Question

Using the Frobenius method, solve the following ODE:

x²y"+4xy'+(x²+2)y=0

Answer 1

Let the solution of above differential equation be

$$y=a_0x^m+a_1x^{m+1}+a_2x^{m+2}+...+a_nx^{m+n}$$

$$\dfrac{dy}{dx}=ma_0x^{m-1}+(m+1)a_1x^m$$

$$+(m+2)a_2x^{m+1}+...+(m+n)a_nx^{m+n-1}$$

$$\dfrac{d^2y}{dx^2}=m(m-1)a_0x^{m-2}+(m+1)ma_1x^{m-1}$$

$$+(m+2)(m+1)a_2x^{m}+...$$

$$+(m+n)(m+n-1)a_nx^{m+n-2}$$

$$x^2\dfrac{d^2y}{dx^2}+4x\dfrac{dy}{dx}+(x^2+2)y=0$$

$$x^2\big[m(m-1)a_0x^{m-2}+(m+1)ma_1x^{m-1}$$

$$+(m+2)(m+1)a_2x^{m}+...$$

$$+(m+n)(m+n-1)a_nx^{m+n-2}\big]$$

$$+4x\big[ma_0x^{m-1}+(m+1)a_1x^m$$

$$+(m+2)a_2x^{m+1}+...+(m+n)a_nx^{m+n-1}\big]$$

$$+(x^2+2)\big[a_0x^m+a_1x^{m+1}+a_2x^{m+2}+...$$

$$+a_nx^{m+n}\big]=0$$

$$(m^2+3m+2)a_0x^m+$$

$$+\displaystyle\sum_{n=1}^{\infin}((n+m)(n+m-1)+4(m+n)+2)a_nx^{m+n}$$

$$+\displaystyle\sum_{n=1}^{\infin}(n+m+1)a_{n-1}x^{m+n}=0$$

Equating the coefficient of lowest degree in $x$ (which is $x^m$ ) equal to zero

$$m^2+3m+2=0$$

$$m=-2,-1$$

$m=-2$

$$\displaystyle\sum_{n=1}^{\infin}((n-2)(n-3)+4(n-2)+2)a_n$$

$$+\displaystyle\sum_{n=1}^{\infin}(n-1)a_{n-1}x^{n-2}=0, n\geq1$$

$$a_n=-\dfrac{a_{n-1}}{n}, n\geq1$$

$$a_1=-a_0$$

$$a_2=-\dfrac{a_{1}}{2}=\dfrac{a_{0}}{2}$$

$$a_3=-\dfrac{a_{2}}{3}=-\dfrac{a_{0}}{6}$$

$$y_1(x)=a_0x^{-2}(1-x+\dfrac{x^2}{2}-\dfrac{x^3}{6}+...)$$

$m=-1$

$$\displaystyle\sum_{n=1}^{\infin}((n-1)(n-2)+4(n-1)+2)a_n$$

$$+\displaystyle\sum_{n=1}^{\infin}na_{n-1}=0, n\geq1$$

$$a_n=-\dfrac{a_{n-1}}{n+1}, n\geq1$$

$$a_1=-\dfrac{a_{0}}{2}$$

$$a_2=-\dfrac{a_{1}}{3}=\dfrac{a_{0}}{6}$$

$$a_3=-\dfrac{a_{2}}{4}=-\dfrac{a_{0}}{24}$$

$$y_2(x)=a_0x^{-1}(1-\dfrac{x}{2}+\dfrac{x^2}{6}-\dfrac{x^3}{24}+...)$$

$$y=C_3y_1(x)+C_4y_2(x)$$

$$=C_1x^{-2}(1-x+\dfrac{x^2}{2}-\dfrac{x^3}{6}+...)$$

$$+C_2x^{-1}(1-\dfrac{x}{2}+\dfrac{x^2}{6}-\dfrac{x^3}{24}+...)$$