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

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

One comment

January 19, 2025 at 1:42 pm

Anonymous says:

There’s no need to use the Cauchy convolution formula for the product of series. You can use the uniqueness theorem of power series coefficients and just take derivatives using the initial condition and follow the recursion.

Stumbling Robot

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

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