Answer to Question #224754 in Differential Equations for Lexam

Question

Use the Taylor series expansion to find the general solution of each of the following

2x²y''-3xy'+(x+1)y=0

y(1)=4 y^'(1)=1

Answer 1

The Taylor series expansion of the function $f(x)$ has the form: $f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+...$

In our case the point is $a=1$. We receive: $f(x)=f(1)+f'(1)(x-1)+\frac{f''(1)}{2!}(x-1)^2+\frac{f'''(1)}{3!}(x-1)^3+...$

We take derivatives and get:

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

$f''(x)=f''(1)+f'''(1)(x-1)+...$

The initial conditions take the form: $f(1)=4$, $f'(1)=1$. We substitute the expansion and receive:

$2x^2(f''(1)+f'''(1)(x-1)+...)-3x(f'(1)+f''(1)(x-1)+\frac{f'''(1)}{2!}(x-1)^2+...)+(x+1)(f(1)+f'(1)(x-1)+\frac{f''(1)}{2!}(x-1)^2+\frac{f'''(1)}{3!}(x-1)^3+...)=0$

We denote: $z=x-1$ and receive: $2(z+1)^2(f''(1)+f'''(1)z+...)-3(z+1)(f'(1)+f''(1)z+\frac{f'''(1)}{2!}z^2+...)+(z+2)(f(1)+f'(1)z+\frac{f''(1)}{2!}z^2+\frac{f'''(1)}{3!}z^3+...)=0$

We rewrite and receive:

$2(z^2+2z+1)(f''(1)+f'''(1)z+...)-3(z+1)(f'(1)+f''(1)z+\frac{f'''(1)}{2!}z^2+...)+(z+2)(f(1)+f'(1)z+\frac{f''(1)}{2!}z^2+\frac{f'''(1)}{3!}z^3+...)=0$

We collect the terms near $1,z,z^2,z^3,...$ and get: $2f''(1)-3f'(1)+2f(1)+z(4f''(1)+2f'''(1)-3f''(1)-3f'(1)+f(1)+2f'(1))+...$

We set: $f(1)=4$ and $f'(1)=1$ and get: $2f''(1)-3+8=0$. It implies $f''(1)=-2.5$. The term near $z$ yields:$-10+2f'''(1)-7.5-3+4+2=2f'''(1)-20.5+6=-14.5$. Thus, $f'''(1)=-7.25$. We obtained first terms of the expansion: $f(1)=4$, $f'(1)=1$, $f''(1)=-2.5$, $f'''(1)=-14.5$. In case we calculate the coefficients near $z^4,z^5$ and other terms, we will receive all the other derivatives ($f^{(4)}(1),f^{(5)}(1),...$ ) and the explicit form of $f(x)$.