Question

Find a solution using the power series method

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

Answer 1

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)(n)/(x)+a_(n)t)x^n\\\\\rightarrow y'+t y=0\\\\\rightarrow a_(n-1)(n)/(x)+a_(n)t=0\\\\a_(n-1)=(-xta_(n))/(n)`

For, n=1

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

So,the series can be written as

`y_(n)=\sum_(n=1)^(\infty)(-1)^n(n!a_(0))/(x^nt^n)`

Answer 2

Answer: The required solution is

`y=(t-(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=-(a_0)/(2),\\\\\\3a_3+a_1=0\\\\\Rightarrow a_3=-(a_1)/(3),\\\\\\\vdots~~~\vdots~~~\vdots`

Therefore, we get

`y=a_0(1-(1)/(2)t^2+~~.~~.~~.)+a_1(t-(t^3)/(2)+~~.~~.~~.).`

The condition y(0) = 1 gives

`a_0=0.`

So, the required solution is

`y=t-(t^3)/(2)+~~.~~.~~.`