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pantera1 [17]
3 years ago
13

Solve for y where y(2)=2 and y'(2)=0 by representing y as a power series centered at x=a

Mathematics
1 answer:
Crank3 years ago
6 0

I'll assume the ODE is actually

y''+(x-2)y'+y=0

Look for a series solution centered at x=2, with

y=\displaystyle\sum_{n\ge0}c_n(x-2)^n

\implies y'=\displaystyle\sum_{n\ge0}(n+1)c_{n+1}(x-2)^n

\implies y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n

with c_0=y(2)=2 and c_1=y'(2)=0.

Substituting the series into the ODE gives

\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge0}(n+1)c_{n+1}(x-2)^{n+1}+\sum_{n\ge0}c_n(x-2)^n=0

\displaystyle\sum_{n\ge0}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge0}c_n(x-2)^n=0

\displaystyle2c_2+c_0+\sum_{n\ge1}(n+2)(n+1)c_{n+2}(x-2)^n+\sum_{n\ge1}nc_n(x-2)^n+\sum_{n\ge1}c_n(x-2)^n=0

\displaystyle2c_2+c_0+\sum_{n\ge1}\bigg((n+2)(n+1)c_{n+2}+(n+1)c_n\bigg)(x-2)^n=0

\implies\begin{cases}c_0=2\\c_1=0\\(n+2)c_{n+2}+c_n=0&\text{for }n>0\end{cases}

  • If n=2k for integers k\ge0, then

k=0\implies n=0\implies c_0=c_0

k=1\implies n=2\implies c_2=-\dfrac{c_0}2=(-1)^1\dfrac{c_0}{2^1(1)}

k=2\implies n=4\implies c_4=-\dfrac{c_2}4=(-1)^2\dfrac{c_0}{2^2(2\cdot1)}

k=3\implies n=6\implies c_6=-\dfrac{c_4}6=(-1)^3\dfrac{c_0}{2^3(3\cdot2\cdot1)}

and so on, with

c_{2k}=(-1)^k\dfrac{c_0}{2^kk!}

  • If n=2k+1, we have c_{2k+1}=0 for all k\ge0 because c_1=0 causes every odd-indexed coefficient to vanish.

So we have

y(x)=\displaystyle\sum_{k\ge0}c_{2k}(x-2)^{2k}=\sum_{k\ge0}(-1)^k\frac{(x-2)^{2k}}{2^{k-1}k!}

Recall that

e^x=\displaystyle\sum_{n\ge0}\frac{x^k}{k!}

The solution we found can then be written as

y(x)=\displaystyle2\sum_{k\ge0}\frac1{k!}\left(-\frac{(x-2)^2}2\right)^k

\implies\boxed{y(x)=2e^{-(x-2)^2/2}}

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Step-by-step explanation:

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