Power series solution (Why the constant of the recurrence relation can be chosen arbitrarily?)

Question

Please help me understand this:

Solve $(x+1)y''-(2-x)y'+y =0$

First, since $x_0=0$ is an ordinary point, it can be guaranteed that we can find two independent power series solutions centered at $x_0=0$. We substitute $y=\sum_{n=0}^{\infty} c_nx^n$ into the equation and we will have:

$$ (x+1)y''-(2-x)y'+y =0 $$ $$ \sum_{n=2}^{\infty} [n(n-1)c_nx^{n-1}+n(n-1)c_{n}x^{n-2}]-2\sum_{n=1}^{\infty} nc_nx^{n-1}+\sum_{n=1}^{\infty} nc_{n}x^n +\sum_{n=0}^{\infty} c_{n}x^n =0 $$ $$ \sum_{k=1}^{\infty} (k+1)kc_{k+1}x^{k} + \sum_{k=0}^{\infty} (k+1)(k+2)kc_{k+2}x^{k} - 2\sum_{k=0}^{\infty} (k+1)c_{k+1}x^{k} + \sum_{k=1}^{\infty} kc_{k}x^{k} + \sum_{k=0}^{\infty} c_{k}x^{k}$$ $$ (2c_{2}-2c_{1}+c_{0}) + \sum_{k=1}^{\infty} [(k+2)(k+1)c_{k+2} + (k-2)(k+1)c_{k+1} + (k+1)c_{k}]x^k = 0 $$

Thus $$ 2c_{2}-2c_{1}+c_{0}=0 \tag{1}$$ $$ (k+2)(k+1)c_{k+2}+(k+1)(k-2)c_{k+1}+(k+1)c_{k} = 0 \tag{2}$$ and the last equation gives $$ c_{k+2} = - \frac{(k-2)c_{k+1}+c_{k}}{k+2}, \quad \textrm{k=1,2,3...} \tag{3}$$

Now, the part that I don't understand is the part that follows from this. Most textbooks usually proceed from equation$(3)$ by choosing $c_0=1$ and $c_1=0$, without telling any reason, which will give $$c_{2}=-\frac{1}{2}, c_{3}=-\frac{1}{6}, c_{4}=\frac{1}{12},...$$ thus the first solution is $y_{1}=1-\frac{1}{2}x^2 - \frac{1}{6}x^3 + \frac{1}{12}x^4+...$

Then they choose $c_{0}=0$ and $c_{1}=1$ and they will obtain $$c_{2}=1, c_{3}=0, c_{4}=\frac{1}{4},...$$ thus the second solution is $y_{2}=x+x^2-\frac{1}{4}x^4+...$

And the general solution is $$y=C_{1}(1-\frac{1}{2}x^2 - \frac{1}{6}x^3 + \frac{1}{12}x^4+...)+C_{2}(x+x^2-\frac{1}{4}x^4+...) \tag{4}$$

My question is why can we choose $c_0$ and $c_1$ arbitrarily like this? Does this work in a higher or more term of recurrence relation? If I modify $c_0$ and $c_1$, wouldn't it change the subsequent coefficients $c_{k}$? And therefore we will have the solution which is different? (In other words, choosing $c_0=0$ and $c_1=1$ will give one solution, but choosing $c_0=0$ and $c_1=123$ will give another different solution?)

Any help is greatly appreciated.

Answer 1

The general theory about second order homogeneous differential equations says that (i) there is an open interval $I$ on the $x$-axis containing containing $0$ in its interior such that the given ODE has a two-dimensional vector space ${\cal L}$ of solutions $y(\cdot):\>I\to{\mathbb R}$, and (ii) these solutions are even real analytic, i.e. have convergent power series representations in $I$.

It follows that if you have, by whatever means, found two linearly independent solutions $y_1(\cdot)$ and $y_2(\cdot)$ in the form of power series then the general solution is given by $y(x)=C_1y_1(x)+C_2y_2(x)$ with real constants $C_1$, $C_2$.

By choosing the initial conditions for $y_1(\cdot)$ and $y_2(\cdot)$ in the way you have done you can be sure that these solutions are linearly independent, because each of them has a property that cannot be produced by a linear combination of the others (resp., by a multiple of the other): You have $y_2'(0)=1$, whereas $y_1'(0)=0$.

Answer 2

It's a second order differential equation, so there are $2$ arbitrary constants in the general solution. As you could see, there was a constraint on $c_2$, but there is none on $c_0$ and $c_1$, and there is no mutual constraint between the latter two constants. It seems most convenient to separate things out into $2$ series, one where all the exponents are divisible by $3$, and the other where the exponents are $\equiv1\pmod3$.

Answer 3

We can see that each solution is completed characterized by two numbers $c_0$ and $c_1$. Also, the terms that are influeced by $c_0$ do not overlap with the terms that are influenced by the terms influenced by $c_1$.

In particular observe the relationship:

$$y(c_0,c_1)=c_0y(1,0)+c_1y(0,1)$$

Answer 4

As a complement to the previous answers, observe that you can give initial conditions to select a unique solutions. The choice of initial data corresponds with the choice of the constanta $c_0$ and $c_1$:

1. $c_0=1$ and $c_1=0$ corresponds to $y(0)=1$ and $y'(0)=0$.
2. $c_0=0$ and $c_1=1$ corresponds to $y(0)=0$ and $y'(0)=1$.