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Find the first 5 terms of the expansion in a power series $$y′=xe^{x}+2y^{2}$$

I've got a riccati equation $$ x e^{x}+2y^{2}, y(0)=0$$ After solving: $$y=e^{x}(x-1)+\frac{2}{3}y^{3} - 1$$

And I don't know how to go forward. Please help me.

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Your "After solving" is not correct. Most terms are correct, but the integral of $y^2\; dx$ is not $ y^3/3$ – Ross Millikan May 20 '13 at 19:04
hmm..I thought it correct, but y'=xe^x + 2y^2 can be solved without second summand, can't it? – SevenDays May 20 '13 at 20:13
No, you are integrating everything with respect to $x$. Take the simpler $y'=y$, which is solved by $y=Ae^x$. Your technique would give us $y=\frac {y^2}2$ which is not at all the same. – Ross Millikan May 20 '13 at 20:16
Maybe I didn't understand that, but with or without mistakes I've got a correct answer. Thanks for a correction, I'll be more attentive. – SevenDays May 20 '13 at 20:21
up vote 4 down vote accepted

You assume that there exist $a_i$ such that $$y = a_0 +a_1x + a_2x^2 + \cdots.$$
From here it is very easy to calculate power series expansions for $y', xe^x,$ and $2y^2$. Then you equate like powers of $x$ on both sides of the equality to get a relation between the $a_i$, and get a sequence of fairly simple equations in the $a_i$. The initial condition $y(0)=0$ tells you immediately that $a_0 = 0$ and from there you can usually get the other coefficients. Once you have $a_0,\ldots a_4$, you have the answer to the question.

To take a much simpler example that gives an idea of the method, suppose the equation were $$y' = y + 1$$ with the initial condition $y(0) = 0$.

We start with:

$$\begin{array}{rlrrr} y &= &a_0 &+a_1x &+ a_2x^2 &+ \cdots \\ y+1 &=& 1+ a_0 &+a_1x &+ a_2x^2 &+ \cdots \\ y' &=& a_1 &+2a_2x &+ 3a_3x^2 &+ \cdots \end{array}$$

Then equating the last two we have $$a_1 + 2a_2x + 3a_3x^2 + \cdots = 1 + a_0 + a_1x + a_2x^2 + \cdots$$

and equating coefficients of like powers gives equations in the $a_i$: $$ \begin{align} a_1 & =& 1+a_0\\ 2a_2 & = &a_1 \\ 3a_3 & = & a_2\\ &\vdots& \end{align} $$

Now $y(0) = 0$ tells us immediately that $a_0 = 0$. And the equations above then give us $a_1 = 1, a_2 = \frac12, a_3 = \frac16,\ldots$. So the series for $y$ is $$y = x + \frac12x^2 + \frac16x^3 + \cdots$$

which is in fact the power series for $y=e^x-1$.

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Thank you very much! You have completely solved my problem. – SevenDays May 20 '13 at 18:56
I am glad that I could help. – MJD May 20 '13 at 19:00

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