Question

what do you understand by ordinary paint of the equation

a₀(x)d²y/dx²+a₁(x)dy/dx+a₂(x)y=0? hence using taylor series expansion find a series solution in powers of x for the equation(2x³-3)d²y/dx²-2xdy/dx+y=0 y(0)=-1

y¹(0)=5

Answer 1

Ordinary point of equation -

$a_{0}(x)\dfrac{d^{2}y}{dx^{2}}+a_1(x)\dfrac{dy}{dx}+a_2(x)y=0$

A point $x_{0}$ is an ordinary point of an differential equation if both $p(x)\ and \ q(x)$ are analytical at $x=0$ .If point is not ordinary the it is a singular point . In other words , we can say that , a point

$x=x_{0}$ is called ordinary point of equation if $a_{o}(x)\ne0$ at $x=x_{0}$ or other than is is singular point.

2

$=(2x^{3}-3)\dfrac{d^{2}y}{dx^{2}}-2x\dfrac{dy}{dx}+y=0$

$y(0)=-1$ $y^{'}(0)=-5$

Now , we have to expand this series by taylor's series method , so we will expand it like this -

we know that general form of taylor's series , which is given by -

$=$ Let $y(x)$ be a solution of differential equation -

now taylor's series can be written as , we have to expand our series between point $x=0$ ,

$=y(x)=y(0)+y^{'}(0)(x-0)+\dfrac{{y^{''}(0)}(x-0)^{2}}{2}$ $+.........$

We can write our given differential equation in this form also-

$=y^{''}(x)=\dfrac{2xy^{'}{(x)}}{(2x^{3}-3)}$ $-\dfrac{y(x)}{(2x^{3}-3)}$

$=$ Now putting $x=0, we\ get-$

$=y^{''}(0)=$ $\dfrac{20y^{'}{(0)}}{(20^{3}-3)}-\dfrac{y(0)}{(20^{3}-3)}$

$=$ $y^{''}=\dfrac{-1}{3}$

now putting all the values in general form of taylor's series ,we get -

$=$ $y(x)=y(0)+y^{'}(0)(x-0)+\dfrac{{y^{''}(0)}(x-0)^{2}}{2}+......$

$=y(x)=-1+5x-\dfrac{x^{2}}{6}+......$

This is our general form of taylor.s series .