Question

Even though x 5 0 is an ordinary point of the differential equation, explain why it is not a good idea to try to find a solution of the

$y"+xy\text{'}+y=0,y\left(1\right)=-6,y\text{'}\left(1\right)=3$

of the form $y=\sum _{n=0}^x{C}_n{X}^n$.Using power series, find a better way to solve the problem.

Short answer

$y=-6\left(1-\frac{1}{2}{(x-1)}^2+\frac{1}{6}{(x-1)}^3+\frac{1}{12}{(x-1)}^4+...\right)+3\left((x-1)-\frac{1}{2}{(x-1)}^2-\frac{1}{6}{(x-1)}^3+\frac{1}{6}{(x-1)}^4+...\right)$

Step-by-step solution

Definition of Power series solutions

If $X=X_0$is an ordinary point of the differential equation , we can always find twolinearly independent solutions in the form of a power series centered at X₀, that is,

$y=\sum _{n=0}^x{C}_n{(x-x_0)}^{n.}$

A power series solution converges at least on some interval defined by$\left|X-X_0\right|<R$, where R is the distance from X₀to the closest singular point.

Using initial value problems .

We were given the next equation

$y\text{'}+xy+y=0,y\left(1\right)=-6,y\left(1\right)=3$

X= 0 is an ordinary point of the DE (1), but it is not good idea to try to find a solution of the given IVP of the form localid="1665050500747" width="99">y=∑n=0∞CnXn . That is because the initial conditions are given at the point X = 1 and because of that, we would get the constants in the terms of infinite series.

That is why we will use the change of variables X= t + 1 . The given IVP is then equivalent to

$\frac{d^{2}y}{d{t}^2}+(t+1)\frac{dy}{dt}+y=0,y\left(0\right)=-6,y\text{'}\left(0\right)=3$

By Theorem 6.2.1, we can find two linearly independent solutions in the next form

$y=\sum _{n-0}^{\infty }{C}_n{t}^n$

The derivations of (2) are

$y=\sum _{n-1}^{\infty }{C}_{n}n{t}^{n-1},y=\sum _{n=2}^{\infty }{C}_{n}n(n-1)t^{n-2}$

Substitute the initial conditions

Substitute this into (1') to obtain.

$0=\sum _{n=2}^{\infty }{C}_{n}n(n-1)t^{n-2}+\sum _{n=1}^{\infty }{C}_{n}n{t}^n+\sum _{n=1}^{\infty }{C}_{n}n{t}^{n-1}+\sum _{n=0}^{\infty }{C}_n{t}^n$

We can rewrite (4) changing index of the first sum in k = n - 2 and index of the third sum in k = n - 1:

$$\begin{gathered} 0=\sum _{k=0}^{\infty }{C}_{k+2}(k+2)(k+1)t^k+\sum _{k=1}^{\infty }{C}_{k}k{t}^k+\sum _{k=0}^{\infty }{C}_{k+1}(k+1)t^k+\sum _{k=0}^{\infty }{C}_k{I}^k \\ =2c_2+c_1+c_0+\sum _{k=1}^{\infty }\left[\left(k+2\right)(k+1)C_{k+2}+(k+1)C_{k+1}+(k+1)C_k\right]t^k \end{gathered}$$

From this, we can see that

$=2c_2+c_1+c_0=0\Rightarrow c_2=-\frac{c_0+c_1}{2}$

and

$C_{k+2}=-\frac{C_{k=1}+C_k}{k+2},k\in N$

Let C₀ = 0 and C₁= 1 . From this, we find

$C_2=-\frac{1}{2},C_3=-\frac{1}{6},C_4=\frac{1}{6},.....$

Let C₀ = 1 and C₁=0. From this, we find

$C_2=-\frac{1}{2},C_3=\frac{1}{6},C_4=\frac{1}{12},.....$

Therefore, the general solution is

$y=C_0\left(1-\frac{1}{2}{t}^2+\frac{1}{6}{t}^3+\frac{1}{12}{t}^4+,.....\right)+C_1\left(-1-\frac{1}{2}{t}^2+\frac{2}{3}{t}^3+....\right)$

Its derivation is

$y=C_0\left(-t+\frac{1}{2}{t}^2+\frac{1}{3}{t}^3+,..\right)+C_1\left(-1-\frac{1}{2}{t}^2+\frac{2}{3}{t}^3+....\right)$

The initial conditions y (0) = -6 and y(0) = 3 give us C₀ = -6 and C₁ = 3.'Therefore, the solution of the IVP (1') is

$y=-6\left(1-\frac{1}{2}{(x-1)}^2+\frac{1}{6}{(x-1)}^3+\frac{1}{12}{(x-1)}^4+,..\right)+3\left((x-1)-\frac{1}{2}{(x-2)}^2-\frac{1}{6}{(x-1)}^3+\frac{1}{6}{(x-1)}^4+....\right)$

Thus, the solution of the given IVP is