Answer to Question #129210 in Differential Equations for khyzer montebon

Question

x²y"-(2x+1)y'+y²=0

Answer 1

$$Solution$$

Write the differential equation in standard form: $y''-\frac{2x+1}{x^2}y'+\frac{y^2}{x^2}=0$

We have $P(x)=-\frac{2x+1}{x^2}$ and $Q(x)=\frac{1}{x^2}$ , so $x= 0$ is a singular point. Taking limits we get $lim_{x→0}xP(x) = 1$ and $lim_{x→0}x^2Q(x) = 0$ . Therefore $x= 0$ is a regular singular point. We look for a solution of the form $y(x) =∑^∞_{n=0}a_nx^{n+r}$ . Plug the series into the equation to get

$y(x) =∑^∞_{n=0}a_nx^{n+r}\\ y' =∑^∞_{n=0}(n+r)a_nx^{n+r-1}\\ y'' =∑^∞_{n=0}(n+r)(n+r-1)a_nx^{n+r-2}$

$y'' =x^2∑^∞_{n=0}(n+r)(n+r-1)a_nx^{n+r-2}-2x∑^∞_{n=0}(n+r)a_nx^{n+r-1}-∑^∞_{n=0}(n+r)a_nx^{n+r-1}+(∑^∞_{n=0}a_nx^{n+r})^2=0$

$\implies y'' =∑^∞_{n=0}(n+r)(n+r-1)a_nx^{n+r}-∑^∞_{n=0}2(n+r)a_nx^{n+r}-∑^∞_{n=0}(n+r)a_nx^{n+r-1}+(∑^∞_{n=0}a_nx^{n+r})^2=0$

Upon re-arrangement we get

$(r(r-1)-2r)a_0x^r+\sum_{n=0}^\infin \{[(n+r+1)(n+r+3)]q_{n+1}-(n+r-1)a_n\}x^{n+r}$

Equating the powers of x yields

$$(r^2-3r)a_0=0,\ \ a_{n+1}=\frac{a_n}{n+r+3}, for\ n=0,\ 1,\ ....$$

But $a_0 \ne 0$ implies that $r= 0,3$ . Solving for the $a_n’s$ we get

For r=0

$a_{n+1}=\frac{a_n}{n+3}\\ a_1=\frac{a_0}{3}\\ a_2=\frac{a_1}{4}=\frac{a_0}{4 \cdot 3}=\frac{a_0 \cdot 2 \cdot 1}{4 \cdot 3 \cdot 2 \cdot 1}=\frac{2!}{4!}a_0 \\ a_3=\frac{a_2}{5}=\frac{a_0}{5 \cdot 4 \cdot 3}=\frac{a_0 \cdot 2 \cdot 1}{5 \cdot4 \cdot 3 \cdot 2 \cdot 1}=\frac{2!}{5!}a_0 \\ .\\ .\\ .\\ a_n=\frac{2!}{(n+3)!}a_0$

For r=3

$a_{n+1}=\frac{a_n}{n+6}\\ a_1=\frac{a_0}{6}\\ a_2=\frac{a_1}{7}=\frac{a_0}{7 \cdot 6}=\frac{a_0}{7 \cdot 6!}\\ a_3=\frac{a_2}{8}=\frac{a_0}{8 \cdot 7 \cdot 6}=\frac{a_0}{8 \cdot7 \cdot 6!} \\ .\\ .\\ .\\ a_n=\frac{a_0}{(n+6)!}$

Two linearly independent Frobenius series (with $a_0= 1$ ) are given by

$$y_1(x)=\sum_{n=0}^\infin a_nx^n=\sum_{n=0}^\infin \frac{2!}{(n+3)!}x_n,\ y_2(x)=\sum_{n=0}^\infin a_nx^{n+3}=x^3\sum_{n=0}^\infin \frac{x_n}{(n+6)!}$$