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Find the first four nonzero terms of the power series solution of y′ = 1 + xy2

Consider the differential equation

    \[ y' = 1 + xy^2 \]

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} &= 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*}

From the initial condition y =0 when x = 0 we know a_0 = 0. Therefore, equating like powers of x 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*}

(Note: The book gives the value a_{10} = \frac{23}{1120}. 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

    \[ f(x) = x + \frac{1}{4} x^4 + \frac{1}{14} x^7 + \frac{23}{112} x^{10} + \cdots. \]

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