By Power Series Method, the solution of the differential equation is
#y=c_0sum_{n=0}^infty{(x^2/2)^n}/{n!}=c_0e^{x^2/2}#,
where #c_0# is any constant.
Let us look at some details.
Let #y=sum_{n=0}^\inftyc_nx^n#.
By taking the derivative term by term,
#y'=sum_{n=1}^infty nc_nx^{n-1}#
Now, let us look at the differential equation.
#y'=xy#
by substituting the above power series in the equation,
#Rightarrow sum_{n=1}^infty nc_nx^{n-1}=x cdot sum_{n=0}^\inftyc_nx^n#
by pulling the first term from the summation on the left,
#Rightarrow c_1+sum_{n=2}^inftync_nx^{n-1}=sum_{n=0}^infty c_nx^{n+1}#
by shifting the indices of the summation on the left by 2,
#Rightarrow c_1+sum_{n=0}^infty(n+2)c_{n+2}x^{n+1}=sum_{n=0}^infty c_nx^{n+1}#
By matching coefficients,
#c_1=0#
and
#(n+2)c_{n+2}=c_n Rightarrow c_{n+2}=c_n/{n+2}#
Let us observe the odd terms.
#c_3=c_1/3=0/3=0#
#c_5=c_3/5=0/5=0#
.
.
.
#c_{2n+1}=0#
Let us observe the even terms.
#c_2=c_0/2#
#c_4=c_2/4=c_0/{4cdot2}=c_0/{2^2cdot2!}#
#c_6=c_4/6=c_0/{6cdot4cdot2}=c_0/{2^3cdot3!}#
.
.
.
#c_{2n}=c_0/{2^ncdot n!}#
Hence,
#y=sum_{n=0}^infty c_0/{2^n cdot n!}x^{2n}=c_0 sum_{n=0}^infty{{x^{2n}}/{2^n}}/{n!}
=c_0sum_{n=0}^infty{(x^2/2)^n}/{n!}
=c_0e^{x^2/2}#
(by replacing #x# by #x^2/2# in #e^x=sum_{n=0}^infty{x^n}/{n!}#)
