Question

Using Taylor series expression, find a power series solution for the equation

2x²d²y/dx²-xdy/dx+(x-5)y=0 in powers of (x-1) if y(1)=4 and y'(1)=2

Answer 1

Solution;

Taylors series expansion of f(x) at x=a is given as follows;

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

Given;x=1

We have;

$f(x)=f(1)+f'(1)(x-1)+\frac{f''(1)}{2!}(x-1)^2+\frac{f'''(1)}{3!}(x-1)+...$

Since;

y(1)=4

y'(1)=2

Also;

$2x^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}+(x-5)y=0$

By direct substitution;

$2(1^2)y''-1(2)+(1-5)(4)=0$

$2y''-2-16=0$

$y''(1)=9$

Now we find the triple derivative at x=1;

$2x^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}+(x-5)y=0$

Factorise to obtain;

$2x^2\frac{d^2y}{dx^2}-x\frac{dy}{dx}+xy-5y=0$

Differentiate by applying the product rule;

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

By direct substitution;

$2(1^2)y'''+(9)(4×1)-(1)(9)-2+(1)(2)+4=0$

Simplify to obtaining;

$y'''(1)=-\frac{31}{2}$

Hence we have the power solution series as;

$f(x)=4+2(x-1)+\frac9{2!}(x-1)^2-\frac{31}{2×3!}(x-1)^3+...$

Simplify;

$f(x)=4+2(x-1)+\frac92(x-1)^2-\frac{31}{12}(x-1)^3+...$