Find a general solution using power swries: ((x^2)+1)y'' + 6xy' + 6y =0

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cosinusix | College Teacher | (Level 3) Assistant Educator

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If `y= sum_(n=0)^(oo)a_nx^n` is solution to the equation `(x^2+1)y'' + 6xy' + 6y =0`












If we plug it in the differential equation,

`sum_(n=2)^(oo)n(n-1)a_nx^(n-2)+sum_(n=2)^(oo)n(n-1)a_nx^n+6sum_(n=1)^(oo)na_nx^(n)+ 6sum_(n=0)^(oo)a_nx^n=0`

In the first sum, make the substitution `n-2->n`

The differential equation becomes

`sum_(n=0)^(oo)(n+2)(n+1)a_(n+2)x^(n)+sum_(n=2)^(oo)n(n-1)a_nx^n+6sum_(n=1)^(oo)na_nx^(n)+ 6sum_(n=0)^(oo)a_nx^n=0`


let's combine everything in one sum. Some sums start for n=0, for n=1, for n=2 therefore, iI will take care of n=0 and n=1 separately.



If a power serie is identically 0, il means that all its coefficients are 0.

Therefore `2a_2+6a_0=0 <=> a_2=-3a_0/2`

`6a_3+12a_1=0 <=>a_3=-2a_1`

`(n+2)(n+1)a_(n+2)+(n^2+5n+6)a_n=0 lt=gt a_(n+2)=(n^2+5n+6)/((n+2)(n+1)) a_n`

Remark that `(n^2+5n+6)=(n+2)(n+3)`

<=> `a_(n+2)=-(n+3)/(n+1) a_n`

Remark that for `n=0` , the previous relation gives `a_2=-3a_0 ` which is the first relation we found.


For `n=1` , `a_3=-4/2a_1=-2a_1` wchich is a relation we found too. Therefore the  3 conditions on the coefficients we  found, can all be  summarize in one single relation



A general solution to the differential equation is

`y=sum_(n=0)^(oo)a_nx^n` where `a_(n+2)=-(n+3)/(n+1) a_n` for some `(a_0 ,a_1) in RR^2`