# Polynomial of at least degree 3 for this Cauchy Problem

I'm given this :

$$y''(x)+\sin(x)y'(x)+x^3y(x)=1+x$$

$$y'(0)=0,\;y(0)=1$$

and I'm asked to calculate a polynomial of at least degree $3$ that approximates the solution of that problem. May I try a series for $y(x)$ and then just chop at degree $3$?. If this is correct, then

$$y(x)=\sum_{k=0}^{\infty}c_{k}x^{k}, y'(x)=\sum_{k=1}^{\infty}kc_{k}x^{k-1},y''(x)=\sum_{k=2}^{\infty}k(k-1)c_{k}x^{k-2}$$

Substitution into the ODE gives

$$\sum_{k=2}^{\infty}k(k-1)c_{k}x^{k-2}+\sin(x)\sum_{k=1}^{\infty}kc_{k}x^{k-1}+x^3\sum_{k=0}^{\infty}c_{k}x^{k}=1+x$$

and

$$\sum_{k=2}^{\infty}k(k-1)c_{k}x^{k-2}+\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!}\sum_{k=1}^{\infty}kc_{k}x^{k-1}+\sum_{k=0}^{\infty}c_{k}x^{k+3}=1+x$$

Now, how do I match the powers to set the coefficients of $x^0=1$, the coefficients of $x^1=1$ , and the rest equal to zero? I'm stuck in the index change and in the product of $y'(x)$ and $sin(x)$ series.

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Using $y'(0)=0, y(0)=1$, you can tell $c_0 = 1$ and $c_1=0$
Looking at constant terms in your final line, you have $2c_2=1$ so $c_2=1/2$.
Looking at coefficients of $x$ in the final line you have $6 c_3 x +c_1 x = x$ so $c_3 = 1/6$ meaning $$y(x)=1+\frac{x^2}{2} + \frac{x^3}{6} \cdots$$
If you wanted to go further and look at coefficients of $x$ in the final line you have $12 c_4 x^2 + 2c_2 x^2 =0$ so $c_4=-1/12$, and it is not difficult to continue doing this again and again.
Hi, sorry for being late. So the point is, looking for the constant, $x^1$, $x^2$, until required, instead of trying to derive a general expresion for $c_k$ and let $k=0,..$ until required. Thank you again, it was very helpful. – Jorge Jan 25 '12 at 20:38