find a solution using the power series methody'+ty=0, y(0)=1

Answer:  The required solution is[tex]y=(t-\dfrac{t^3}{2}+~~.~~.~~.).[/tex]Step-by-step explanation:  We are given to find the solution of the following differential equation using power series method :[tex]y^\prime+ty=0,~~y(0)=1~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]Let [tex]y=\sum_{n=0}^{\infty}a_nq^n[/tex] be the solution of the given equation.Then, we have[tex]y^\prime=\sum_{n=1}^{\infty}a_nnt^{n-1}.[/tex]From equation (i), we get[tex]\sum_{n=1}^{\infty}a_nnt^{n-1}+t\sum_{n=0}^{\infty}a_nt^n=0\\\\\\\Rightarrow \sum_{n=1}^{\infty}a_nnt^{n-1}+t\sum_{n=-1}^{\infty}a_nt^{n+1}=0.[/tex]Comparing the coefficients of [tex]t^n,~t^{n+1},~,~~.~~.~~.[/tex] from both sides of the above, we get [tex]2a_2+a_0\\\\\Rightarrow a_2=-\dfrac{a_0}{2},\\\\\\3a_3+a_1=0\\\\\Rightarrow a_3=-\dfrac{a_1}{3},\\\\\\\vdots~~~\vdots~~~\vdots[/tex]Therefore, we get[tex]y=a_0(1-\dfrac{1}{2}t^2+~~.~~.~~.)+a_1(t-\dfrac{t^3}{2}+~~.~~.~~.).[/tex]The condition y(0) = 1 gives[tex]a_0=0.[/tex]So, the required solution is[tex]y=t-\dfrac{t^3}{2}+~~.~~.~~.[/tex]