Answer to Question #218948 in Differential Equations for john

Question

Find a series solution in powers of x of the equation

2x²d²y/dx²+xdy/dx+(x²-1)y=0

Answer 1

$\text{We choose $x_0=0$. Since $2x^2=0$, we know that $x_0$ is a singular point. Next}\\\text{we check if the point is regular by computing }\\\lim_{x \to x_0}a(x)(x-x_0) \text{ and } \lim_{x \to x_0}b(x)(x-x_0)^2\\\text{where a(x) = $\frac{1}{2x}$ and b(x) = $\frac{x^2-1}{2x^2}$}\\\text{$\lim_{x \to 0}\frac{1}{2x}(x)=\frac{1}{2}$ and $\lim_{x \to 0}x^2(\frac{x^2-1}{2x^2})=\frac{-1}{2}$}\\\text{Since both limits are finite, we say $x_0$ is a regular singular point.}\\\text{Using Frobenius method, let y = $\sum^{\infty}_{n=0}a_nx^{n+r}$}\\\text{therefore $y'=\sum^{\infty}_{n=0}(n+r)a_nx^{n+r-1}$ and $y''=\sum^{\infty}_{n=0}(n+r)(n+r-1)a_nx^{n+r-2}$}\\\text{Next, we input the above values in the original differential equation} \\\text{$2x^2\sum^{\infty}_{n=0}(n+r)(n+r-1)a_nx^{n+r-2}$ + $x\sum^{\infty}_{n=0}(n+r)a_nx^{n+r-1}$}\\+x^2\sum^{\infty}_{n=0}a_nx^{n+r}-\sum^{\infty}_{n=0}a_nx^{n+r} \\=\sum^{\infty}_{n=0}(a_n(2(n+r)(n+r-1)+(n+r)-1)+a_{n-2})x^{n+r}-(1)\\\text{Let n = 0, therefore we have the indicial equation}\\2r(r-1)+r-1)=0\\\implies r=1 \text{ and } r=-\frac{1}{2} .\\\text{Also, $a_1(2(1+r)r+(1+r)-1)=0\\$}\\\implies a_1(r(2r+3))=0\text{, which gives the recurrence relation}\\a_n=\frac{-a_n-2}{(n+r-1)(2(n+r)+1)}, n =2,3,4,5,...\\\text{using the equation above we have that, when n=1, $a_1=0$}\\\text{when n=2, $a_2=\frac{-a_0}{14}$}\\\text{when n = 4, $a_4=\frac{-a_0}{616}$}\\\text{Since $a_1=0$, we have that, }a_1=a_3=a_5=...=0\\\text{Recall that $y_1=\sum^{\infty}_{n=0}a_nx^{n+r}$}\\\text{Therefore, we have that}\\y_1=x-\frac{x^3}{14}+\frac{x^5}{616}+... \\\text{Repeating the process for r = $\frac{-1}{2}$, we have that}\\y_2=x^{\frac{-1}{2}}(1-\frac{x^2}{2}+\frac{x^4}{40}+...)$