Question

Consider the following differential equation. 6xy" + 7xy' + 13e+y=0 (a) Find all the regular singular points of the given differential equation. If necessary, enter your answers separated by commas. 2= QC (b) Determine the indicial equation and the exponents at the singularity of each regular singular point. Enter the values of ry and r2 in increasing order. Fr) QC T1 = 12 QC Click if you would like to Show Work for this question: Open Show Work

Answer 1

6xy''+7xy'+13e^xy=0

=>. y''+(7/6)y''+(13e^x/6x)y=0

So

$p(x)=7/6$

$q(x)=\frac{13e^x}{6x}$

q(x) becomes infinite when X tends to zero.

And xp(x) and x^2q(x) are finite so X=0 is regular singular point

So

X=0. First answer

(B)we compare the given differential equation to

y"+(p(x)/x )y'+(q(x)/x^2)y=0

then we have p(x)=7x/6

and $q(x)=\frac{13xe^x}{6}$

the indicial equation is

r(r-1)+p(0)r+q(0)=0

=>. r(r-1)+0+0=0

=>. r=0,1

So we get

F(r)= r(r-1)=0

$r_{1}=0$

$r_{2}=1$

answer