Q:

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

Accepted Solution

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