Question

find a solution using the power series method

y'+ty=0, y(0)=1

Answer 1

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

$y=t-\dfrac{t^3}{2}+~~.~~.~~.$

Answer 2

Answer with explanation:

The diiferential equation of first order is,

  y'+t y=0

$y=\sum_{n=0}^{\infty}a_{n}x^n\\\\y'=\sum_{n=1}^{\infty}na_{n}x^{n-1}\\\\y'+ty=\sum_{n=0}^{\infty}na_{n-1}x^{n-1}+t\sum_{n=0}^{\infty}a_{n}x^n\\\\=\sum_{n=0}^{\infty}(a_{n-1}\frac{n}{x}+a_{n}t)x^n\\\\\rightarrow y'+t y=0\\\\\rightarrow a_{n-1}\frac{n}{x}+a_{n}t=0\\\\a_{n-1}=\frac{-xta_{n}}{n}$

For, n=1

$a_{0}=\frac{-xta_{1}}{1}\\\\a_{0}=-txa_{1}\\\\a_{1}=\frac{-a_{0}}{tx}\\\\\text{for}, n=2\\\\a_{1}=\frac{-xta_{2}}{2}\\\\a_{2}=\frac{-2a_{1}}{xt}\\\\=\frac{2a_{0}}{x^2t^2}\\\\\text{for}, n=3\\\\a_{2}=\frac{-xta_{3}}{3}\\\\a_{3}=\frac{-6a_{0}}{x^3t^3}\\\\a_{4}=\frac{24a_{0}}{x^4t^4}\\\\a_{n}=(-1)^n\frac{n!a_{0}}{x^nt^n}$

So,the series can be written as

    $y_{n}=\sum_{n=1}^{\infty}(-1)^n\frac{n!a_{0}}{x^nt^n}$