Consider the differential equation
![\[ y' = 1 + xy^2 \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-516030259d3c73748831c42ea0cc3231_l3.png "Rendered by QuickLaTeX.com")
with initial conditions 
when 
. Assume this differential equation has a power-series solution and compute the first four nonzero terms of the expansion.
Let
![\[ f(x) = \sum_{n=0}^{\infty} a_n x^n \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-d72916b5ebec41e642700512f64fb753_l3.png "Rendered by QuickLaTeX.com")
be the power series solution of the differential equation. Then we must have
![\begin{align*} && \sum_{n=1}^{\infty} na_n x^{n-1} &= 1 + x \left( \sum_{n=0}^{\infty} a_n x^n \right)^2 \\[9pt] \implies && \sum_{n=1}^{\infty} na_n x^{n-1} &= 1 + x \left( \sum_{n=0}^{\infty} \left( \sum_{k=0}^n a_k a_{n-k} \right) x^n \right) \\[9pt] \implies && a_1 + 2a_2 x + 3a_3 x^2 + \cdots & = 1 + a_0^2 x + (2a_0 a_1)x^2 + (a_1^2 + 2a_0a_2)x^3 + \cdots. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-78e8571c567faf292a92e76c430ccdfc_l3.png "Rendered by QuickLaTeX.com")
From the initial condition 
when we know 
. Therefore, equating like powers of 
we have
![\begin{align*} a_1 &= 1 \\[9pt] 2a_2 &= a_0^2 & \implies && a_2 &= 0 \\[9pt] 3a_3 &= 2a_0 a_1 & \implies && a_3 &= 0 \\[9pt] 4a_4 &= a_1^2 + 2a_0 a_2 & \implies && a_4 &= \frac{1}{4} \\[9pt] 5a_5 &= 2(a_0 a_3 + a_1 a_2) & \implies && a_5 &= 0 \\[9pt] 6a_6 &= a_2^2 + 2(a_0 a_4 + a_1 a_3) & \implies && a_6 &= 0 \\[9pt] 7a_7 &= 2(a_0 a_5 + a_1 a_4 + a_2 a_3) & \implies && a_7 &= \frac{1}{14} \\[9pt] 8a_8 &= a_3^2 + 2(a_0 a_6 + a_1 a_5 + a_2 a_4) & \implies && a_8 &= 0 \\[9pt] 9a_9 &= 2(a_0 a_7 + a_1 a_6 + a_2 a_5 + a_3 a_4) & \implies && a_9 &= 0 \\[9pt] 10a_{10} &= a_4^2 + 2(a_0 a_8 + a_1 a_7 + a_2 a_6 + a_3 a_5) & \implies && a_{10} &= \frac{23}{112}. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-9bce6f870699e8fbcccf9f0e29b21e50_l3.png "Rendered by QuickLaTeX.com")
(Note: The book gives the value 
. I think the answer we have above is correct. I’m marking this as errata unless someone convinces me that Apostol is correct.)
Therefore, we have
![\[ f(x) = x + \frac{1}{4} x^4 + \frac{1}{14} x^7 + \frac{23}{112} x^{10} + \cdots. \]](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-664ff09a8f2d5ff929b103d78c7d6b04_l3.png "Rendered by QuickLaTeX.com")
It should be 23/1120… You have the equation
. But we are looking for the coefficient of the original series, so we need 
, not 
..