<p>got a question from my math hw…(i’m an engineering student)</p>
<p>given (1-x)y’-y-x=0 , you get</p>
<p>(n+1)SUM a<em>n+1 *x^(n-1) -(n+1)SUM a</em>n *x^(n-1) - 1 =0</p>
<p>what do you do with that ‘1’ when you want to find the recursive coefficient thing?</p>
<p>thanks!!</p>
<p>i’m not quite sure what you’re asking, but consider this: each power series has a coefficient a_0 corresponding to the term x^0, namely, the constant term.</p>
<p>Adding or subtracting a constant, then, is just changing the a_0 term in the final power series.</p>
<p>hope this helps!</p>
<p>break it into two cases:</p>
<p>for n=1, incorporate the “1” in your formula.</p>
<p>for n=2->infty, ignore the “1”</p>
<p>"break it into two cases:</p>
<p>for n=1, incorporate the “1” in your formula.</p>
<p>for n=2->infty, ignore the “1"”</p>
<p>That makes sense… since this is a first-order system the response of a constant input is going to be linear.</p>
<p>Too bad I just turned in the assignment… I gave up and put y = SUM (1-2/n!) *x^n</p>