Question Consider the following IVP 4ty''+2y'+y=0, t > 0. a. Show that to=0 is a regular singular point of the ODE. b. Use the Frobenius method to construct the general solution of the ODE

Given IVP is4ty^('')+2y^(')+y=0,quad t > 0(a) To show: t_(0)=0 is a negular singular point.4ty^('')+2y^(')+y=0On, quady^('')+(1)/(2t)y^(')+(1)/(4t)y=0Compare with y^('')+p(t)y^(')+9(t)y=0 we get,P(t)=(1)/(2t) and q(t)=(1)/(4t).Now, lim_(t rarr0)(t-0)P(t)=lim_(t rarr0)(t)/(2t)=(1)/(2)and lim_(t rarr0)(t-0)^(2)q(t)=lim_(t rarr0)(t^(2))/(4t)=lim_(t rarr0)(t)/(4)=0lim_(t rarr0)(t-0)p(t) and lim_(t rarr0)(t-0)^(2)q(t) is exists.Thenefore t_(0)=0 is regular singuloos point:(b) Let y=sum_(n=0)^(oo)c_(n)t^(n) be a solution of the equation (i).y^(')=sum_(1)^(oo)nc_(n)t^((n-1))" and "y^('')=sum_(2)^(oo)n(n-1)c_(n)t^((n-2))substitues these, values in the equation (i) we. have,4tsum_(2)^(oo)n(n-1)c_(n)t^((n-2))+2sum_(1)^(oo)nc_(n)*t^((n-1))+sum_(0)^(oo)c_(n)t^(n)=0or, sum_(2)^(oo)4n(n-1)C_(n)t^((n-1))+sum_(1)^(oo)2nC_(n)*t^((n-1))+sum_(0)^(oo)c_(n)t^(n)=0Substitute quad n-y=m" For "quad{:[:.n=m+1],[" and "n=1","quad m=1]:}we have,{ ... See the full answer