Once again, write down the condition of the problem
{ x ⋅ d 2 y d x 2 + d y d x + x y = 0 y ( 0 ) = 1 y ′ ( 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. ⎩ ⎨ ⎧ x ⋅ d x 2 d 2 y + d x d y + x y = 0 y ( 0 ) = 1 y ′ ( 0 ) = 0
We are offered to use the approximating formulas :
d 2 y d x 2 ≈ y n + 1 − 2 y n + y n − 1 h 2 d y d x ≈ y n + 1 − y n − 1 2 h \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} d x 2 d 2 y ≈ h 2 y n + 1 − 2 y n + y n − 1 d x d y ≈ 2 h y n + 1 − y n − 1
To fulfill item (i) we must substitute the indicated formulas in the original equation :
x ⋅ d 2 y d x 2 + d y d x + x y = 0 → x n ⋅ y n + 1 − 2 y n + y n − 1 h 2 + y n + 1 − y n − 1 2 h + x n y n = 0 ∣ × ( 2 h 2 ) 2 x n ( y n + 1 − 2 y n + y n − 1 ) + h ( y n + 1 − y n − 1 ) + 2 h 2 x n y n = 0 2 x n y n + 1 + h y n + 1 = 4 x n y n − 2 h 2 x n y n + h y n − 1 − 2 x n y n − 1 y n + 1 ( h + 2 x n ) = 2 x n y n ( 2 − h 2 ) + y n − 1 ( h − 2 x n ) ∣ ÷ ( h + 2 x n ) y n + 1 = 2 x n y n ( 2 − h 2 ) + y n − 1 ( h − 2 x n ) h + 2 x n 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}} x ⋅ d x 2 d 2 y + d x d y + x y = 0 → x n ⋅ h 2 y n + 1 − 2 y n + y n − 1 + 2 h y n + 1 − y n − 1 + x n y n = 0 ∣ ∣ × ( 2 h 2 ) 2 x n ( y n + 1 − 2 y n + y n − 1 ) + h ( y n + 1 − y n − 1 ) + 2 h 2 x n y n = 0 2 x n y n + 1 + h y n + 1 = 4 x n y n − 2 h 2 x n y n + h y n − 1 − 2 x n y n − 1 y n + 1 ( h + 2 x n ) = 2 x n y n ( 2 − h 2 ) + y n − 1 ( h − 2 x n ) ∣ ∣ ÷ ( h + 2 x n ) y n + 1 = h + 2 x n 2 x n y n ( 2 − h 2 ) + y n − 1 ( h − 2 x n )
Q.E.D.
To fulfill point (ii), we must understand that our splitting occurs as follows :
x 0 = 0 → x n = n ⋅ h and y ( 0 ) = 1 ↔ y 0 = 1 x_0=0\to x_n=n\cdot h\quad\text{and}\quad y(0)=1\leftrightarrow y_0=1 x 0 = 0 → x n = n ⋅ h and y ( 0 ) = 1 ↔ y 0 = 1
Now substitute in the formula that we have received the item (i), n = 0 n=0 n = 0
y + 1 = 2 x 0 y 0 ( 2 − h 2 ) + y − 1 ( h − 2 x 0 ) h + 2 x 0 → y + 1 = 2 ⋅ 0 ⋅ 1 ( 2 − h 2 ) + y − 1 ⋅ h h → y + 1 = y − 1 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] y + 1 = h + 2 x 0 2 x 0 y 0 ( 2 − h 2 ) + y − 1 ( h − 2 x 0 ) → y + 1 = h 2 ⋅ 0 ⋅ 1 ( 2 − h 2 ) + y − 1 ⋅ h → y + 1 = y − 1
Q.E.D.
To find the value of y ( 0.6 ) y\left(0.6\right) y ( 0.6 )
h = 0.2 and x = n h → n = x h = 0.6 0.2 = 3 h=0.2\quad\text{and}\quad x=nh\to n=\frac{x}{h}=\frac{0.6}{0.2}=3 h = 0.2 and x = nh → n = h x = 0.2 0.6 = 3
We somehow additionally need to calculate y 1 y_1 y 1 y 1 = y − 1 y_1=y_{-1} y 1 = y − 1
As we know
y ′ ≡ d y d x ≈ y n + 1 − y n h → y ′ ( 0 ) = 0 = y 1 − 1 0.2 → y 1 = 1 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} y ′ ≡ d x d y ≈ h y n + 1 − y n → y ′ ( 0 ) = 0 = 0.2 y 1 − 1 → y 1 = 1
Then,
y n + 1 = 2 x n y n ( 2 − h 2 ) + y n − 1 ( h − 2 x n ) h + 2 x n n = 1 : y 2 = 2 x 1 y 1 ( 2 − h 2 ) + y 0 ( h − 2 x 1 ) h + 2 x 1 → y 2 = 2 ⋅ 0.2 ⋅ 1 ( 2 − 0. 2 2 ) + 0 ( 0.2 − 2 ⋅ 0.2 ) 0.2 + 2 ⋅ 0.2 = 0.4 ⋅ 1.96 0.6 y 2 ≈ 1.307 n = 2 : y 3 = 2 x 2 y 2 ( 2 − h 2 ) + y 1 ( h − 2 x 2 ) h + 2 x 2 → y 3 = 2 ⋅ 0.4 ⋅ 1.307 ( 2 − 0. 2 2 ) + 1 ( 0.2 − 2 ⋅ 0.4 ) 0.2 + 2 ⋅ 0.4 = 0.8 ⋅ 1.307 ⋅ 1.96 − 0.6 1 y ( 0.6 ) = y 3 ≈ 1.449 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} y n + 1 = h + 2 x n 2 x n y n ( 2 − h 2 ) + y n − 1 ( h − 2 x n ) n = 1 : y 2 = h + 2 x 1 2 x 1 y 1 ( 2 − h 2 ) + y 0 ( h − 2 x 1 ) → y 2 = 0.2 + 2 ⋅ 0.2 2 ⋅ 0.2 ⋅ 1 ( 2 − 0. 2 2 ) + 0 ( 0.2 − 2 ⋅ 0.2 ) = 0.6 0.4 ⋅ 1.96 y 2 ≈ 1.307 n = 2 : y 3 = h + 2 x 2 2 x 2 y 2 ( 2 − h 2 ) + y 1 ( h − 2 x 2 ) → y 3 = 0.2 + 2 ⋅ 0.4 2 ⋅ 0.4 ⋅ 1.307 ( 2 − 0. 2 2 ) + 1 ( 0.2 − 2 ⋅ 0.4 ) = 1 0.8 ⋅ 1.307 ⋅ 1.96 − 0.6 y ( 0.6 ) = y 3 ≈ 1.449
ANSWER
y ( 0.6 ) ≈ 1.449 y(0.6)\approx1.449 y ( 0.6 ) ≈ 1.449
#332078 on Oct 2022
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