Should answer obtained by Picard's iteration match that obtained by any other method?

Question

I have some problems regarding solutions of first order differential equations obtained by Picard's iteration and by other methods. Like this example: $y'=xy+1$, $y(0)=1$.

I applied Picard's iteration as follows:

$$\begin{align} &y_1(x)=1+\int_0^x[1+t(1)]dt=1+x+\frac{x^2}{2}\\ &y_2(x)=1+\int_0^x\left[1+t\left(1+t+\frac{t^2}{2}\right)\right]dt=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{8}\\ &y_3(x)=1+x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{8}+\frac{x^5}{15}+\frac{x^6}{48}\end{align}$$ and so on.

Now, I tried to solve it using integration factor $e^{-\int xdx}=e^{-x^2/2}$

Then, $$\begin{align}ye^{-x^2/2}& =\int e^{-x^2/2}dx\\ &=\int1.e^{-x^2/2}dx\\ &=e^{-x^2/2}\int1dx-\int\left(-xe^{-x^2/2}\int 1dx \right)dx\\ &=xe^{-x^2/2}+\int x^2e^{-x^2/2}dx\\ &=xe^{-x^2/2}+e^{-x^2/2}\int x^2dx-\int\left(-xe^{-x^2/2}\int x^2dx \right)dx\\ &=xe^{-x^2/2}+\frac{x^3}{3}e^{-x^2/2}+\int \frac{x^4}{3}e^{-x^2/2}dx\end{align}$$

I am getting only odd powers of $x$ through these integrations. Why is that? Shouldn't both the answers match? Moreover, are the series obtained through the iterations always expansions of some function? I cannot identify such a 'function' here. And if not, how do I know the series converges?

Answer 1

You forgot about the initial value resp. that the indefinite integral contains an integration constant. If you start with $$ ye^{-x^2/2}-1=\int_0^xe^{-t^2/2}dt $$ then the right side gives, as observed, the odd degree terms. The initial condition gives an in the solution an additional term $$ 1·e^{x^2/2}=1+\frac{x^2}2+\frac{x^4}{2·2^2}+\frac{x^6}{6·2^3}+… $$ which are exactly the even order term of the power series.