goalie2012 Group Title 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 one year ago one year ago

1. oldrin.bataku Group Title

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 Group Title

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

3. oldrin.bataku Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

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

8. oldrin.bataku Group Title

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

9. goalie2012 Group Title

ya. that's where I got a little lost

10. oldrin.bataku Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

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

16. oldrin.bataku Group Title

$$\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}=0$$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 Group Title

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 Group Title

$$\left[(n+2)a_{n+2}-2na_{n+1}-a_{n+1}\right](n+1)+a_{n-1}=0$$ 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 Group Title

Can you take it from here? ;-)

20. goalie2012 Group Title

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

21. oldrin.bataku Group Title

I can try!

22. goalie2012 Group Title

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

23. goalie2012 Group Title

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