0
$\begingroup$

The question was as follows;
Solve; $$(x+2)y'+ (x+1)y = 0$$ using power series methods. Verify that the resulting power series solution can be written in the form $$y=A(x+2)e^{-x}$$ where $A$ is a constant (consider terms up to $x^5$).
My working out was: $$y'=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$$ so$$(x+y)y'+ (x+1)y=0$$=$$\sum_{n=0}^\infty[(n+1)a_{n+1}x^{n+1}+2(n+1)a_{n+1}x^n +a_nx^{n+1}+a_nx^n]=0$$ so;$$\sum_{n=0}^\infty[x^n((n+1)a_{n+1}(x+2)+a_n(x+1)]=0$$ So then $$(n+1)a_{n+1}(x+2)+a_n(x+1)=0$$ would this be right so far?
It's here when i'm finding diffuclty...
I then went on to make $a_{n+1}$ the subject so, $a_{n+1}=\frac{-a_n(x+1)}{(x+2)(n+1)}$. Then i tried plugging different values of n, to try to get a formula for $a_n$ in terms of $a_0$ (ie the arbitrary constant $A$)...
So, what I'm trying to get at is, is my method so far correct, and also, how should i go about finding the solution?
I also think that $e^{-x}$ is, in sum form;$$\sum_{n=0}^\infty(-1)^n\frac{x^n}{n!}$$

Any help is much appreciated.
EDIT
Whilst waiting, I tried to evaluate $a_{n+1}$ and I got to; $$a_{n}=\frac{a_0(-1)^n(x+1)^n}{n!(x+2)^n}$$
EDIT 2
$y=\sum_{n=0}^\infty a_nx^n$ $=\sum_{n=0}^\infty [\frac{a_0(-1)^n(x+1)^nx^n}{n!(x+2)^n}]$ $=a_0 \sum_{n=0}^\infty [\frac{(-1)^nx^n}{n!}]\sum_{n=0}^\infty [\frac{(x+1)^n}{(x+2)^n}]$ $$=A\sum_{n=0}^\infty [\frac{(x+1)^n}{(x+2)^n}]e^{-x}$$
I'm not sure what to do next...
Nevermind, I've figured out the answer!

$\endgroup$
  • $\begingroup$ Are you sure that your equation is correct? Could it be that the equation is $$(x+2) y' + (x+1)y = 0$$ $\endgroup$ – Futurologist Oct 9 '16 at 7:10
  • $\begingroup$ @Futurologist yh, thats what i meant...thanks, but the working out was for $$(x+2)y'+(x+1)y=0$$ $\endgroup$ – Codefailure Oct 9 '16 at 10:55
  • $\begingroup$ @Moo Initial conditions? no, the qs is just as it is there... $\endgroup$ – Codefailure Oct 9 '16 at 10:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.