Answer to Question #156203 in Quantitative Methods for Keleko

Question

Given that y satisfies the differential equation x(d^2y/dx^2) + dy/dx + xy = 0 and that y = 1, dy/dx = 0 when x = 0.
Using approximations, h^2( d^2y/dx^2 )base n ~= ybase(n+1) - 2ybase n + ybase(n-1) and 2h(dy/dx) ~= ybase(n+1) - ybase (n-1) and a step-length of 0.2,
Show that
(i) ybase(n+1) ~= ((2xnyn(2 - h^2) + yn - 1( h - 2xn )) / (h + 2xn) where n = base n
(ii) ybase(-1) = ybase(+1).
Hence find the value of y when x = 0.6

Answer 1

Once again, write down the condition of the problem

$$\left\{\begin{array}{l} x\cdot\displaystyle\frac{d^2y}{dx^2}+\displaystyle\frac{dy}{dx}+xy=0\\[0.3cm] y(0)=1\\[0.3cm] y'(0)=0 \end{array}\right.$$

We are offered to use the approximating formulas :

$$\frac{d^2y}{dx^2}\approx\frac{y_{n+1}-2y_n+y_{n-1}}{h^2}\\[0.3cm] \frac{dy}{dx}\approx\frac{y_{n+1}-y_{n-1}}{2h}$$

To fulfill item (i) we must substitute the indicated formulas in the original equation :

$$x\cdot\frac{d^2y}{dx^2}+\frac{dy}{dx}+xy=0\to\\[0.3cm] \left.x_n\cdot\frac{y_{n+1}-2y_n+y_{n-1}}{h^2}+\frac{y_{n+1}-y_{n-1}}{2h}+x_ny_n=0\right|\times\left(2h^2\right)\\[0.3cm] 2x_n\left(y_{n+1}-2y_n+y_{n-1}\right)+h\left(y_{n+1}-y_{n-1}\right)+2h^2x_ny_n=0\\[0.3cm] 2x_ny_{n+1}+hy_{n+1}=4x_ny_n-2h^2x_ny_n+hy_{n-1}-2x_ny_{n-1}\\[0.3cm] \left.y_{n+1}\left(h+2x_n\right)=2x_ny_n\left(2-h^2\right)+y_{n-1}\left(h-2x_n\right)\right|\div\left(h+2x_n\right)\\[0.3cm] \boxed{y_{n+1}=\frac{2x_ny_n\left(2-h^2\right)+y_{n-1}\left(h-2x_n\right)}{h+2x_n}}$$

Q.E.D.

To fulfill point (ii), we must understand that our splitting occurs as follows :

$$x_0=0\to x_n=n\cdot h\quad\text{and}\quad y(0)=1\leftrightarrow y_0=1$$

Now substitute in the formula that we have received the item (i), $n=0$ :

$$y_{+1}=\frac{2x_0y_0\left(2-h^2\right)+y_{-1}\left(h-2x_0\right)}{h+2x_0}\to\\[0.3cm] y_{+1}=\frac{2\cdot0\cdot1\left(2-h^2\right)+y_{-1}\cdot h}{h}\to\boxed{y_{+1}=y_{-1}}\\[0.3cm]$$

Q.E.D.

To find the value of $y\left(0.6\right)$ , we must understand that

$$h=0.2\quad\text{and}\quad x=nh\to n=\frac{x}{h}=\frac{0.6}{0.2}=3$$

We somehow additionally need to calculate $y_1$ , since the last obtained formula $y_1=y_{-1}$ does not help us in calculations.

As we know

$$y'\equiv\frac{dy}{dx}\approx\frac{y_{n+1}-y_n}{h}\to y'(0)=0=\frac{y_1-1}{0.2}\to\boxed{y_1=1}$$

Then,

$$y_{n+1}=\frac{2x_ny_n\left(2-h^2\right)+y_{n-1}\left(h-2x_n\right)}{h+2x_n}\\[0.3cm] n=1 : y_{2}=\frac{2x_1y_1\left(2-h^2\right)+y_{0}\left(h-2x_1\right)}{h+2x_1}\to\\[0.3cm] y_2=\frac{2\cdot0.2\cdot1\left(2-0.2^2\right)+0\left(0.2-2\cdot0.2\right)}{0.2+2\cdot0.2}=\frac{0.4\cdot1.96}{0.6}\\[0.3cm] \boxed{y_2\approx1.307}\\[0.3cm] n=2 : y_{3}=\frac{2x_2y_2\left(2-h^2\right)+y_{1}\left(h-2x_2\right)}{h+2x_2}\to\\[0.3cm] y_3=\frac{2\cdot0.4\cdot1.307\left(2-0.2^2\right)+1\left(0.2-2\cdot0.4\right)}{0.2+2\cdot0.4}=\frac{0.8\cdot1.307\cdot1.96-0.6}{1}\\[0.3cm] \boxed{y(0.6)=y_3\approx1.449}$$

ANSWER

$$y(0.6)\approx1.449$$