Question

Determine the power series solution to the equation d²y/dx²+6xdy/dx-4y=0

about the point x₀=0

Answer 1

Given equation

$\dfrac{d^{2}y}{dx^{2}}+6x\dfrac{dy}{dx}-4y=0$ about the point $x_{0}=0$

Let the solution of given series will be

$y=a_0x^{m}+a_1x^{m+1}+a_2x^{m+2}+a_3x^{m+3}+....$

$\dfrac{dy}{dx}=ma_0x^{m-1}+a_1(m+2)x^{m+1}+(m+3)a_2x^{m+2}+.....$

$\dfrac{d^{2}y}{dx^{2}}=m(m-1)a_0x^{m-2}+a_1(m+2)(m+1)x^{m}+(m+3)a_2x^{m+1}+....$

Now putting , in the above equation we get -

$m(m-1)a_0x^{m-2}+a_1(m+2)(m+1)x^{m}+(m+3)a_3x^{m+1}+....6x[( ma_0x^{m-1}+a_1(m+2)x^{m+1}+(m+3)a_2x^{m+2}+.....)]-4[(a_0x^{m}+a_1x^{m+1}+a_2x^{m+2}+a_3x^{m+3}+....)]=0$

$a_0x^{m-2}[m(m-1)]+x^{m}[a_1(m+2)(m+1)+6ma_0-4a_0]+x^{m+1}[a_3(m+3)-4a_1]+x^{m+2}[6a_1(m+2)-4a_2]+x^{m+3}[6(m+3)a_2-4a_3]=0$

On equating the coefficient the lower power of $x,$ we get

The lowest power of $x$ is $x^{m-2}$ $,$

On equation the lowest power of $x$ which is $x^{m-2}$ on both side of equation we get ,

$a_0m(m-1)=0$ , which gives

$a_0=0$ $m=0,1$

The solution of indicial equation , is given as

$\therefore y=c_1y_1+c_2y_2$

Now , on equating the coefficient of $x^{m}$ , we get

$=$ $[a_1(m+2)(m+1)+6ma_0-4a_0]$ $=0$

$=$ $[a_1(m+2)(m+1)+a_0(6m-4)]=0$

$a_1=-\dfrac{a_0(6m-4)}{(m+2)(m+1)}$

Now , on equation the coefficient of $x^{m+1},$ we get

$a_3(m+3)-4a_1=0$

$a_3=\dfrac{4a_1}{m+3}$

$a_3=-\dfrac{4}{(m+3)}\dfrac{a_0(6m-4)}{(m+2)(m+1)}$

Now , on equating the coefficient of $x^{m+2},$ we get

$6a_1(m+2)-4a_2=0$

$a_2=\dfrac{6a_1(m+2)}{4}$

$a_2=\dfrac{6(m+2)}{4}\dfrac{a_0(6m-4)}{(m+2)(m+1)}$

Thus , for $m=0$ , we get

$y_{(m=0)}=y_1=[x^{m}(a_0+a_1x+a_2x^{2}+.....)]_{m=0}$

$=a_0-2a_0x-6a_0x^{2}+....$

$y_{m=0}=a_0(1-2x-6x^{2}+...)$

$y_{m=1}=y_2=[x^{m}(a_0+a_1x+a_2x^{2}+...)$

$y_2=x(a_0-\dfrac{a_0}{3}x+\dfrac{9}{2}\dfrac{a_0}{3}x^{2}+....)$

$y_2=a_0x(1-\dfrac{a_0}{3}x+\dfrac{3}{2}a_0x^{2}+...)$

The solution of given , equation is

$y=c_1a_0(1-2x-6x^{2})+c_2a_0x((1-\dfrac{a_0}{3}x+\dfrac{3}{2}a_0x^{2}+...)$