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

Chapter 6, Problem 30E

(choose chapter or problem)

To prove Abel's identity  for , proceed as follows:
(a) Let \(W(x):=W\left[y_{1}, y_{2}, y_{3}\right](x)\). Use the product rule for differentiation to show

\(W^{\prime}(x)=\left|y_{1}^{\prime} y_{2}^{\prime} y_{3}^{\prime} y_{1}^{\prime} y_{2}^{\prime} y_{3}^{\prime} y_{1}^{\prime \prime} y_{2}^{\prime} y_{3}^{\prime}\right|+\left|y_{1} y_{2} y_{3} y_{1}^{\prime \prime} y_{2}^{\prime \prime} y_{3} y_{1}^{\prime \prime} y_{2}^{\prime} y_{3}^{\prime \prime} \mathrm{I}+\right| y_{1} y_{2} y_{3} y_{1}^{\prime} y_{2}^{\prime} y_{3}^{\prime} y_{1}^{\prime \prime} y_{2}^{\prime \prime} y_{3}^{\prime \prime}\)

(b) Show that the above expression reduces to

(32) \(W^{\prime}(x)=\left\lfloor y_{1} y_{2} y_{3} y_{1}^{\prime} y_{2}^{\prime} y_{3}^{\prime} y_{1}^{-1} y_{2}^{2 m} y_{3}^{-\prime} \mid\right.\)

(c) Since each  satisfies , show that

(33) \(y_{1}^{(3)}(x)=\sum_{k=1}^{3} P_{k}(x) y_{1}^{3-k j}(x)(t=1,2,3)\)

(d) Substituting the expressions in (33) into (32), show that
(34) \(W^{\prime}(x)=p_{1}(x) W(x)\)

(e) Deduce Abel’s identity by solving the first-order

differential equation (34).

Equation transcription:

Text transcription:

W(x):=W\left[y_{1}, y_{2}, y_{3}\right](x)

W^{\prime}(x)=\left|y_{1}^{\prime} y_{2}^{\prime} y_{3}^{\prime} y_{1}^{\prime} y_{2}^{\prime} y_{3}^{\prime} y_{1}^{\prime \prime} y_{2}^{\prime} y_{3}^{\prime}\right|+\left|y_{1} y_{2} y_{3} y_{1}^{\prime \prime} y_{2}^{\prime \prime} y_{3} y_{1}^{\prime \prime} y_{2}^{\prime} y_{3}^{\prime \prime} \mathrm{I}+\right| y_{1} y_{2} y_{3} y_{1}^{\prime} y_{2}^{\prime} y_{3}^{\prime} y_{1}^{\prime \prime} y_{2}^{\prime \prime} y_{3}^{\prime \prime}

W^{\prime}(x)=\left\lfloor y_{1} y_{2} y_{3} y_{1}^{\prime} y_{2}^{\prime} y_{3}^{\prime} y_{1}^{-1} y_{2}^{2 m} y_{3}^{-\prime} \mid\right.

y_{1}^{(3)}(x)=\sum_{k=1}^{3} P_{k}(x) y_{1}^{3-k j}(x)(t=1,2,3)

W^{\prime}(x)=p_{1}(x) W(x)