Theory of Equation

Expansion of an Implicit Function of 𝑥

Let 𝑦^𝛼(𝐴𝑥^𝑎+)+𝑦^𝛽(𝐵𝑥^𝑏+)+⋯+𝑦^𝜎(𝑆𝑥^𝑠+)=01 be an equation arranged in descending powers of 𝑦, the coefficients being functions of 𝑥, the highest powers only of 𝑥 in each coefficient being written.
It is required to obtain 𝑦 in a series of descending powers of 𝑥.
First form the fractions −𝛼−𝑏𝛼−𝛽, −𝛼−𝑐𝛼−𝛾, −𝛼−𝑑𝛼−𝛿, ⋯, −𝛼−𝑠𝛼−𝜎2 Let −𝛼−𝑘𝛼−𝑛=𝑡 be the greatest of these algebraically, or if several are equal and greater than the rest, let it be the last of such. Then, with the letters corresponding to these equal and greatest fractions, form the equation 𝐴𝑢^𝛼+⋯+𝐾𝑢^𝜅=03 550 Each value of 𝑢 in this equation corresponds to a value of 𝑦, commencing with 𝑢𝑥^𝑡
Next select the greatest of the fractions −𝑘−𝑙𝜅−𝜆, −𝑘−𝑚𝜅−𝜇, ⋯, −𝑘−𝑠𝜅−𝜎4 Let −𝑘−𝑛𝜅−𝜈=𝑡′ be the last of the greatest ones. Form the corresponding equation 𝐾𝑢^𝜅+⋯+𝑁𝑢^𝜈=05 Then each value of 𝑢 in this equation gives a corresponding value of 𝑦, commencing with 𝑢𝑥^𝑡.
Proceed in this way until the last fraction of the series [2] is reached.
To obtain the second term in the expansion of 𝑦, put 𝑦=𝑥^𝑡(𝑢+𝑦₁) in [1]6 employing the different values of 𝑢, and again of 𝑡′ and 𝑢, 𝑡″ and 𝜈, ⋯ in succession; and in each case this substitution will produce an equation in 𝑦, 𝑥 similar to the original equation in 𝑦.
Repeat the foregoing process with the new equation in 𝑦, observing the following additional rule:
When all the values of 𝑡, 𝑡′, 𝑡″, ⋯, have been obtained, the negative ones only must be employed in forming the equations in 𝑢. 7 552 To obtain 𝑦 in a series of ascending powers of 𝑥.
Arrange equation [1] so that 𝛼, 𝛽, 𝛾, ⋯ may be in ascending order of magnitude, and 𝑎, 𝑏, 𝑐, ⋯ the lowest powers of 𝑥 in the respective coefficients.
Select 𝑡, the greatest of the fractions in [2], and proceed exactly as before, with the one exception of substituting the word positive for negative in [7]. 553 Example: Take the equation (𝑥³+𝑥⁴)+(3𝑥²−5𝑥³)𝑦+(−4𝑥+7𝑥²+𝑥³)𝑦²−𝑦⁵=0 It is required to expand 𝑦 in ascending powers of 𝑥.
The fractions [2] are −3−20−1, −3−10−2, −3−00−5; or 1, 1, and 35.
The first two being equal and greatest, we have 𝑡=1.
The fractions [4] reduce to −1−02−5=13=𝑡′.
Equation [3] is 1+3𝑢−4𝑢²=0 which gives 𝑢=1 and −14, with 𝑡=1 Equation [5] −4𝑢²−𝑢⁵=0 and from this 𝑢=0 and −4¹², with 𝑡′=13 We have now to substitute for 𝑦, according in [6], either 𝑥(1+𝑦₁), 𝑥(−14+𝑦₁), 𝑥¹³𝑦, or 𝑥¹³(−4¹³+𝑦₁) Put 𝑦=𝑥(1+𝑦₁), the first of these values, in the original equation, and arrange n ascending powers of 𝑦, thus −4𝑥⁴+(−5𝑥³+)𝑦₁+(−4𝑥³+)𝑦²₁−10𝑥³𝑦³₁−5𝑥³𝑦⁴₁−𝑥⁵𝑦⁵₁=0 The lowest power only of 𝑥 in each coefficient is here written.
The fractions [2] now become −4−30−1, −4−30−2, −4−50−3, −4−50−4, −4−50−5, or 1, 12, −13, −14, −15, From these 𝑡=1, and equation [3] becomes −4−5𝑢=0; ∴𝑢=−45 Hence one of the values of 𝑦₁ is, as in [6], 𝑦₁=𝑥(−45+𝑦₂)
Therefore 𝑦=𝑥{1+𝑥(−45+𝑦₂)}=𝑥−45𝑥²+⋯ Thus the first two terms of one of the expansions have been obtained.

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive