Find the first four nonzero terms of the power series solution of y′ = x2 + y2

Consider the differential equation

$y' = x^2 + y^2$

with initial conditions $y = 1$ 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

$\begin{align*} && f'(x) &= x^2 + (f(x))^2 \\[9pt] \implies && \sum_{n=1}^{\infty} na_n x^{n-1} &= x^2 + \sum_{n=0}^{\infty} \left( \sum_{k=0}^n a_k a_{n-k} \right) x^n \\[9pt] \implies && a_1 + 2a_2 x + 3a_3 x^2 + \cdots &= a_0^2 + 2a_0 a_1 x + (a_1^2 + 2a_0 a_2 + 1) x^2 + \cdots. \end{align*}$

From the initial condition $y = 1$ when $x = 0$ we have $a_0 = 1$ . Therefore, equating like powers of $x$ we have the following equations

$\begin{align*} a_0 &= 1 \\[9pt] a_1 &= a_0^2 = 1 \\[9pt] 2a_2 &= 2a_0 a_1 = 2 & \implies && a_2 &= 1 \\[9pt] 3a_3 &= a_1^2 + 2a_0 a_2 + 1 = 4 & \implies && a_3 &= \frac{4}{3}. \end{align*}$

Therefore, we have

$y = 1 + x + x^2 + \frac{4}{3} x^3 + \cdots.$

Stumbling Robot

Find the first four nonzero terms of the power series solution of y′ = x² + y²

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