5.2

If we wanted to solve \(y^{\prime \prime}-4 y^{\prime}+4 y=0\) then we could use the quick 3.4 method and guess \(y=e^{r t}\) and plug into equation to find \(r^2-4 r+4=0\). Thus \((r-2)^2=0\). Hence \(r=2\). Therefore general solution is \(y=c_1 e^{2 x}+c_2 x e^{2 x}\).

Alternatively, we could use LONG 5.2 method (normally use this method only when other shorter methods don't exist) to find solution for values near \(x_0=0\).

Suppose the solution \(y=f(x)\) is analytic at \(x_0=0\). That is \(f(x)=\Sigma_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x-0)^n\) for \(x\) near \(x_0=0\). Thus there are constants \(a_n=\frac{f^{(n)}(0)}{n!}\) such that,
\[ \notag
f(x)=\sum_{n=0}^{\infty} a_n(x-0)^n=\Sigma_{n=0}^{\infty} a_n x^n .
\]

Note

If wish to approximate solution near \(x_0=p\), then translate equation (let \(u=x-p\) ), solve, then translate back to avoid needing to multiply out \((x-p)^n\)

We then want to find a recursive formula for the constants of the series solution to \(y^{\prime \prime}-4 y^{\prime}+4 y=0\) near \(x_0=0\)

We will determine these constants \(a_n\) by plugging \(f\) into the ODE.
\[ \notag
\begin{array}{l}
f(x)=\sum_{n=0}^{\infty} a_n x^n, f^{\prime}(x)=\sum_{n=1}^{\infty} a_n n x^{n-1}, f^{\prime \prime}(x)=\sum_{n=2}^{\infty} a_n n(n-1) x^{n-2} . \\
\sum_{n=2}^{\infty} a_n n(n-1) x^{n-2}-4 \Sigma_{n=1}^{\infty} a_n n x^{n-1}+4 \Sigma_{n=0}^{\infty} a_n x^n=0 . \\
\sum_{n=0}^{\infty} a_{n+2}(n+2)(n+1) x^n-4 \Sigma_{n=0}^{\infty} a_{n+1}(n+1) x^n+4 \Sigma_{n=0}^{\infty} a_n x^n=0 . \\
\sum_{n=0}^{\infty}\left[a_{n+2}(n+2)(n+1)-4 a_{n+1}(n+1)+4 a_n\right] x^n=0 . \\
a_{n+2}(n+2)(n+1)-4 a_{n+1}(n+1)+4 a_n=0 . \\
a_{n+2}=\frac{4 a_{n+1}(n+1)-4 a_n}{(n+2)(n+1)} .
\end{array}
\]

Hence the recursive formula (if we know previous terms, can determine later terms) is
\[ \notag
a_{n+2}=4\left(\frac{(n+1) a_{n+1}-a_n}{(n+2)(n+1)}\right)
\]

Given the recursive formula, \(a_{n+2}=4\left(\frac{(n+1) a_{n+1}-a_n}{(n+2)(n+1)}\right)\), determine \(a_n\). Determine formula for \(a_k\) by noticing patterns. Note: It is easier to notice patterns if you do NOT simplify too much.
Find the first 6 terms of the series solution
\[ \notag
\begin{array}{ll}
n=0: & a_2=4\left(\frac{a_1-a_0}{(2)(1)}\right) \\
n=1: & a_3=4\left(\frac{2 a_2-a_1}{(3)(2)}\right)=4\left(\frac{(2)(4)\left(\frac{a_1-a_0}{(2)(1)}\right)-a_1}{(3)(2)}\right)=4\left(\frac{4\left(a_1-a_0\right)-a_1}{(3)(2)}\right)=4\left(\frac{3 a_1-4 a_0}{3!}\right)\\
n=2: & a_4=4\left(\frac{3 a_3-a_2}{(4)(3)}\right)=4\left(\frac{3(4)\left(\frac{3 a_1-4 a_0}{3!}\right)-4\left(\frac{a_1-a_0}{2!}\right)}{(4)(3)}\right)=4\left(\frac{3\left(\frac{3 a_1-4 a_0}{3!}\right)-\left(\frac{a_1-a_0}{2!}\right)}{3}\right)=4\left(\frac{\left(\frac{3 a_1-4 a_0}{2!}\right)-\left(\frac{a_1-a_0}{2!}\right)}{3}\right)=4\left(\frac{\left(3 a_1-4 a_0\right)-\left(a_1-a_0\right)}{3!}\right)=4\left(\frac{2 a_1-3 a_0}{(3!)}\right)\\
n=3: & a_5=4\left(\frac{(4) a_4-a_3}{(5)(4)}\right)=4\left(\frac{(4) 4\left(\frac{\left(2 a_1-3 a_0\right.}{3!}\right)-4\left(\frac{3 a_1-4 a_0}{3!}\right)}{(5)(4)}\right) =4\left(\frac{4\left(\frac{2 a_1-3 a_0}{3!}\right)-\left(\frac{3 a_1-4 a_0}{3!}\right)}{5}\right)=4\left(\frac{4\left(2 a_1-3 a_0\right)-\left(3 a_1-4 a_0\right)}{5(3!)}\right)=4\left(\frac{5 a_1-8 a_0}{5(3!)}\right)
\end{array}
\]

\[ \notag
f(x) \sim a_0+a_1 x+4\left(\frac{a_1-a_0}{2!}\right) x^2+4\left(\frac{3 a_1-4 a_0}{3!}\right) x^3+4\left(\frac{2 a_1-3 a_0}{(3!)}\right) x^4+4\left(\frac{5 a_1-8 a_0}{5(3!)}\right) x^5
\]

Recall \(f(x)=a_0 \phi_0(x)+a_1 \phi_1(x)\) for linearly independent solutions \(\phi_0\) and \(\phi_1\) to equation \(y^{\prime \prime}-4 y^{\prime}+4 y=0\)
Find the first 5 terms in each of the 2 solns \(y=\phi_0(x)\) and \(y=\phi_1(x)\)
\[ \notag
\begin{aligned}
\phi_0 & \sim 1+4\left(\frac{-1}{2!}\right) x^2+4\left(\frac{-4}{3!}\right) x^3+4\left(\frac{-3}{(3!)}\right) x^4+4\left(\frac{-8}{5(3!)}\right) x^5 \\
\phi_1 & \sim x+4\left(\frac{1}{2!}\right) x^2+4\left(\frac{3}{3!}\right) x^3+4\left(\frac{2}{(3!)}\right) x^4+4\left(\frac{5}{5(3!)}\right) x^5
\end{aligned}
\]

\[ \notag
\begin{array}{ll}
n=0: & a_2=4\left(\frac{a_1-a_0}{(2)(1)}\right)=2\left(\frac{2 a_1-2 a_0}{2!}\right) \\
n=1: & a_3==4\left(\frac{3 a_1-4 a_0}{3!}\right)=2^2\left(\frac{3 a_1-4 a_0}{3!}\right) \\
n=2: & a_4=4\left(\frac{2 a_1-3 a_0}{3!}\right)=16\left(\frac{2 a_1-3 a_0}{4!}\right)=8\left(\frac{4 a_1-6 a_0}{4!}\right)=2^3\left(\frac{4 a_1-6 a_0}{4!}\right) \\
n=3: & a_5=4\left(\frac{5 a_1-8 a_0}{5(3!)}\right)=16\left(\frac{5 a_1-8 a_0}{5!}\right)=2^4\left(\frac{5 a_1-8 a_0}{5!}\right)
\end{array}
\]

Hence it appears \(a_k=\frac{2^{k-1}\left(k a_1-2(k-1) a_0\right)}{k!}\)

Exercise \(\PageIndex{1}\)

Prove that if \(a_{n+2}=4\left(\frac{(n+1) a_{n+1}-a_n}{(n+2)(n+1)}\right)\), then \(a_k=\frac{2^{k-1}\left(k a_1-2(k-1) a_0\right)}{k!}\)

Answer

We need to prove \(a_k=\frac{2^{k-1}\left(k a_1-2(k-1) a_0\right)}{k!}\) for \(k \geq 0\) and we are given that \(a_{n+2}=4\left(\frac{(n+1) a_{n+1}-a_n}{(n+2)(n+1)}\right)\) for \(n \geq 2\). We proceed by induction on \(k\). First suppose \(k=0\), then \(\frac{2^{0-1}\left(0\left(a_1\right)-2(-1) a_0\right)}{0!}=\frac{1}{2}\left(2 a_0\right)=a_0.\) Next suppose \(k=1\), Thus \(\frac{2^{1-1}\left(1\left(a_1\right)-2(1-1) a_0\right)}{1!}=a_1.\) Next suppose \(a_k=\frac{2^{k-1}\left(k a_1-2(k-1) a_0\right)}{k!}\) for \(k=n, n+1. \)Thus \(a_n=\frac{2^{n-1}\left(n a_1-2(n-1) a_0\right)}{n!}\) and \(a_{n+1}=\frac{2^n\left((n+1) a_1-2 n a_0\right)}{(n+1)!}.\) We then claim the following:
\[ \notag a_{n+2}=\frac{2^{n+1}\left((n+2) a_1-2(n+1) a_0\right)}{(n+2)!}.\]

We see this is true as,

\[\notag
\begin{array}{l}
a_{n+2}=4\left(\frac{(n+1) a_{n+1}-a_n}{(n+2)(n+1)}\right)=4\left(\frac{(n+1)\left[\frac{2^n\left((n+1) a_1-2 n a_0\right)}{(n+1)!}\right]-\left[\frac{2^{n-1}\left(n a_1-2(n-1) a_0\right)}{n!}\right]}{(n+2)(n+1)}\right) \\
=4\left(\frac{\left[\frac{2^n\left((n+1) a_1-2 n a_0\right)}{n!}\right]-\left[\frac{2^{n-1}\left(n a_1-2(n-1) a_0\right)}{n!}\right]}{(n+2)(n+1)}\right) \\
=4(2)^{n-1}\left(\frac{\left[2\left((n+1) a_1-2 n a_0\right)\right]-\left[n a_1-2(n-1) a_0\right]}{n!(n+2)(n+1)}\right) \\
=2^{n+1}\left(\frac{2(n+1) a_1-4 n a_0-n a_1+2(n-1) a_0}{n!(n+2)(n+1)}\right)=2^{n+1}\left(\frac{\left.(n+2) a_1-2(n+1) a_0\right)}{(n+2)!}\right)
\end{array}\]

Thus \(f(x)=\Sigma_{n=0}^{\infty} \frac{2^{n-1}\left(n a_1-2(n-1) a_0\right)}{n!} x^n\)
\[ \notag
\begin{aligned}
& =a_1 \sum_{n=0}^{\infty} \frac{2^{n-1}(n)}{n!} x^n-2 a_0 \sum_{n=0}^{\infty} \frac{2^{n-1}(n-1)}{n!} x^n \\
= & a_0(-2) \sum_{n=0}^{\infty} \frac{2^{n-1}(n-1)}{n!} x^n+a_1 \sum_{n=1}^{\infty} \frac{2^{n-1}}{(n-1)!} x^n
\end{aligned}
\]

if these two series converge.

Exercise \(\PageIndex{2}\)

For what values of \(x\) does \(\sum_{n=0}^{\infty} \frac{(n-1) 2^{n-1}}{n!} x^n\) converge

Answer

Ratio test: Suppose we have the series \(\Sigma b_n\). Let \(L=\lim _{n \rightarrow \infty}\left|\frac{b_{n+1}}{b_n}\right|\)
Then, if \(L<1\), the series is absolutely convergent (and hence convergent).
If \(L>1\), the series is divergent.
If \(L=1\), the series may be divergent, conditionally convergent, or absolutely convergent.
\[ \notag
\lim _{n \rightarrow \infty}\left|\frac{\frac{n 2^n}{(n+1)!} x^{n+1}}{\frac{(n-1))^{n-1}}{n!} x^n}\right|=\lim _{n \rightarrow \infty}\left|\frac{2 n x}{(n+1)(n-1)}\right| =2 x \lim _{n \rightarrow \infty}\left|\frac{n}{(n+1)(n-1)}\right|=0
\]

Hence the series converges for all \(x\)
For what values of \(x\) does \(\sum_{n=1}^{\infty} \frac{2^{n-1}}{(n-1)!} x^n\) converge

Hence the series converges for all \(x\)

Exercise \(\PageIndex{3}\)

For what values of \(x\) does \(\Sigma_{n=1}^{\infty} \frac{2^{n-1}}{(n-1)!} x^n\) converge

Answer

Again using the ratio test, we have

\[ \notag
\lim _{n \rightarrow \infty}\left|\frac{\frac{2^n}{n!} x^{n+1}}{\frac{2 n-1}{(n-1)!} x^n}\right|=\lim _{n \rightarrow \infty}\left|\frac{2 x}{n}\right|=2 x \lim _{n \rightarrow \infty}\left|\frac{1}{n}\right|=0.
\]

Hence the series converges for all \(x\)

Using the two exercises above, we see the solution is
\[ \notag
f(x)=a_0(-2) \sum_{n=0}^{\infty} \frac{2^{n-1}(n-1)}{n!} x^n+a_1 \sum_{n=1}^{\infty} \frac{2^{n-1}}{(n-1)!} x^n
\]
and the domain is all real numbers.

I.e., the general solution is \(f(x)=a_0 \phi_0(x)+a_1 \phi_1(x)\)
where \(\phi_0(x)=(-2) \Sigma_{n=0}^{\infty} \frac{2^{n-1}(n-1)}{n!} x^n\) and \(\phi_1(x)=\Sigma_{n=1}^{\infty} \frac{2^{n-1}}{(n-1)!} x^n\)

Note

We could have replaced the constant \(a_0\) with \(-2 a_0\), but the \(a_i\) 's have meaning: \(a_n=\frac{f^{(n)}(0)}{n!}\). Thus our initial values are \(a_0=f(0)\) and \(a_1=f^{\prime}(0)\)

 

In general, to determine if there is a unique solution to the IVP, \(y^{\prime \prime}-4 y^{\prime}+\) \(4 y=0, y\left(x_0\right)=y_0, y^{\prime}\left(x_0\right)=y_1\), we solve for unknowns \(a_0\) and \(a_1\).
\[ \notag
\begin{array}{l}
y\left(x_0\right)=a_0 \phi_0\left(x_0\right)+a_1 \phi_1\left(x_0\right) \\
y^{\prime}\left(x_0\right)=a_0 \phi_0^{\prime}\left(x_0\right)+a_1 \phi_1^{\prime}\left(x_0\right)
\end{array}
\]

Note

The above system of two equations has a unique solution for the two unknowns \(a_0\) and \(a_1\) if and only if \(\operatorname{det}\left(\begin{array}{ll}\phi_0\left(x_0\right) & \phi_1\left(x_0\right) \\ \phi_0^{\prime}\left(x_0\right) & \phi_1^{\prime}\left(x_0\right)\end{array}\right) \neq 0\)

In other words the IVP has a unique solution if and only if the Wronskian of \(\phi_0\) and \(\phi_1\) evaluated at \(x_0\) is not zero. Recall that by theorem, this also implies that \(\phi_0\) and \(\phi_1\) are linearly independent and hence the general solution is \(y=a_0 \phi_0(x)+a_1 \phi_1(x)\) by theorem.

Exercise \(\PageIndex{4}\)

Show that \(\phi_0(x)=(-2) \sum_{n=0}^{\infty} \frac{2^{n-1}((n-1))}{n!} x^n\) and \(\phi_1(x)=\sum_{n=1}^{\infty} \frac{2^{n-1}}{(n-1)!} x^n\) are linearly independent by calculating the Wronskian of these two functions evaluated at \(x_0=0\).

Answer

\[\notag
\begin{array}{l}
W\left(\phi_1, \phi_2\right)(x)=\left(\begin{array}{ll}
\phi_1(x) & \phi_2(x) \\
\phi_1^{\prime}(x) & \phi_2^{\prime}(x)
\end{array}\right)=\left(\begin{array}{cc}
(-2) \sum_{n=0}^{\infty} \frac{2^{n-1}(n-1)}{n!} x^n & \sum_{n=1}^{\infty} \frac{2^{n-1}}{(n-1)!} x^n \\
(-2) \sum_{n=1}^{\infty} \frac{2^{n-1}(n-1)}{(n-1)!x^{n-1}} & \sum_{n=1}^{\infty} \frac{n 2^{n-1}}{(n-1)!} x^{n-1}
\end{array}\right) \\
W\left(\phi_1, \phi_2\right)(0)=\left(\begin{array}{cc}
(-2) 2^{0-1}(-1) & 0 \\
0 & 1
\end{array}\right)=\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)=1 \neq 0 \\
\end{array}
\]

Hence \(\phi_0(x)=(-2) \sum_{n=0}^{\infty} \frac{2^{n-1}((n-1))}{n!} x^n\) and \(\phi_1(x)=\Sigma_{n=1}^{\infty} \frac{2^{n-1}}{(n-1)!} x^n\) are linearly independent

 

When possible identify the functions giving the series solutions. Recall that by Taylor's theorem and the ratio test, \(e^{2 x}=\Sigma_{n=0}^{\infty} \frac{f^{(n)}(x)}{n!} x^n=\Sigma_{n=0}^{\infty} \frac{2^n}{n!} x^n\) for all \(x\).\[
\begin{array}{l}
f(x)=a_1 \sum_{n=0}^{\infty} \frac{n 2^{n-1}}{n!} x^n-2 a_0 \sum_{n=0}^{\infty} \frac{2^{n-1}(n-1)}{n!} x^n \\
=a_1 \sum_{n=0}^{\infty} \frac{n 2^{n-1}}{n!} x^n-2 a_0 \sum_{n=0}^{\infty} \frac{n 2^{n-1}}{n!} x^n+2 a_0 \sum_{n=0}^{\infty} \frac{2^{n-1}}{n!} x^n \\
=\left(a_1-2 a_0\right) \sum_{n=0}^{\infty} \frac{n 2^{n-1}}{n!} x^n+a_0 \Sigma_{n=0}^{\infty} \frac{2^n}{n!} x^n\\
=\left(a_1-2 a_0\right) x \sum_{n=1}^{\infty} \frac{2^{n-1}}{(n-1)!} x^{n-1}+a_0 \sum_{n=0}^{\infty} \frac{2^n}{n!} x^n \\
=\left(a_1-2 a_0\right) x \sum_{n=0}^{\infty} \frac{2^n}{n!} x^n+a_0 \sum_{n=0}^{\infty} \frac{2^n}{n!} x^n \\
=\left(a_1-2 a_0\right) x e^{2 x}+a_0 e^{2 x}
\end{array}
\]

Note

We have recovered the solution we found using the 3.4 method.

Note

A power series solutions exists in a neighborhood of \(x_0\) when the solution is analytic at \(x_0\). I.e, the solution is of the form \(y=\sum_{n=0}^{\infty} a_n\left(x-x_0\right)^n\) where this series has a nonzero radius of convergence about \(x_0\).