\bf{y}}_{\bf{1}}}{\bf{,}}\,{{\bf{y}}_{\bf{2}}}{\bf{,}}...,\,{{\bf{y}}_{\bf{n}}}\)be n solutions on (a, b) of

\({{\bf{y}}^{\left( {\bf{n}} \right)}}\left( {\bf{x}} \right){\bf{ + }}{{\bf{p}}_{\bf{1}}}\left( {\bf{x}} \right){{\bf{y}}^{\left( {{\bf{n - 1}}} \right)}}\left( {\bf{x}} \right){\bf{ + }}...{\bf{ + }}{{\bf{p}}_{\bf{n}}}\left( {\bf{x}} \right){\bf{y}}\left( {\bf{x}} \right){\bf{ = 0}}\)

We can write as,

\({{\bf{y}}_{\bf{i}}}^{\left( {\bf{n}} \right)}\left( {\bf{x}} \right){\bf{ = - }}\sum\limits_{{\bf{k = 1}}}^{\bf{n}} {{{\bf{p}}_{\bf{k}}}\left( {\bf{x}} \right){{\bf{y}}_{\bf{i}}}^{\left( {{\bf{n - k}}} \right)}\left( {\bf{x}} \right)} \)

We need to show that for n=3,

\({{\bf{y}}_{\bf{i}}}^{\left( {\bf{3}} \right)}\left( {\bf{x}} \right){\bf{ = - }}\sum\limits_{{\bf{k = 1}}}^{\bf{3}} {{{\bf{p}}_{\bf{k}}}\left( {\bf{x}} \right){{\bf{y}}_{\bf{i}}}^{\left( {{\bf{3 - k}}} \right)}\left( {\bf{x}} \right),\,\,\,\,\,\,\,\,\,\,\left( {{\bf{i = 1,2,3}}} \right)} \)

04

(d) Step 4: We need to show that,\({\bf{W'}}\left( {\bf{x}} \right){\bf{ =  - }}{{\bf{p}}_{\bf{1}}}\left( {\bf{x}} \right){\bf{W}}\left( {\bf{x}} \right)\).

From part (c),

\({{\bf{y}}_{\bf{i}}}^{\left( {\bf{3}} \right)}\left( {\bf{x}} \right){\bf{ = - }}\sum\limits_{{\bf{k = 1}}}^{\bf{3}} {{{\bf{p}}_{\bf{k}}}\left( {\bf{x}} \right){{\bf{y}}_{\bf{i}}}^{\left( {{\bf{3 - k}}} \right)}\left( {\bf{x}} \right),\,\,\,\,\,\,\,\,\,\,\left( {{\bf{i = 1,2,3}}} \right)} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( {\bf{1}} \right)\)

And from part (b),\({\bf{W'}}\left( {\bf{x}} \right){\bf{ = }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}}&{{{\bf{y}}_{\bf{2}}}}&{{{\bf{y}}_{\bf{3}}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'''}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'''}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'''}}}\end{array}} \right|\,\,\,\,\,\,\,......\left( {\bf{2}} \right)\)

Substituting the expressions in (1) into (2),

Substitute \({{\bf{y}}_{\bf{i}}}{\bf{'''}}\left( {\bf{x}} \right){\bf{ = }}{{\bf{p}}_{\bf{1}}}\left( {\bf{x}} \right){{\bf{y}}_{\bf{i}}}^{{\bf{''}}}\left( {\bf{x}} \right)\)in the equation (2),

\(\begin{array}{l}{\bf{W'}}\left( {\bf{x}} \right){\bf{ = }}\left| {\begin{array}{*{20}{c}}{{{\bf{y}}_{\bf{1}}}}&{{{\bf{y}}_{\bf{2}}}}&{{{\bf{y}}_{\bf{3}}}}\\{{{\bf{y}}_{\bf{1}}}{\bf{'}}}&{{{\bf{y}}_{\bf{2}}}{\bf{'}}}&{{{\bf{y}}_{\bf{3}}}{\bf{'}}}\\{{\bf{ - }}{{\bf{p}}_{\bf{1}}}{{\bf{y}}_{\bf{1}}}{\bf{''}}}&{{\bf{ - }}{{\bf{p}}_{\bf{1}}}{{\bf{y}}_{\bf{2}}}{\bf{''}}}&{{\bf{ - }}{{\bf{p}}_{\bf{1}}}{{\bf{y}}_{\bf{3}}}{\bf{''}}}\end{array}} \right|\\{\bf{W'}}\left( {\bf{x}} \right){\bf{ = - }}{{\bf{p}}_{\bf{1}}}{\bf{W}}\\{\bf{W'}}\left( {\bf{x}} \right){\bf{ = - }}{{\bf{p}}_{\bf{1}}}\left( {\bf{x}} \right){\bf{W}}\left( {\bf{x}} \right)\end{array}\)

05

(e) Step 5: Use the part (d) and integrate the given equation

From part (d),

\({\bf{W'}}\left( {\bf{x}} \right){\bf{ = - }}{{\bf{p}}_{\bf{1}}}\left( {\bf{x}} \right){\bf{W}}\left( {\bf{x}} \right)\)

Cross multiplication of both sides in the above equation

\(\frac{{{\bf{W'}}\left( {\bf{x}} \right)}}{{{\bf{W}}\left( {\bf{x}} \right)}}{\bf{ = - }}{{\bf{p}}_{\bf{1}}}\left( {\bf{x}} \right)\)

Taking integration of both sides in the above equation

\(\begin{array}{c}\int {\frac{{{\bf{W'}}\left( {\bf{x}} \right)}}{{{\bf{W}}\left( {\bf{x}} \right)}}} {\bf{ = - }}\int {{{\bf{p}}_{\bf{1}}}\left( {\bf{x}} \right){\bf{dx}}} \\{\bf{ln}}\left( {\bf{W}} \right){\bf{ = - }}{{\bf{p}}_{\bf{1}}}{\bf{x + ln}}\left( {\bf{A}} \right)\end{array}\)

Simplify the above equation

\(\begin{array}{l}\frac{{{\bf{ln}}\left( {\bf{W}} \right)}}{{{\bf{ln}}\left( {\bf{A}} \right)}}{\bf{ = - }}{{\bf{p}}_{\bf{1}}}{\bf{x}}\\\frac{{\bf{W}}}{{\bf{A}}}{\bf{ = }}{{\bf{e}}^{{\bf{ - }}{{\bf{p}}_{\bf{1}}}{\bf{x}}}}\\{\bf{W = A}}{{\bf{e}}^{{\bf{ - }}{{\bf{p}}_{\bf{1}}}{\bf{x}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,......\left( {\bf{3}} \right)\end{array}\)

At x=0,

\(\begin{array}{c}{\bf{W}}\left( {\bf{0}} \right){\bf{ = A}}{{\bf{e}}^{{\bf{ - }}{{\bf{p}}_{\bf{1}}}\left( {\bf{0}} \right)}}\\{\bf{W}}\left( {\bf{0}} \right){\bf{ = A}}\end{array}\)

Substitute the value of A in the equation (3),

\({\bf{W = W}}\left( {\bf{0}} \right){{\bf{e}}^{{\bf{ - }}{{\bf{p}}_{\bf{1}}}{\bf{x}}}}\)

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