ODE in terms of simple fraction

[renegade05]

Hmm.. Solving this bad boy:

\(\displaystyle (2-x)y''-xy'-y=0, y(0)=1; y'(0)=1/2\)

And I found the recurrence relation to be \(\displaystyle a_m=2^{-m}\)

And the solution to the power series around 0 to be: \(\displaystyle y(x)=1+x/2+x^2/4+x^3/8+...\)

Now the question asks to write down the solution to the ODE in terms of a simple fraction involving (2-x).

I am stuck on this part. Can someone give me a pointer? I know the closed form of this series will probably involve an e^x as well...

 

[Ishuda]

Hmm.. Solving this bad boy:

\(\displaystyle (2-x)y''-xy'-y=0, y(0)=1; y'(0)=1/2\)

And I found the recurrence relation to be \(\displaystyle a_m=2^{-m}\)

And the solution to the power series around 0 to be: \(\displaystyle y(x)=1+x/2+x^2/4+x^3/8+...\)

Now the question asks to write down the solution to the ODE in terms of a simple fraction involving (2-x).

I am stuck on this part. Can someone give me a pointer? I know the closed form of this series will probably involve an e^x as well...

If you let t=x/2, then the partial series becomes
y_n = 1 + t + t² + t³ + ... + tⁿ
Do you recognize this series [hint multiple by 1-t and see what the result is]. What happens if the magnitude of t is less than 1 as n goes to infinity? If you can get y_n to go to y someway, substitute x/2 back in for t, and simplify I think you might get what you want.

 

[HallsofIvy]

Equivalently, that is the geometric series, of the form \(\displaystyle \sum ar^n\) with a= 1 and r= x/2.

It is well known that \(\displaystyle \sum ar^n= \frac{a}{1- r}\).
(which is why Ishuda suggests that you multiply by 1- r.)

 

[renegade05]

Equivalently, that is the geometric series, of the form \(\displaystyle \sum ar^n\) with a= 1 and r= x/2.

It is well known that \(\displaystyle \sum ar^n= \frac{a}{1- r}\).
(which is why Ishuda suggests that you multiply by 1- r.)

Thank you:

\(\displaystyle y(x)=\frac{2}{2-x}\)