If any differential equation is given by $f''(x)+f'(x)+f^2(x) = x^2\;,$ Then $f(x)=$

Solution 1:

Using the power series trick we get that:

\begin{align} &f(x) = \sum_{k=0}^\infty a_kx^k\\ &f'(x) = \sum_{k=0}^\infty a_{k+1}(k+1) x^k\\ &f''(x) = \sum_{k=0}^\infty a_{k+2}(k+1)(k+2)x^k\\ &f(x)^2 = \sum_{k=0}^\infty \left(\sum_{i=0}^k a_ia_{k-i} \right)x^k \end{align}

Inserting this into the equation we have that: $$ (k+1)(k+2)a_{k+2}+(k+1)a_{k+1}+\sum_{i=0}^k a_ia_{k-i} = \delta_{k,2}, \quad \mbox{for } k\geq 0. $$

Noting that $a_0 = f(0)$ we may successively solve the recurrence equation to find a solution. As a side remark we know (since the coefficients in the ODE are entire) that the radius of convergence is infinite (Fuchs theorem - if i recall correctly).