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Find the first four nonzero terms of the power series solution of y′ = x + y2

Consider the differential equation

    \[ y' = x + y^2 \]

with initial conditions y = 0 when x = 0. Assume this differential equation has a power-series solution and compute the first four nonzero terms of the expansion.


Let

    \[ f(x) = \sum_{n=0}^{\infty} a_n x^n.\]

be the power-series solution of the differential equation. Then we must have

    \begin{align*}  && \sum_{n=1}^{\infty} na_n x^{n-1} &= x + \left( \sum_{n=0}^{\infty} a_n x^n \right)^2 \\[9pt]  \implies && \sum_{n=1}^{\infty} na_n x^{n-1} &= x + \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 &= a_0^2 + (2a_0 a_1 + 1)x + (a_1^2 + 2a_0a_2)x^2 + \cdots. \end{align*}

From the initial conditions we know a_0 = 0. Then, equating like powers of x we can solve for the first four nonzero terms in the power series expansion:

    \begin{align*}  a_0 &= 0 \\[9pt]  a_1 &= a_0^2 & \implies && a_1 &= 0 \\[9pt]  2a_2 &= 2a_0 a_1 + 1 & \implies && a_2 & = \frac{1}{2} \\[9pt]  3a_3 &= a_1^2 + 2a_0 a_2 & \implies && a_3 &= 0 \\[9pt]  4a_4 &= 2(a_0 a_3 + a_1 a_2) & \implies && a_4 &= 0 \\[9pt]  5a_5 &= a_2^2 + 2(a_0 a_4 + a_1 a_3) & \implies && a_5 &= \frac{1}{20} \\[9pt]  6a_6 &= 2(a_0 a_5 + a_1 a_4 + a_2 a_3) & \implies && a_6 &= 0 \\[9pt]  7a_7 &= a_3^2 + 2(a_0 a_6 + a_1 a_5 + a_2 a_4) & \implies && a_7 &= 0 \\[9pt]  8a_8 &= 2(a_0 a_7 + a_1 a_6 + a_2 a_5 + a_3 a_4) & \implies && a_8 &= \frac{1}{160} \\[9pt]  9a_9 &= a_4^2 + 2(a_0 a_8 + a_1 a_7 + a_2 a_6 + a_3 a_5) & \implies && a_9 &= 0 \\[9pt]  10a_{10} &= 2(a_0 a_9 + a_1 a_8 + a_2 a_7 + a_3 a_6 + a_4 a_5) & \implies && a_{10} &= 0 \\[9pt]  11a_{11} &= a_5^2 + 2(a_0 a_{10} + a_1 a_9 + a_2 a_8 + a_3 a_7 + a_4 a_6) & \implies && a_{11} &= \frac{1}{8800} \end{align*}

(Note: I think the solution in the back of Apostol is wrong on this. Apostol has a_5 = \frac{1}{12}, a_8 = \frac{1}{060}, and a_{11} = \frac{7}{8800}. I’m going to mark this as errata until someone convinces me Apostol is actually correct.)

Therefore, we have

    \[ y = \frac{1}{2} x^2 + \frac{1}{20} x^5 + \frac{1}{160} x^8 + \frac{1}{8800} x^{11} + \cdots. \]

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