Question

Using taylor series expansion obtain a power series solution to the initial value problem

(2x²-3)d²y/dx²-2xdy/dx+y=0 y(0)=1 y'(0)=7

Answer 1

Solution;

The Taylor's series expansion of f(x) about x=a is;

$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...$

The series solution about x=0 can be written as;

$y(x)=y(0)+y'(0)x+\frac{y''(0)}{2!}x^2+\frac{y'''}{3!}x^3+...$

Given y(0)=1 and y'(0)=0;

Also;

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

At x=0,

$(0-3)\frac{d^2y}{dx^2}-(0)(7)+1=0$

$-3\frac{d^2y}{dx^2}=-1$

$y''=\frac13$

By distribution,we write the equation as,

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

Differentiate by applying product rule;

$[2x^2\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}(4x)]-3\frac{d^3y}{dx^3}-[2x\frac{d^2y}{dx^2}+2\frac{dy}{dx}]+\frac{dy}{dx}=0$

At x=0;

$(0+0)-3\frac{d^3y}{dx^3}-(0+(2×7))+7=0$

$-3\frac{d^3y}{dx^3}=-21$

y'''=7

Hence ,the power series solution give by ;

$y(x)=y(0)+y'(0)x+\frac{y"(0)}{2!}x^2+\frac{y'''(0)}{3!}x^3+...$

By substitution ,will be;

$y(x)=1+7x+\frac16x^2+\frac76x^3+...$