## goalie2012 2 years ago solve the differential equation with initial conditions; compute the first 6 coefficients(a0-a5); find the general pattern: (1-2x)y''-y'+xy=0 y(0)=0, y'(0)=1

1. oldrin.bataku

Have you read the FAQ? It's generally frowned upon to post the same question multiple times... additionally, you are expected to reward helpful answers with medals (re: your previous question)

2. goalie2012

sorry about that. Kind of new. thanks for the tip

3. oldrin.bataku

Anyways, again consider an analytic solution $$y=\sum\limits_{n=0}^\infty a_nx^n$$ with derivatives $$y'=\sum_{n=0}^\infty (n+1)a_{n+1}x^n\\y''=\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n$$

4. oldrin.bataku

Plugging into our equation we have:$$(1-2x)\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=0}^\infty(n+1)a_{n+1}x^n+x\sum_{n=0}^\infty a_nx^n=0$$

5. oldrin.bataku

For our last term, note we can multiply into our sum $$x$$ and shift our indices of summation:$$x\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty a_nx^{n+1}=\sum_{n=1}^\infty a_{n-1}x^n$$

6. oldrin.bataku

For our first term, observe we can distribute $$1-2x$$ and do something similar:\begin{align*}(1-2x)&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-2x\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=0}^\infty2(n+1)(n+2)a_{n+2}x^{n+1}\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=1}^\infty2n(n+1)a_{n+1}x^n\end{align*}

7. oldrin.bataku

@goalie2012 you should award the medal on the other question... I haven't answered this one yet!

8. oldrin.bataku

Now, I haven't combined them into 1 yet because dealing with the starting indices would be gross...

9. goalie2012

ya. that's where I got a little lost

10. oldrin.bataku

Well, we can safely rewrite that summation again with 'nicer' indices since starting at $$n=0$$ will just add $$2(0)(1)a_1x^0=0$$:\begin{align*}&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=1}^\infty2n(n+1)a_{n+1}x^n\\=&\sum_{n=0}^\infty(n+1)(n+2)a_{n+2}x^n-\sum_{n=0}^\infty2n(n+1)a_{n+1}x^n\\=&\sum_{n=0}^\infty[(n+2)a_{n+2}-2na_{n+1}](n+1)x^n\end{align*}

11. oldrin.bataku

We can also extend our previous summation:$$\sum_{n=1}^\infty a_{n-1}x^n=\sum_{n=0}^\infty a_{n-1}x^n$$where we can impose $$a_m=0$$ for $$m<0$$ (this way we don't change the value of the summation); this is okay because any $$a_m$$ for $$m<0$$ do not appear in our solution.

12. oldrin.bataku

Anyways, let's put it all together:$$\sum_{n=0}^\infty[(n+2)a_{n+2}-2na_{n+1}](n+1)x^n-\sum_{n=0}^\infty (n+1)a_{n+1}x^n+\sum_{n=0}^\infty a_{n-1}x^n=0\\\sum_{n=0}^\infty\left(\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}\right)x^n=0$$Because our right must be identically $$0$$, we require the coefficient of each power of $$x$$ to be $$0$$:$$\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}=0$$

13. oldrin.bataku

Now, consider our initial conditions $$y(0)=0,y'(0)=1$$:$$y(0)=0\\\sum_{n=0}^\infty a_nx^n=0$$Since $$x^n=0$$ for $$n>0$$, we're left with only the term for $$n=0$$:$$a_0x^0=0\\a_0=0$$

14. oldrin.bataku

Moving on to our second condition $$y'(0)=1$$, we find:$$\sum_{n=0}^\infty(n+1)a_{n+1}x^n=1$$Again, $$x=0$$ leaves all but our $$n=0$$ term:$$a_1x^0=1\\a_1=1$$... so we've solved for first two coefficients, $$a_0=0,a_1=1$$.

15. oldrin.bataku

For our last three, we go back to that expression!

16. oldrin.bataku

$$\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}=0$$Letting $$n=0$$ we know $$a_{-1}=0,a_0=0,a_1=1$$ so:$$(2a_2-a_1)+a_{-1}=0\\2a_2-1=0\\a_2=\frac12$$

17. oldrin.bataku

For $$n=1$$ we find:$$2(3a_3-2a_2-a_2)+a_0=0\\6a_3-6a_2+a_0=0$$Knowing $$a_0=0,a_1=1$$ and having just concluded $$a_2=\frac12$$:$$6a_3-3=0\\6a_3=3\\a_3=\frac12$$

18. oldrin.bataku

$$\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}=0$$ For $$n=2$$ we find:$$3(4a_4-2(2)a_3-a_3)+a_1=0\\12a_4-15a_3+a_1=0$$Again, knowing $$a_1=1$$ and having concluded $$a_3=\frac12$$:$$12a_4-\frac{15}2+1=0\\24a_4-15+2=0\\24a_4-13=0\\a_4=\frac{13}{24}$$

19. oldrin.bataku

Can you take it from here? ;-)

20. goalie2012

yep. thank you very much. I have two more. would you be willing to help?

21. oldrin.bataku

I can try!

22. goalie2012

one won't take long and the other I'm not sure about

23. goalie2012

I have to close this question to write another one, right?