Determine the power series solution to the equation d2y/dx2+6xdy/dx-4y=0
about the point x0=0
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}+...)"
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