Answer to Question #224756 in Differential Equations for Handsome

Question

Find the Taylor series solution of following

(X²+1)y^''+4xy^'+(x-5)y=0 y(2)=-1

y^'(2)=3

Answer 1

Solution;

By definition, Taylor's series expansion of f(x) at x=a is given by;

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

We are given X=2,hence;

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

Since we have;

y(2)=-1

y'(2)=3

Also;

(x²+1)y''+4xy'+(x-5)y=0

By substitution;

(2²+1)y''+4(2)(3)+(2-5)(-1)=0

5y''=-27

y''(2)=$\frac{-27}{5}$

Now ,find the triple derivative at x=2;

(x²+1)y''+4xy'+(x-5)y=0

Distribute;

$x^2\frac{d^2y}{dx}+\frac{d^2y}{dx^2}+4x\frac{dy}{dx}+xy-5y=0$

Differentiate;

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

Rewrite as;

$[x^2y'''+2xy'']+y'''+[4xy''+4y']+[xy'+y]=0$

Substitute all known values;

$4y'''-\frac{27}{5}(4)+y'''-\frac{27}{5}(8)+4(3)+2(3)-1=0$

Simplify;

$5y'''-\frac{324}{5}+17=0$

Equate to obtain;

y'''(2)=$\frac{239}{25}$

Hence;

$f(x)=-1+3(x-2)-\frac{27}{10}(x-2)^2+\frac{239}{150}(x-2)^3+...$

Distribute;$f(x)=-1+3(x-2)-\frac{27}{10}(x^2-4x+4)+\frac{239}{150}(x^3-6x^2+12x-8)+...$

Simplify;

$f(x)=-\frac{2291}{75}+\frac{823}{25}x-\frac{613}{50}x^2+\frac{239}{150}x^3+...$