Question #156203
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
1
Expert's answer
2021-01-26T03:15:44-0500

Once again, write down the condition of the problem



{xd2ydx2+dydx+xy=0y(0)=1y(0)=0\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 :



d2ydx2yn+12yn+yn1h2dydxyn+1yn12h\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 :



xd2ydx2+dydx+xy=0xnyn+12yn+yn1h2+yn+1yn12h+xnyn=0×(2h2)2xn(yn+12yn+yn1)+h(yn+1yn1)+2h2xnyn=02xnyn+1+hyn+1=4xnyn2h2xnyn+hyn12xnyn1yn+1(h+2xn)=2xnyn(2h2)+yn1(h2xn)÷(h+2xn)yn+1=2xnyn(2h2)+yn1(h2xn)h+2xnx\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 :



x0=0xn=nhandy(0)=1y0=1x_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=0n=0 :



y+1=2x0y0(2h2)+y1(h2x0)h+2x0y+1=201(2h2)+y1hhy+1=y1y_{+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(0.6)y\left(0.6\right) , we must understand that



h=0.2andx=nhn=xh=0.60.2=3h=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 y1y_1 , since the last obtained formula y1=y1y_1=y_{-1} does not help us in calculations.

As we know



ydydxyn+1ynhy(0)=0=y110.2y1=1y'\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,


yn+1=2xnyn(2h2)+yn1(h2xn)h+2xnn=1:y2=2x1y1(2h2)+y0(h2x1)h+2x1y2=20.21(20.22)+0(0.220.2)0.2+20.2=0.41.960.6y21.307n=2:y3=2x2y2(2h2)+y1(h2x2)h+2x2y3=20.41.307(20.22)+1(0.220.4)0.2+20.4=0.81.3071.960.61y(0.6)=y31.449y_{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)1.449y(0.6)\approx1.449


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