题目
Problem
2 x d 2 y d x 2 + y d y d x − 6 x y = 0 \begin{align*}
2x\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
+y\frac{\mathrm{d}y}{\mathrm{d}x}
-6xy=0
\end{align*} 2 x d x 2 d 2 y + y d x d y − 6 x y = 0 Given that d y d x = 3 \frac{\mathrm{d}y}{\mathrm{d}x}=3 d x d y = 3 y = 1 2 y=\frac12 y = 2 1 x = 2 x=2 x = 2
(a) determine the value of d 3 y d x 3 \frac{\mathrm{d}^3y}{\mathrm{d}x^3} d x 3 d 3 y x = 2 x=2 x = 2
(6)
(b) Hence determine the series expansion for y y y x = 2 x=2 x = 2 ( x − 2 ) (x-2) ( x − 2 ) ( x − 2 ) 3 (x-2)^3 ( x − 2 ) 3
(2)
解答
(a)
解法一
思路
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先把 x = 2 , y = 1 2 , y ′ = 3 x=2,y=\frac12,y'=3 x = 2 , y = 2 1 , y ′ = 3 y ′ ′ y'' y ′′ y y ′ yy' y y ′ ( y ′ ) 2 + y y ′ ′ (y')^2+yy'' ( y ′ ) 2 + y y ′′ y ′ ′ ′ y''' y ′′′
答题过程
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At x = 2 x=2 x = 2
y = 1 2 , d y d x = 3 \begin{align*}
y=\frac12,
\qquad
\frac{\mathrm{d}y}{\mathrm{d}x}=3
\end{align*} y = 2 1 , d x d y = 3 Substitute into the differential equation:
2 ( 2 ) d 2 y d x 2 ( 1 2 ) ( 3 ) − 6 ( 2 ) ( 1 2 ) = 0 4 d 2 y d x 2 3 2 − 6 = 0 4 d 2 y d x 2 = 9 2 \begin{align*}
2(2)\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
\left(\frac12\right)(3)
-6(2)\left(\frac12\right)
={}&
0\\[3mm]
4\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
\frac32-6
={}&
0\\[3mm]
4\frac{\mathrm{d}^2y}{\mathrm{d}x^2}
={}&
\frac92
\end{align*} 2 ( 2 ) d x 2 d 2 y ( 2 1 ) ( 3 ) − 6 ( 2 ) ( 2 1 ) = 4 d x 2 d 2 y 2 3 − 6 = 4 d x 2 d 2 y = 0 0 2 9 So
d 2 y d x 2 ∣ x = 2 = 9 8 \begin{align*}
\left.\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right|_{x=2}
={}&
\frac98
\end{align*} d x 2 d 2 y x = 2 = 8 9 Differentiate
2 x y ′ ′ + y y ′ − 6 x y = 0 \begin{align*}
2x y''+yy'-6xy=0
\end{align*} 2 x y ′′ + y y ′ − 6 x y = 0 with respect to x x x
2 y ′ ′ + 2 x y ′ ′ ′ ( ( y ′ ) 2 + y y ′ ′ ) − 6 y − 6 x y ′ = 0 \begin{align*}
2y''+2xy'''
\left((y')^2+yy''\right)
-6y-6xy'
={}&
0
\end{align*} 2 y ′′ + 2 x y ′′′ ( ( y ′ ) 2 + y y ′′ ) − 6 y − 6 x y ′ = 0 Substitute x = 2 x=2 x = 2 y = 1 2 y=\frac12 y = 2 1 y ′ = 3 y'=3 y ′ = 3 y ′ ′ = 9 8 y''=\frac98 y ′′ = 8 9
2 ( 9 8 ) + 2 ( 2 ) y ′ ′ ′ + 3 2 + 1 2 ( 9 8 ) − 6 ( 1 2 ) − 6 ( 2 ) ( 3 ) = 0 9 4 + 4 y ′ ′ ′ + 9 + 9 16 − 3 − 36 = 0 4 y ′ ′ ′ − 435 16 = 0 \begin{align*}
2\left(\frac98\right)
+2(2)y'''
+3^2
+\frac12\left(\frac98\right)
-6\left(\frac12\right)
-6(2)(3)
={}&
0\\[3mm]
\frac94+4y'''+9+\frac9{16}-3-36
={}&
0\\[3mm]
4y'''-\frac{435}{16}
={}&
0
\end{align*} 2 ( 8 9 ) + 2 ( 2 ) y ′′′ + 3 2 + 2 1 ( 8 9 ) − 6 ( 2 1 ) − 6 ( 2 ) ( 3 ) = 4 9 + 4 y ′′′ + 9 + 16 9 − 3 − 36 = 4 y ′′′ − 16 435 = 0 0 0 Therefore
d 3 y d x 3 ∣ x = 2 = 435 64 \begin{align*}
\boxed{
\left.\frac{\mathrm{d}^3y}{\mathrm{d}x^3}\right|_{x=2}
=\frac{435}{64}
}
\end{align*} d x 3 d 3 y x = 2 = 64 435
(b)
解法一
思路
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Taylor series about x = 2 x=2 x = 2
y ( 2 ) + y ′ ( 2 ) ( x − 2 ) + y ′ ′ ( 2 ) 2 ! ( x − 2 ) 2 + y ′ ′ ′ ( 2 ) 3 ! ( x − 2 ) 3 + ⋯ \begin{align*}
y(2)+y'(2)(x-2)+\frac{y''(2)}{2!}(x-2)^2
+\frac{y'''(2)}{3!}(x-2)^3+\cdots
\end{align*} y ( 2 ) + y ′ ( 2 ) ( x − 2 ) + 2 ! y ′′ ( 2 ) ( x − 2 ) 2 + 3 ! y ′′′ ( 2 ) ( x − 2 ) 3 + ⋯ 把 (a) 中求出的导数值代入并化简系数。
答题过程
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Using Taylor’s formula about x = 2 x=2 x = 2
y = y ( 2 ) + y ′ ( 2 ) ( x − 2 ) + y ′ ′ ( 2 ) 2 ! ( x − 2 ) 2 + y ′ ′ ′ ( 2 ) 3 ! ( x − 2 ) 3 + ⋯ \begin{align*}
y
={}&
y(2)+y'(2)(x-2)
+\frac{y''(2)}{2!}(x-2)^2\\[2mm]
&\,\hspace{2pt}+\frac{y'''(2)}{3!}(x-2)^3+\cdots
\end{align*} y = y ( 2 ) + y ′ ( 2 ) ( x − 2 ) + 2 ! y ′′ ( 2 ) ( x − 2 ) 2 + 3 ! y ′′′ ( 2 ) ( x − 2 ) 3 + ⋯ Substitute
y ( 2 ) = 1 2 , y ′ ( 2 ) = 3 , y ′ ′ ( 2 ) = 9 8 , y ′ ′ ′ ( 2 ) = 435 64 \begin{align*}
y(2)=\frac12,\quad
y'(2)=3,\quad
y''(2)=\frac98,\quad
y'''(2)=\frac{435}{64}
\end{align*} y ( 2 ) = 2 1 , y ′ ( 2 ) = 3 , y ′′ ( 2 ) = 8 9 , y ′′′ ( 2 ) = 64 435 Hence
y = 1 2 + 3 ( x − 2 ) + 9 8 2 ( x − 2 ) 2 + 435 64 6 ( x − 2 ) 3 = 1 2 + 3 ( x − 2 ) + 9 16 ( x − 2 ) 2 + 145 128 ( x − 2 ) 3 \begin{align*}
y
={}&
\frac12+3(x-2)
+\frac{\frac98}{2}(x-2)^2
+\frac{\frac{435}{64}}{6}(x-2)^3\\[3mm]
={}&
\boxed{
\frac12+3(x-2)
+\frac9{16}(x-2)^2
+\frac{145}{128}(x-2)^3
}
\end{align*} y = = 2 1 + 3 ( x − 2 ) + 2 8 9 ( x − 2 ) 2 + 6 64 435 ( x − 2 ) 3 2 1 + 3 ( x − 2 ) + 16 9 ( x − 2 ) 2 + 128 145 ( x − 2 ) 3 