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高等数学（第七版）同济大学 习题12-5 个人解答

12 个人 习题 解答 高等数学 同济大学 第七版
2023-09-14 09:06:59 时间

高等数学（第七版）同济大学 习题12-5

 

1.  利用函数的幂级数展开式求下列各数的近似值： \begin{aligned}&1. \ 利用函数的幂级数展开式求下列各数的近似值：&\end{aligned} ​1. 利用函数的幂级数展开式求下列各数的近似值：​​

   ( 1 )    l n   3  （误差不超过 0.0001 ）；    ( 2 )    e  （误差不超过 0.001 ）；    ( 3 )    522 9 （误差不超过 0.00001 ）；    ( 4 )    c o s   2 ∘  （误差不超过 0.0001 ） . \begin{aligned} &\ \ (1)\ \ ln\ 3\ （误差不超过0.0001）；\\\\ &\ \ (2)\ \ \sqrt{e}\ （误差不超过0.001）；\\\\ &\ \ (3)\ \ \sqrt[9]{522}（误差不超过0.00001）；\\\\ &\ \ (4)\ \ cos\ 2^{\circ}\ （误差不超过0.0001）. & \end{aligned} ​  (1)  ln 3 （误差不超过0.0001）；  (2)  e ​ （误差不超过0.001）；  (3)  9522 ​（误差不超过0.00001）；  (4)  cos 2∘ （误差不超过0.0001）.​​

解：

   ( 1 )   l n 1 + x 1 − x = 2 ( x + x 3 3 + x 5 5 + ⋅ ⋅ ⋅ + x 2 n − 1 2 n − 1 + ⋅ ⋅ ⋅ ) ， x ∈ ( − 1 ,   1 ) ，令 1 + x 1 − x = 3 ，得 x = 1 2 ，则          l n   3 = l n   1 + 1 2 1 − 1 2 = 2 [ 1 2 + 1 3 ⋅ 2 3 + 1 5 ⋅ 2 5 + ⋅ ⋅ ⋅ + 1 ( 2 n − 1 ) 2 n − 1 + ⋅ ⋅ ⋅ ] ，          ∣ r n ∣ = 2 [ 1 ( 2 n + 1 ) 2 2 n + 1 + 1 ( 2 n + 3 ) 2 2 n + 3 + ⋅ ⋅ ⋅ ] = 2 ( 2 n + 1 ) 2 2 n + 1 [ 1 + ( 2 n + 1 ) 2 2 n + 1 ( 2 n + 3 ) 2 2 n + 3 + ( 2 n + 1 ) 2 2 n + 1 ( 2 n + 5 ) 2 2 n + 5 + ⋅ ⋅ ⋅ ] <          2 ( 2 n + 1 ) 2 2 n + 1 ( 1 + 1 2 2 + 1 2 4 + ⋅ ⋅ ⋅ ) = 2 ( 2 n + 1 ) 2 2 n + 1 ⋅ 1 1 − 1 4 = 1 3 ( 2 n + 1 ) 2 2 n − 2 ，          ∣ r 5 ∣ < 1 3 ⋅ 11 ⋅ 2 8 ≈ 0.00012 ， ∣ r 6 ∣ < 1 3 ⋅ 13 ⋅ 2 10 ≈ 0.00003 < 1 0 − 4 ，取 n = 6 ，         则 l n   3 ≈ 2 ( 1 2 + 1 3 ⋅ 2 3 + 1 5 ⋅ 2 5 + ⋅ ⋅ ⋅ + 1 11 ⋅ 2 11 ) ，考虑到舍入误差，取五位小数，得 l n   3 ≈ 1.0986.    ( 2 )   e x = 1 + x + x 2 2 ! + ⋅ ⋅ ⋅ + x n n ! + ⋅ ⋅ ⋅ ， x ∈ ( − ∞ ,   + ∞ ) ，令 x = 1 2 ，得 e = 1 + 1 2 + 1 2 ! 2 2 + ⋅ ⋅ ⋅ + 1 n ! 2 n + ⋅ ⋅ ⋅ ，           r n = 1 ( n + 1 ) ! 2 n + 1 + 1 ( n + 2 ) ! 2 n + 2 + ⋅ ⋅ ⋅ = 1 ( n + 1 ) ! 2 n + 1 [ 1 + 1 ( n + 2 ) ⋅ 2 + 1 ( n + 2 ) ( n + 3 ) ⋅ 2 2 + ⋅ ⋅ ⋅ ] <           1 ( n + 1 ) ! 2 n + 1 ( 1 + 1 2 + 1 2 2 + ⋅ ⋅ ⋅ ) = 1 ( n + 1 ) ! 2 n + 1 ⋅ 1 1 − 1 2 = 1 ( n + 1 ) ! 2 n ， r 4 < 1 5 ! 2 4 ≈ 0.0005 < 1 0 − 3 ，取 n = 4 ，          考虑到舍入误差，取四位小数，得 e ≈ 1 + 1 2 + 1 2 ! 2 2 + 1 3 ! 2 3 + 1 4 ! 2 4 ≈ 1.648.    ( 3 )   522 9 = 2 9 + 10 9 = 2 ( 1 + 10 2 9 ) 1 9 ，因为 ( 1 + x ) m =          1 + m x + m ( m − 1 ) 2 ! x 2 + ⋅ ⋅ ⋅ + m ( m − 1 ) ⋅   ⋅ ⋅ ⋅   ⋅ ( m − n + 1 ) n ! x n + ⋅ ⋅ ⋅   ( − 1 < x < 1 ) ，         所以 522 9 = 2 ( 1 + 10 2 9 ) 1 9 = 2 [ 1 + 1 9 ⋅ 10 2 9 + 1 9 ( 1 9 − 1 ) 2 ! ⋅ 1 0 2 2 18 + ⋅ ⋅ ⋅ + 1 9 ( 1 9 − 1 ) ⋅   ⋅ ⋅ ⋅   ⋅ ( 1 9 − n + 1 ) n ! ⋅ 1 0 n 2 9 n + ⋅ ⋅ ⋅ ] =          2 ⋅ ( 1 + 1 9 ⋅ 10 2 9 − 1 9 ⋅ 8 9 2 ! ⋅ 1 0 2 2 18 + ⋅ ⋅ ⋅ ) = 2 + 2 9 ⋅ 10 2 9 − 1 9 ⋅ 8 9 ⋅ 1 0 2 2 18 + ⋅ ⋅ ⋅ ，上式右端第 2 项起为一交错级数，         则有 ∣ r 3 ∣ ≤ u 4 = 8 ⋅ 17 3 ⋅ 9 3 ⋅ 1 0 3 2 27 < 1 0 − 6 ，取 3 项，取六位小数，得 522 9 ≈ 2 + 2 9 ⋅ 10 2 9 − 8 9 2 ⋅ 1 0 2 2 18 ≈ 2.00430.    ( 4 )   c o s   2 ∘ = c o s   π 90 = 1 − 1 2 ! ( π 90 ) 2 + 1 4 ! ( π 90 ) 4 − ⋅ ⋅ ⋅ ，上式为交错级数， ∣ r 2 ∣ ≤ u 3 = 1 4 ! ( π 90 ) 4 < 1 0 − 7 ，         取 2 项，取五位小数，得 c o s   2 ∘ ≈ 1 − 1 2 ! ( π 90 ) 2 ≈ 0.9994. \begin{aligned} &\ \ (1)\ ln\frac{1+x}{1-x}=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdot\cdot\cdot+\frac{x^{2n-1}}{2n-1}+\cdot\cdot\cdot\right)，x \in (-1, \ 1)，令\frac{1+x}{1-x}=3，得x=\frac{1}{2}，则\\\\ &\ \ \ \ \ \ \ \ ln\ 3=ln\ \frac{1+\frac{1}{2}}{1-\frac{1}{2}}=2\left[\frac{1}{2}+\frac{1}{3\cdot 2^3}+\frac{1}{5\cdot 2^5}+\cdot\cdot\cdot+\frac{1}{(2n-1)^{2n-1}}+\cdot\cdot\cdot\right]，\\\\ &\ \ \ \ \ \ \ \ |r_n|=2\left[\frac{1}{(2n+1)2^{2n+1}}+\frac{1}{(2n+3)2^{2n+3}}+\cdot\cdot\cdot\right]=\frac{2}{(2n+1)2^{2n+1}}\left[1+\frac{(2n+1)2^{2n+1}}{(2n+3)2^{2n+3}}+\frac{(2n+1)2^{2n+1}}{(2n+5)2^{2n+5}}+\cdot\cdot\cdot\right] \lt \\\\ &\ \ \ \ \ \ \ \ \frac{2}{(2n+1)2^{2n+1}}\left(1+\frac{1}{2^2}+\frac{1}{2^4}+\cdot\cdot\cdot\right)=\frac{2}{(2n+1)2^{2n+1}}\cdot\frac{1}{1-\frac{1}{4}}=\frac{1}{3(2n+1)2^{2n-2}}，\\\\ &\ \ \ \ \ \ \ \ |r_5| \lt \frac{1}{3\cdot 11\cdot 2^8} \approx 0.00012，|r_6| \lt \frac{1}{3\cdot 13\cdot 2^{10}} \approx 0.00003 \lt 10^{-4}，取n=6，\\\\ &\ \ \ \ \ \ \ \ 则ln\ 3 \approx 2\left(\frac{1}{2}+\frac{1}{3\cdot 2^3}+\frac{1}{5\cdot 2^5}+\cdot\cdot\cdot+\frac{1}{11\cdot 2^{11}}\right)，考虑到舍入误差，取五位小数，得ln\ 3 \approx 1.0986.\\\\ &\ \ (2)\ e^x=1+x+\frac{x^2}{2!}+\cdot\cdot\cdot+\frac{x^n}{n!}+\cdot\cdot\cdot，x \in (-\infty, \ +\infty)，令x=\frac{1}{2}，得\sqrt{e}=1+\frac{1}{2}+\frac{1}{2!2^2}+\cdot\cdot\cdot+\frac{1}{n!2^n}+\cdot\cdot\cdot，\\\\ &\ \ \ \ \ \ \ \ \ r_n=\frac{1}{(n+1)!2^{n+1}}+\frac{1}{(n+2)!2^{n+2}}+\cdot\cdot\cdot=\frac{1}{(n+1)!2^{n+1}}\left[1+\frac{1}{(n+2)\cdot 2}+\frac{1}{(n+2)(n+3)\cdot 2^2}+\cdot\cdot\cdot\right] \lt \\\\ &\ \ \ \ \ \ \ \ \ \frac{1}{(n+1)!2^{n+1}}\left(1+\frac{1}{2}+\frac{1}{2^2}+\cdot\cdot\cdot\right)=\frac{1}{(n+1)!2^{n+1}}\cdot \frac{1}{1-\frac{1}{2}}=\frac{1}{(n+1)!2^n}，r_4 \lt \frac{1}{5!2^4} \approx 0.0005 \lt 10^{-3}，取n=4，\\\\ &\ \ \ \ \ \ \ \ \ 考虑到舍入误差，取四位小数，得\sqrt{e} \approx 1+\frac{1}{2}+\frac{1}{2!2^2}+\frac{1}{3!2^3}+\frac{1}{4!2^4} \approx 1.648.\\\\ &\ \ (3)\ \sqrt[9]{522}=\sqrt[9]{2^9+10}=2\left(1+\frac{10}{2^9}\right)^{\frac{1}{9}}，因为(1+x)^m=\\\\ &\ \ \ \ \ \ \ \ 1+mx+\frac{m(m-1)}{2!}x^2+\cdot\cdot\cdot+\frac{m(m-1)\cdot\ \cdot\cdot\cdot\ \cdot(m-n+1)}{n!}x^n+\cdot\cdot\cdot\ (-1 \lt x \lt 1)，\\\\ &\ \ \ \ \ \ \ \ 所以\sqrt[9]{522}=2\left(1+\frac{10}{2^9}\right)^{\frac{1}{9}}=2\left[1+\frac{1}{9}\cdot \frac{10}{2^9}+\frac{\frac{1}{9}\left(\frac{1}{9}-1\right)}{2!}\cdot \frac{10^2}{2^{18}}+\cdot\cdot\cdot+\frac{\frac{1}{9}\left(\frac{1}{9}-1\right)\cdot\ \cdot\cdot\cdot\ \cdot \left(\frac{1}{9}-n+1\right)}{n!}\cdot\frac{10^n}{2^{9n}}+\cdot\cdot\cdot\right]=\\\\ &\ \ \ \ \ \ \ \ 2\cdot \left(1+\frac{1}{9}\cdot \frac{10}{2^9}-\frac{\frac{1}{9}\cdot\frac{8}{9}}{2!}\cdot\frac{10^2}{2^{18}}+\cdot\cdot\cdot\right)=2+\frac{2}{9}\cdot\frac{10}{2^9}-\frac{1}{9}\cdot\frac{8}{9}\cdot\frac{10^2}{2^{18}}+\cdot\cdot\cdot，上式右端第2项起为一交错级数，\\\\ &\ \ \ \ \ \ \ \ 则有|r_3| \le u_4=\frac{8\cdot17}{3\cdot 9^3}\cdot \frac{10^3}{2^{27}} \lt 10^{-6}，取3项，取六位小数，得\sqrt[9]{522} \approx 2+\frac{2}{9}\cdot \frac{10}{2^9}-\frac{8}{9^2}\cdot\frac{10^2}{2^{18}}\approx 2.00430.\\\\ &\ \ (4)\ cos\ 2^{\circ}=cos\ \frac{\pi}{90}=1-\frac{1}{2!}\left(\frac{\pi}{90}\right)^2+\frac{1}{4!}\left(\frac{\pi}{90}\right)^4-\cdot\cdot\cdot，上式为交错级数，|r_2|\le u_3=\frac{1}{4!}\left(\frac{\pi}{90}\right)^4 \lt 10^{-7}，\\\\ &\ \ \ \ \ \ \ \ 取2项，取五位小数，得cos\ 2^{\circ} \approx 1-\frac{1}{2!}\left(\frac{\pi}{90}\right)^2 \approx 0.9994. & \end{aligned} ​  (1) ln1−x1+x​=2(x+3x3​+5x5​+⋅⋅⋅+2n−1x2n−1​+⋅⋅⋅)，x∈(−1, 1)，令1−x1+x​=3，得x=21​，则        ln 3=ln 1−21​1+21​​=2[21​+3⋅231​+5⋅251​+⋅⋅⋅+(2n−1)2n−11​+⋅⋅⋅]，        ∣rn​∣=2[(2n+1)22n+11​+(2n+3)22n+31​+⋅⋅⋅]=(2n+1)22n+12​[1+(2n+3)22n+3(2n+1)22n+1​+(2n+5)22n+5(2n+1)22n+1​+⋅⋅⋅]<        (2n+1)22n+12​(1+221​+241​+⋅⋅⋅)=(2n+1)22n+12​⋅1−41​1​=3(2n+1)22n−21​，        ∣r5​∣<3⋅11⋅281​≈0.00012，∣r6​∣<3⋅13⋅2101​≈0.00003<10−4，取n=6，        则ln 3≈2(21​+3⋅231​+5⋅251​+⋅⋅⋅+11⋅2111​)，考虑到舍入误差，取五位小数，得ln 3≈1.0986.  (2) ex=1+x+2!x2​+⋅⋅⋅+n!xn​+⋅⋅⋅，x∈(−∞, +∞)，令x=21​，得e ​=1+21​+2!221​+⋅⋅⋅+n!2n1​+⋅⋅⋅，         rn​=(n+1)!2n+11​+(n+2)!2n+21​+⋅⋅⋅=(n+1)!2n+11​[1+(n+2)⋅21​+(n+2)(n+3)⋅221​+⋅⋅⋅]<         (n+1)!2n+11​(1+21​+221​+⋅⋅⋅)=(n+1)!2n+11​⋅1−21​1​=(n+1)!2n1​，r4​<5!241​≈0.0005<10−3，取n=4，         考虑到舍入误差，取四位小数，得e ​≈1+21​+2!221​+3!231​+4!241​≈1.648.  (3) 9522 ​=929+10 ​=2(1+2910​)91​，因为(1+x)m=        1+mx+2!m(m−1)​x2+⋅⋅⋅+n!m(m−1)⋅ ⋅⋅⋅ ⋅(m−n+1)​xn+⋅⋅⋅ (−1<x<1)，        所以9522 ​=2(1+2910​)91​=2[1+91​⋅2910​+2!91​(91​−1)​⋅218102​+⋅⋅⋅+n!91​(91​−1)⋅ ⋅⋅⋅ ⋅(91​−n+1)​⋅29n10n​+⋅⋅⋅]=        2⋅(1+91​⋅2910​−2!91​⋅98​​⋅218102​+⋅⋅⋅)=2+92​⋅2910​−91​⋅98​⋅218102​+⋅⋅⋅，上式右端第2项起为一交错级数，        则有∣r3​∣≤u4​=3⋅938⋅17​⋅227103​<10−6，取3项，取六位小数，得9522 ​≈2+92​⋅2910​−928​⋅218102​≈2.00430.  (4) cos 2∘=cos 90π​=1−2!1​(90π​)2+4!1​(90π​)4−⋅⋅⋅，上式为交错级数，∣r2​∣≤u3​=4!1​(90π​)4<10−7，        取2项，取五位小数，得cos 2∘≈1−2!1​(90π​)2≈0.9994.​​

2.  利用被积函数的幂级数展开式求下列定积分的近似值： \begin{aligned}&2. \ 利用被积函数的幂级数展开式求下列定积分的近似值：&\end{aligned} ​2. 利用被积函数的幂级数展开式求下列定积分的近似值：​​

   ( 1 )    ∫ 0 0.5 1 1 + x 4 d x （误差不超过 0.0001 ）；    ( 2 )    ∫ 0 0.5 a r c t a n   x x d x  （误差不超过 0.001 ） . \begin{aligned} &\ \ (1)\ \ \int_{0}^{0.5}\frac{1}{1+x^4}dx（误差不超过0.0001）；\\\\ &\ \ (2)\ \ \int_{0}^{0.5}\frac{arctan\ x}{x}dx\ （误差不超过0.001）. & \end{aligned} ​  (1)  ∫00.5​1+x41​dx（误差不超过0.0001）；  (2)  ∫00.5​xarctan x​dx （误差不超过0.001）.​​

解：

   ( 1 )   ∫ 0 0.5 1 1 + x 4 d x = ∫ 0 0.5 [ 1 − x 4 + x 8 − x 12 + ⋅ ⋅ ⋅ + ( − 1 ) n x 4 n + ⋅ ⋅ ⋅ ] d x = ( x − 1 5 x 5 + 1 9 x 9 − 1 13 x 13 + ⋅ ⋅ ⋅ ) ∣ 0 0.5 =          1 2 − 1 5 ⋅ 1 2 5 + 1 9 ⋅ 1 2 9 − 1 13 ⋅ 1 2 13 + ⋅ ⋅ ⋅ ，上式右端为一交错级数，有 ∣ r 3 ∣ ≤ u 4 = 1 13 ⋅ 1 2 13 ≈ 0.000009 < 1 0 − 4 ，         取 3 项，取五位小数，得 ∫ 0 0.5 1 1 + x 4 d x ≈ 1 2 − 1 5 ⋅ 1 2 5 + 1 9 ⋅ 1 2 9 ≈ 0.4940.    ( 2 )  因为 a r c t a n   x = x − x 3 3 + x 5 5 − ⋅ ⋅ ⋅ + ( − 1 ) n x 2 n + 1 2 n + 1 + ⋅ ⋅ ⋅   ( − 1 < x < 1 ) ，所以 ∫ 0 0.5 a r c t a n   x x d x =          ∫ 0 0.5 [ 1 − x 3 3 + x 4 5 − ⋅ ⋅ ⋅ + ( − 1 ) n x 2 n 2 n + 1 + ⋅ ⋅ ⋅ ] d x = ( x − x 3 9 + x 5 25 − x 7 49 + ⋅ ⋅ ⋅ ) ∣ 0 0.5 =          1 2 − 1 9 ⋅ 1 2 3 + 1 25 ⋅ 1 2 5 − 1 49 ⋅ 1 2 7 + ⋅ ⋅ ⋅ ，因为 ∣ r 3 ∣ ≤ u 4 = 1 49 ⋅ 1 2 7 ≈ 0.0002 < 1 0 − 3 ，取 3 项，         取四位小数，得 ∫ 0 0.5 a r c t a n   x x d x ≈ 1 2 − 1 9 ⋅ 1 2 3 + 1 2 5 ⋅ 1 25 ≈ 0.487. \begin{aligned} &\ \ (1)\ \int_{0}^{0.5}\frac{1}{1+x^4}dx=\int_{0}^{0.5}[1-x^4+x^8-x^{12}+\cdot\cdot\cdot+(-1)^nx^{4n}+\cdot\cdot\cdot]dx=\left(x-\frac{1}{5}x^5+\frac{1}{9}x^9-\frac{1}{13}x^{13}+\cdot\cdot\cdot\right)\bigg|_{0}^{0.5}=\\\\ &\ \ \ \ \ \ \ \ \frac{1}{2}-\frac{1}{5}\cdot \frac{1}{2^5}+\frac{1}{9}\cdot\frac{1}{2^9}-\frac{1}{13}\cdot\frac{1}{2^{13}}+\cdot\cdot\cdot，上式右端为一交错级数，有|r_3| \le u_4=\frac{1}{13}\cdot\frac{1}{2^{13}}\approx 0.000009 \lt 10^{-4}，\\\\ &\ \ \ \ \ \ \ \ 取3项，取五位小数，得\int_{0}^{0.5}\frac{1}{1+x^4}dx\approx \frac{1}{2}-\frac{1}{5}\cdot \frac{1}{2^5}+\frac{1}{9}\cdot\frac{1}{2^9}\approx 0.4940.\\\\ &\ \ (2)\ 因为arctan\ x=x-\frac{x^3}{3}+\frac{x^5}{5}-\cdot\cdot\cdot+(-1)^n\frac{x^{2n+1}}{2n+1}+\cdot\cdot\cdot\ (-1 \lt x \lt 1)，所以\int_{0}^{0.5}\frac{arctan\ x}{x}dx=\\\\ &\ \ \ \ \ \ \ \ \int_{0}^{0.5}\left[1-\frac{x^3}{3}+\frac{x^4}{5}-\cdot\cdot\cdot+(-1)^n\frac{x^{2n}}{2n+1}+\cdot\cdot\cdot\right]dx=\left(x-\frac{x^3}{9}+\frac{x^5}{25}-\frac{x^7}{49}+\cdot\cdot\cdot\right)\bigg|_{0}^{0.5}=\\\\ &\ \ \ \ \ \ \ \ \frac{1}{2}-\frac{1}{9}\cdot\frac{1}{2^3}+\frac{1}{25}\cdot\frac{1}{2^5}-\frac{1}{49}\cdot\frac{1}{2^7}+\cdot\cdot\cdot，因为|r_3| \le u_4=\frac{1}{49}\cdot\frac{1}{2^7}\approx 0.0002 \lt 10^{-3}，取3项，\\\\ &\ \ \ \ \ \ \ \ 取四位小数，得\int_{0}^{0.5}\frac{arctan\ x}{x}dx\approx \frac{1}{2}-\frac{1}{9}\cdot \frac{1}{2^3}+\frac{1}{2^5}\cdot\frac{1}{25}\approx 0.487. & \end{aligned} ​  (1) ∫00.5​1+x41​dx=∫00.5​[1−x4+x8−x12+⋅⋅⋅+(−1)nx4n+⋅⋅⋅]dx=(x−51​x5+91​x9−131​x13+⋅⋅⋅) ​00.5​=        21​−51​⋅251​+91​⋅291​−131​⋅2131​+⋅⋅⋅，上式右端为一交错级数，有∣r3​∣≤u4​=131​⋅2131​≈0.000009<10−4，        取3项，取五位小数，得∫00.5​1+x41​dx≈21​−51​⋅251​+91​⋅291​≈0.4940.  (2) 因为arctan x=x−3x3​+5x5​−⋅⋅⋅+(−1)n2n+1x2n+1​+⋅⋅⋅ (−1<x<1)，所以∫00.5​xarctan x​dx=        ∫00.5​[1−3x3​+5x4​−⋅⋅⋅+(−1)n2n+1x2n​+⋅⋅⋅]dx=(x−9x3​+25x5​−49x7​+⋅⋅⋅) ​00.5​=        21​−91​⋅231​+251​⋅251​−491​⋅271​+⋅⋅⋅，因为∣r3​∣≤u4​=491​⋅271​≈0.0002<10−3，取3项，        取四位小数，得∫00.5​xarctan x​dx≈21​−91​⋅231​+251​⋅251​≈0.487.​​

3.  试用幂级数求下列各微分方程的解： \begin{aligned}&3. \ 试用幂级数求下列各微分方程的解：&\end{aligned} ​3. 试用幂级数求下列各微分方程的解：​​

   ( 1 )    y ′ − x y − x = 1 ；    ( 2 )    y ′ ′ + x y ′ + y = 0 ；    ( 3 )    ( 1 − x ) y ′ = x 2 − y . \begin{aligned} &\ \ (1)\ \ y'-xy-x=1；\\\\ &\ \ (2)\ \ y''+xy'+y=0；\\\\ &\ \ (3)\ \ (1-x)y'=x^2-y. & \end{aligned} ​  (1)  y′−xy−x=1；  (2)  y′′+xy′+y=0；  (3)  (1−x)y′=x2−y.​​

解：

   ( 1 )  设方程的解为 y = a 0 + a 1 x + a 2 x 2 + ⋅ ⋅ ⋅ + a n x n + ⋅ ⋅ ⋅  （ a 0 为任意常数），代入方程，         有 y ′ = a 1 + 2 a 2 x + 3 a 3 x 2 + ⋅ ⋅ ⋅ + ( n + 1 ) a n + 1 x n + ⋅ ⋅ ⋅ ， − x y = − a 0 x − a 1 x 2 − ⋅ ⋅ ⋅ − a n − 1 x n − ⋅ ⋅ ⋅ ， − x = − x ，          1 = a 1 + ( 2 a 2 − a 0 − 1 ) x + ( 3 a 3 − a 1 ) x 2 + ⋅ ⋅ ⋅ + [ ( n + 1 ) a n + 1 − a n − 1 ] x n + ⋅ ⋅ ⋅ ，比较系数可得， a 1 = 1 ，          a 2 = a 0 + 1 2 ， a 3 = 1 3 ， a 4 = a 2 4 = a 0 + 1 2 ⋅ 4 ， a 5 = a 3 5 = 1 3 ⋅ 5 ， a 6 = a 4 6 = a 0 + 1 2 ⋅ 4 ⋅ 6 ,   ⋅ ⋅ ⋅ ，          a 2 n − 1 = 1 3 ⋅ 5 ⋅   ⋅ ⋅ ⋅   ⋅ ( 2 n − 1 ) ， a 2 n = a 0 + 1 2 ⋅ 4 ⋅ 6 ⋅   ⋅ ⋅ ⋅   ⋅ 2 n = a 0 + 1 n ! 2 n ，         可知 ∑ n = 1 ∞ a 2 n − 1 x 2 n − 1 ， ∑ n = 0 ∞ a 2 n x 2 n 的收敛域是 ( − ∞ ,   + ∞ ) ，         所以 y = ∑ n = 0 ∞ a n x n = ∑ n = 1 ∞ a 2 n − 1 x 2 n − 1 + ∑ n = 0 ∞ a 2 n x 2 n = ∑ n = 1 ∞ x 2 n − 1 3 ⋅ 5 ⋅   ⋅ ⋅ ⋅   ⋅ ( 2 n − 1 ) + ( a 0 + 1 ) ∑ n = 0 ∞ x 2 n n ! 2 n − 1 =          ∑ n = 1 ∞ x 2 n − 1 3 ⋅ 5 ⋅   ⋅ ⋅ ⋅   ⋅ ( 2 n − 1 ) + ( a 0 + 1 ) ∑ n = 0 ∞ 1 n ! ( x 2 2 ) n − 1 ，因为 ∑ n = 0 ∞ 1 n ! ( x 2 2 ) n = e x 2 2 ，         记 a 0 + 1 = C ， 1 ⋅ 3 ⋅ 5 ⋅   ⋅ ⋅ ⋅   ⋅ ( 2 n − 1 ) = ( 2 n − 1 ) ! ! ，所以 y = C e x 2 2 + ∑ n = 1 ∞ 1 ( 2 n − 1 ) ! ! x 2 n − 1 − 1 ， x ∈ ( − ∞ ,   + ∞ ) .    ( 2 )  设方程的解 y = ∑ n = 0 ∞ a n x n ，其中 a 0 ， a 1 是任意常数，则 y ′ = ∑ n = 1 ∞ n a n x n − 1 ，          y ′ ′ = ∑ n = 2 ∞ n ( n − 1 ) a n x n − 2 = ∑ n = 0 ∞ ( n + 2 ) ( n + 1 ) a n + 2 x n ，代入方程 y ′ ′ + x y ′ + y = 0 ，         得 ∑ n = 0 ∞ [ ( n + 2 ) ( n + 1 ) a n + 2 + n a n + a n ] x n = 0 ，则有 ( n + 2 ) ( n + 1 ) a n + 2 + ( n + 1 ) a n = 0 ，          a n + 2 = − a n n + 1   ( n = 0 ， 1 ， 2 ， ⋅ ⋅ ⋅ ) ，当 n = 2 ( k − 1 ) 时， a 2 k = ( − 1 2 k ) a 2 k − 2 =          ( − 1 2 k ) ( − 1 2 k − 2 ) ⋅ ⋅ ⋅ ( − 1 2 ) a 0 = a 0 ( − 1 ) k k ! 2 k ，         当 n = 2 k − 1 时， a 2 k + 1 = ( − 1 2 k + 1 ) a 2 k − 1 = ( − 1 2 k + 1 ) ( − 1 2 k − 1 ) ⋅ ⋅ ⋅ ( − 1 3 ) a 1 = a 1 ( − 1 ) k ( 2 k + 1 ) ! ! ，         因为 ∑ n = 0 ∞ a 2 n x 2 n ， ∑ n = 0 ∞ a 2 n + 1 x 2 n + 1 的收敛域为 ( − ∞ ,   + ∞ ) ，所以 y = ∑ n = 0 ∞ a n x n = ∑ n = 0 ∞ a 2 n x 2 n + ∑ n = 0 ∞ a 2 n + 1 x 2 n + 1 =           ∑ n = 0 ∞ a 0 ( − 1 ) n n ! 2 n x 2 n + ∑ n = 0 ∞ a 1 ( − 1 ) n ( 2 n + 1 ) ! ! x 2 n + 1 ，即 y = a 0 e − x 2 2 + a 1 ∑ n = 0 ∞ ( − 1 ) n ( 2 n + 1 ) ! ! x 2 n + 1 ， x ∈ ( − ∞ ,   + ∞ ) .    ( 3 )  设方程的解 y = ∑ n = 0 ∞ a n x n ，代入方程，得 ( 1 − x ) ∑ n = 1 ∞ n a n x n − 1 = x 2 − ∑ n = 0 ∞ a n x n ，         有 ∑ n = 1 ∞ n a n x n − 1 − ∑ n = 1 ∞ n a n x n + ∑ n = 0 ∞ a n x n = x 2 ，上式第一个级数 ∑ n = 1 ∞ n a n x n − 1 = ∑ n = 0 ∞ ( n + 1 ) a n + 1 x n ，         有 ∑ n = 0 ∞ [ ( n + 1 ) a n + 1 + ( 1 − n ) a n ] x n = x 2 ，比较系数，得 a 1 + a 0 = 0 ， 2 a 2 = 0 ， 3 a 3 − a 2 = 1 ，          ( n + 1 ) a n + 1 + ( 1 − n ) a n = 0   ( n ≥ 3 ) ，即 a 1 = − a 0 ， a 2 = 0 ， a 3 = 1 3 ， a n + 1 = n − 1 n + 1 a n   ( n ≥ 3 ) ，         或 a n = n − 2 n a n − 1 = n − 2 n ⋅ n − 3 n − 1 ⋅ n − 4 n − 2 ⋅   ⋅ ⋅ ⋅   ⋅ 2 4 ⋅ 1 3 = 2 n ( n − 1 )   ( n ≥ 4 ) ，         于是 y = a 0 − a 0 x + 1 3 x 3 + 1 6 x 4 + 1 10 x 5 + ⋅ ⋅ ⋅ + 2 n ( n − 1 ) x n + ⋅ ⋅ ⋅ ，         或 y = a 0 ( 1 − x ) + x 3 [ 1 3 + 1 6 x + 1 10 x 2 + ⋅ ⋅ ⋅ + 2 ( n + 2 ) ( n + 3 ) x n + ⋅ ⋅ ⋅ ] . \begin{aligned} &\ \ (1)\ 设方程的解为y=a_0+a_1x+a_2x^2+\cdot\cdot\cdot+a_nx^n+\cdot\cdot\cdot\ （a_0为任意常数），代入方程，\\\\ &\ \ \ \ \ \ \ \ 有y'=a_1+2a_2x+3a_3x^2+\cdot\cdot\cdot+(n+1)a_{n+1}x^n+\cdot\cdot\cdot，-xy=-a_0x-a_1x^2-\cdot\cdot\cdot-a_{n-1}x^n-\cdot\cdot\cdot，-x=-x，\\\\ &\ \ \ \ \ \ \ \ 1=a_1+(2a_2-a_0-1)x+(3a_3-a_1)x^2+\cdot\cdot\cdot+[(n+1)a_{n+1}-a_{n-1}]x^n+\cdot\cdot\cdot，比较系数可得，a_1=1，\\\\ &\ \ \ \ \ \ \ \ a_2=\frac{a_0+1}{2}，a_3=\frac{1}{3}，a_4=\frac{a_2}{4}=\frac{a_0+1}{2 \cdot 4}，a_5=\frac{a_3}{5}=\frac{1}{3\cdot 5}，a_6=\frac{a_4}{6}=\frac{a_0+1}{2\cdot 4\cdot 6},\ \cdot\cdot\cdot，\\\\ &\ \ \ \ \ \ \ \ a_{2n-1}=\frac{1}{3\cdot 5\cdot\ \cdot\cdot\cdot\ \cdot(2n-1)}，a_{2n}=\frac{a_0+1}{2\cdot 4\cdot 6 \cdot\ \cdot\cdot\cdot\ \cdot 2n}=\frac{a_0+1}{n!2^n}，\\\\ &\ \ \ \ \ \ \ \ 可知\sum_{n=1}^{\infty}a_{2n-1}x^{2n-1}，\sum_{n=0}^{\infty}a_{2n}x^{2n}的收敛域是(-\infty, \ +\infty)，\\\\ &\ \ \ \ \ \ \ \ 所以y=\sum_{n=0}^{\infty}a_nx^n=\sum_{n=1}^{\infty}a_{2n-1}x^{2n-1}+\sum_{n=0}^{\infty}a_{2n}x^{2n}=\sum_{n=1}^{\infty}\frac{x^{2n-1}}{3\cdot 5\cdot\ \cdot\cdot\cdot\ \cdot (2n-1)}+(a_0+1)\sum_{n=0}^{\infty}\frac{x^{2n}}{n!2^n}-1=\\\\ &\ \ \ \ \ \ \ \ \sum_{n=1}^{\infty}\frac{x^{2n-1}}{3\cdot 5\cdot\ \cdot\cdot\cdot\ \cdot (2n-1)}+(a_0+1)\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{x^2}{2}\right)^n-1，因为\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{x^2}{2}\right)^n=e^{\frac{x^2}{2}}，\\\\ &\ \ \ \ \ \ \ \ 记a_0+1=C，1\cdot 3\cdot 5\cdot\ \cdot\cdot\cdot\ \cdot (2n-1)=(2n-1)!!，所以y=Ce^{\frac{x^2}{2}}+\sum_{n=1}^{\infty}\frac{1}{(2n-1)!!}x^{2n-1}-1，x \in (-\infty, \ +\infty).\\\\ &\ \ (2)\ 设方程的解y=\sum_{n=0}^{\infty}a_nx^n，其中a_0，a_1是任意常数，则y'=\sum_{n=1}^{\infty}na_nx^{n-1}，\\\\ &\ \ \ \ \ \ \ \ y''=\sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}=\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^n，代入方程y''+xy'+y=0，\\\\ &\ \ \ \ \ \ \ \ 得\sum_{n=0}^{\infty}[(n+2)(n+1)a_{n+2}+na_n+a_n]x^n=0，则有(n+2)(n+1)a_{n+2}+(n+1)a_n=0，\\\\ &\ \ \ \ \ \ \ \ a_{n+2}=-\frac{a_n}{n+1}\ (n=0，1，2，\cdot\cdot\cdot)，当n=2(k-1)时，a_{2k}=\left(-\frac{1}{2k}\right)a_{2k-2}=\\\\ &\ \ \ \ \ \ \ \ \left(-\frac{1}{2k}\right)\left(-\frac{1}{2k-2}\right)\cdot\cdot\cdot\left(-\frac{1}{2}\right)a_0=\frac{a_0(-1)^k}{k!2^k}，\\\\ &\ \ \ \ \ \ \ \ 当n=2k-1时，a_{2k+1}=\left(-\frac{1}{2k+1}\right)a_{2k-1}=\left(-\frac{1}{2k+1}\right)\left(-\frac{1}{2k-1}\right)\cdot\cdot\cdot\left(-\frac{1}{3}\right)a_1=\frac{a_1(-1)^k}{(2k+1)!!}，\\\\ &\ \ \ \ \ \ \ \ 因为\sum_{n=0}^{\infty}a_{2n}x^{2n}，\sum_{n=0}^{\infty}a_{2n+1}x^{2n+1}的收敛域为(-\infty, \ +\infty)，所以y=\sum_{n=0}^{\infty}a_nx^n=\sum_{n=0}^{\infty}a_{2n}x^{2n}+\sum_{n=0}^{\infty}a_{2n+1}x^{2n+1}=\\\\ &\ \ \ \ \ \ \ \ \ \sum_{n=0}^{\infty}\frac{a_0(-1)^n}{n!2^n}x^{2n}+\sum_{n=0}^{\infty}\frac{a_1(-1)^n}{(2n+1)!!}x^{2n+1}，即y=a_0e^{-\frac{x^2}{2}}+a_1\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!!}x^{2n+1}，x \in (-\infty, \ +\infty).\\\\ &\ \ (3)\ 设方程的解y=\sum_{n=0}^{\infty}a_nx^n，代入方程，得(1-x)\sum_{n=1}^{\infty}na_nx^{n-1}=x^2-\sum_{n=0}^{\infty}a_nx^n，\\\\ &\ \ \ \ \ \ \ \ 有\sum_{n=1}^{\infty}na_nx^{n-1}-\sum_{n=1}^{\infty}na_nx^n+\sum_{n=0}^{\infty}a_nx^n=x^2，上式第一个级数\sum_{n=1}^{\infty}na_nx^{n-1}=\sum_{n=0}^{\infty}(n+1)a_{n+1}x^n，\\\\ &\ \ \ \ \ \ \ \ 有\sum_{n=0}^{\infty}[(n+1)a_{n+1}+(1-n)a_n]x^n=x^2，比较系数，得a_1+a_0=0，2a_2=0，3a_3-a_2=1，\\\\ &\ \ \ \ \ \ \ \ (n+1)a_{n+1}+(1-n)a_n=0\ (n \ge 3)，即a_1=-a_0，a_2=0，a_3=\frac{1}{3}，a_{n+1}=\frac{n-1}{n+1}a_n\ (n \ge 3)，\\\\ &\ \ \ \ \ \ \ \ 或a_n=\frac{n-2}{n}a_{n-1}=\frac{n-2}{n}\cdot\frac{n-3}{n-1}\cdot\frac{n-4}{n-2}\cdot\ \cdot\cdot\cdot\ \cdot \frac{2}{4}\cdot\frac{1}{3}=\frac{2}{n(n-1)}\ (n \ge 4)，\\\\ &\ \ \ \ \ \ \ \ 于是y=a_0-a_0x+\frac{1}{3}x^3+\frac{1}{6}x^4+\frac{1}{10}x^5+\cdot\cdot\cdot+\frac{2}{n(n-1)}x^n+\cdot\cdot\cdot，\\\\ &\ \ \ \ \ \ \ \ 或y=a_0(1-x)+x^3\left[\frac{1}{3}+\frac{1}{6}x+\frac{1}{10}x^2+\cdot\cdot\cdot+\frac{2}{(n+2)(n+3)}x^n+\cdot\cdot\cdot\right]. & \end{aligned} ​  (1) 设方程的解为y=a0​+a1​x+a2​x2+⋅⋅⋅+an​xn+⋅⋅⋅ （a0​为任意常数），代入方程，        有y′=a1​+2a2​x+3a3​x2+⋅⋅⋅+(n+1)an+1​xn+⋅⋅⋅，−xy=−a0​x−a1​x2−⋅⋅⋅−an−1​xn−⋅⋅⋅，−x=−x，        1=a1​+(2a2​−a0​−1)x+(3a3​−a1​)x2+⋅⋅⋅+[(n+1)an+1​−an−1​]xn+⋅⋅⋅，比较系数可得，a1​=1，        a2​=2a0​+1​，a3​=31​，a4​=4a2​​=2⋅4a0​+1​，a5​=5a3​​=3⋅51​，a6​=6a4​​=2⋅4⋅6a0​+1​, ⋅⋅⋅，        a2n−1​=3⋅5⋅ ⋅⋅⋅ ⋅(2n−1)1​，a2n​=2⋅4⋅6⋅ ⋅⋅⋅ ⋅2na0​+1​=n!2na0​+1​，        可知n=1∑∞​a2n−1​x2n−1，n=0∑∞​a2n​x2n的收敛域是(−∞, +∞)，        所以y=n=0∑∞​an​xn=n=1∑∞​a2n−1​x2n−1+n=0∑∞​a2n​x2n=n=1∑∞​3⋅5⋅ ⋅⋅⋅ ⋅(2n−1)x2n−1​+(a0​+1)n=0∑∞​n!2nx2n​−1=        n=1∑∞​3⋅5⋅ ⋅⋅⋅ ⋅(2n−1)x2n−1​+(a0​+1)n=0∑∞​n!1​(2x2​)n−1，因为n=0∑∞​n!1​(2x2​)n=e2x2​，        记a0​+1=C，1⋅3⋅5⋅ ⋅⋅⋅ ⋅(2n−1)=(2n−1)!!，所以y=Ce2x2​+n=1∑∞​(2n−1)!!1​x2n−1−1，x∈(−∞, +∞).  (2) 设方程的解y=n=0∑∞​an​xn，其中a0​，a1​是任意常数，则y′=n=1∑∞​nan​xn−1，        y′′=n=2∑∞​n(n−1)an​xn−2=n=0∑∞​(n+2)(n+1)an+2​xn，代入方程y′′+xy′+y=0，        得n=0∑∞​[(n+2)(n+1)an+2​+nan​+an​]xn=0，则有(n+2)(n+1)an+2​+(n+1)an​=0，        an+2​=−n+1an​​ (n=0，1，2，⋅⋅⋅)，当n=2(k−1)时，a2k​=(−2k1​)a2k−2​=        (−2k1​)(−2k−21​)⋅⋅⋅(−21​)a0​=k!2ka0​(−1)k​，        当n=2k−1时，a2k+1​=(−2k+11​)a2k−1​=(−2k+11​)(−2k−11​)⋅⋅⋅(−31​)a1​=(2k+1)!!a1​(−1)k​，        因为n=0∑∞​a2n​x2n，n=0∑∞​a2n+1​x2n+1的收敛域为(−∞, +∞)，所以y=n=0∑∞​an​xn=n=0∑∞​a2n​x2n+n=0∑∞​a2n+1​x2n+1=         n=0∑∞​n!2na0​(−1)n​x2n+n=0∑∞​(2n+1)!!a1​(−1)n​x2n+1，即y=a0​e−2x2​+a1​n=0∑∞​(2n+1)!!(−1)n​x2n+1，x∈(−∞, +∞).  (3) 设方程的解y=n=0∑∞​an​xn，代入方程，得(1−x)n=1∑∞​nan​xn−1=x2−n=0∑∞​an​xn，        有n=1∑∞​nan​xn−1−n=1∑∞​nan​xn+n=0∑∞​an​xn=x2，上式第一个级数n=1∑∞​nan​xn−1=n=0∑∞​(n+1)an+1​xn，        有n=0∑∞​[(n+1)an+1​+(1−n)an​]xn=x2，比较系数，得a1​+a0​=0，2a2​=0，3a3​−a2​=1，        (n+1)an+1​+(1−n)an​=0 (n≥3)，即a1​=−a0​，a2​=0，a3​=31​，an+1​=n+1n−1​an​ (n≥3)，        或an​=nn−2​an−1​=nn−2​⋅n−1n−3​⋅n−2n−4​⋅ ⋅⋅⋅ ⋅42​⋅31​=n(n−1)2​ (n≥4)，        于是y=a0​−a0​x+31​x3+61​x4+101​x5+⋅⋅⋅+n(n−1)2​xn+⋅⋅⋅，        或y=a0​(1−x)+x3[31​+61​x+101​x2+⋅⋅⋅+(n+2)(n+3)2​xn+⋅⋅⋅].​​

4.  试用幂级数求下列方程满足所给初值条件的特解： \begin{aligned}&4. \ 试用幂级数求下列方程满足所给初值条件的特解：&\end{aligned} ​4. 试用幂级数求下列方程满足所给初值条件的特解：​​

   ( 1 )    y ′ = y 2 + x 3 ， y ∣ x = 0 = 1 2 ；    ( 2 )    ( 1 − x ) y ′ + y = 1 + x ， y ∣ x = 0 = 0. \begin{aligned} &\ \ (1)\ \ y'=y^2+x^3，y|_{x=0}=\frac{1}{2}；\\\\ &\ \ (2)\ \ (1-x)y'+y=1+x，y|_{x=0}=0. & \end{aligned} ​  (1)  y′=y2+x3，y∣x=0​=21​；  (2)  (1−x)y′+y=1+x，y∣x=0​=0.​​

解：

   ( 1 )  因为 y ∣ x = 0 = 1 2 ，所以设方程特解为 y = 1 2 + ∑ n = 1 ∞ a n x n ，有 y ′ = ∑ n = 1 ∞ n a n x n − 1 = a 1 + ∑ n = 1 ∞ ( n + 1 ) a n + 1 x n ，         代入方程，有 a 1 + ∑ n = 1 ∞ ( n + 1 ) a n + 1 x n = x 3 + ( 1 2 + ∑ n = 1 ∞ a n x n ) 2 = x 3 + 1 4 + ∑ n = 1 ∞ a n x n + ( ∑ n = 1 ∞ a n x n ) 2 =          x 3 + 1 4 + ∑ n = 1 ∞ a n x n + [ a 1 2 x 2 + 2 a 1 a 2 x 3 + ( a 2 2 + 2 a 1 a 3 ) x 4 + ⋅ ⋅ ⋅ + ( ∑ i + j = n a i a j ) x n + ⋅ ⋅ ⋅ ] ，即          a 1 + ( 2 a 2 − a 1 ) x + ( 3 a 3 − a 2 − a 1 2 ) x 2 + ( 4 a 4 − a 3 − 2 a 1 a 2 ) x 3 + ⋅ ⋅ ⋅ + [ ( n + 1 ) a n + 1 − a n − ∑ i + j = n a i a j ] x n + ⋅ ⋅ ⋅ =          1 4 + x 3 ，比较系数，得 a 1 = 1 4 ， 2 a 2 − a 1 = 0 ， 3 a 3 − a 2 − a 1 2 = 0 ， 4 a 4 − a 3 − 2 a 1 a 2 = 1 ， ⋅ ⋅ ⋅ ，          ( n + 1 ) a n + 1 − a n − ∑ i + j = n a i a j = 0   ( n ≥ 4 ) ，解得 a 1 = 1 4 ， a 2 = 1 8 ， a 3 = 1 16 ， a 4 = 9 32 ， ⋅ ⋅ ⋅ ，         所以 y = 1 2 + 1 4 x + 1 8 x 2 + 1 16 x 3 + 9 32 x 4 + ⋅ ⋅ ⋅ .    ( 2 )  因为 y ∣ x = 0 = 0 ，所以设 y = ∑ n = 1 ∞ a n x n 是方程的特解，则 y ′ = ∑ n = 1 ∞ n a n x n − 1 ，代入方程，有          ( 1 − x ) ∑ n = 1 ∞ n a n x n − 1 + ∑ n = 1 ∞ a n x n = 1 + x ，即 ∑ n = 1 ∞ n a n x n − 1 − ∑ n = 1 ∞ n a n x n + ∑ n = 1 ∞ a n x n = 1 + x ，         或 a 1 + ∑ n = 1 ∞ [ ( n + 1 ) a n + 1 + ( 1 − n ) a n ] x n = 1 + x ，比较系数，得 a 1 = 1 ， a 2 = 1 2 ， a n + 1 = n − 1 n + 1 a n   ( n ≥ 2 ) ，         或 a n = n − 2 n a n − 1 = ( n − 2 ) ( n − 3 ) ⋅ ⋅ ⋅ 1 n ( n − 1 ) ⋅ ⋅ ⋅ 3 ⋅ 1 2 = 1 n ( n − 1 )   ( n ≥ 3 ) ，         所以 y = x + 1 2 x 2 + 1 6 x 3 + ⋅ ⋅ ⋅ + 1 n ( n − 1 ) x n + ⋅ ⋅ ⋅ . \begin{aligned} &\ \ (1)\ 因为y|_{x=0}=\frac{1}{2}，所以设方程特解为y=\frac{1}{2}+\sum_{n=1}^{\infty}a_nx^n，有y'=\sum_{n=1}^{\infty}na_nx^{n-1}=a_1+\sum_{n=1}^{\infty}(n+1)a_{n+1}x^n，\\\\ &\ \ \ \ \ \ \ \ 代入方程，有a_1+\sum_{n=1}^{\infty}(n+1)a_{n+1}x^n=x^3+\left(\frac{1}{2}+\sum_{n=1}^{\infty}a_nx^n\right)^2=x^3+\frac{1}{4}+\sum_{n=1}^{\infty}a_nx^n+\left(\sum_{n=1}^{\infty}a_nx^n\right)^2=\\\\ &\ \ \ \ \ \ \ \ x^3+\frac{1}{4}+\sum_{n=1}^{\infty}a_nx^n+\left[a_1^2x^2+2a_1a_2x^3+(a_2^2+2a_1a_3)x^4+\cdot\cdot\cdot+\left(\sum_{i+j=n}a_ia_j\right)x^n+\cdot\cdot\cdot\right]，即\\\\ &\ \ \ \ \ \ \ \ a_1+(2a_2-a_1)x+(3a_3-a_2-a_1^2)x^2+(4a_4-a_3-2a_1a_2)x^3+\cdot\cdot\cdot+\left[(n+1)a_{n+1}-a_n-\sum_{i+j=n}a_ia_j\right]x^n+\cdot\cdot\cdot=\\\\ &\ \ \ \ \ \ \ \ \frac{1}{4}+x^3，比较系数，得a_1=\frac{1}{4}，2a_2-a_1=0，3a_3-a_2-a_1^2=0，4a_4-a_3-2a_1a_2=1，\cdot\cdot\cdot，\\\\ &\ \ \ \ \ \ \ \ (n+1)a_{n+1}-a_n-\sum_{i+j=n}a_ia_j=0\ (n \ge 4)，解得a_1=\frac{1}{4}，a_2=\frac{1}{8}，a_3=\frac{1}{16}，a_4=\frac{9}{32}，\cdot\cdot\cdot，\\\\ &\ \ \ \ \ \ \ \ 所以y=\frac{1}{2}+\frac{1}{4}x+\frac{1}{8}x^2+\frac{1}{16}x^3+\frac{9}{32}x^4+\cdot\cdot\cdot.\\\\ &\ \ (2)\ 因为y|_{x=0}=0，所以设y=\sum_{n=1}^{\infty}a_nx^n是方程的特解，则y'=\sum_{n=1}^{\infty}na_nx^{n-1}，代入方程，有\\\\ &\ \ \ \ \ \ \ \ (1-x)\sum_{n=1}^{\infty}na_nx^{n-1}+\sum_{n=1}^{\infty}a_nx^n=1+x，即\sum_{n=1}^{\infty}na_nx^{n-1}-\sum_{n=1}^{\infty}na_nx^n+\sum_{n=1}^{\infty}a_nx^n=1+x，\\\\ &\ \ \ \ \ \ \ \ 或a_1+\sum_{n=1}^{\infty}[(n+1)a_{n+1}+(1-n)a_n]x^n=1+x，比较系数，得a_1=1，a_2=\frac{1}{2}，a_{n+1}=\frac{n-1}{n+1}a_n\ (n \ge 2)，\\\\ &\ \ \ \ \ \ \ \ 或a_n=\frac{n-2}{n}a_{n-1}=\frac{(n-2)(n-3)\cdot\cdot\cdot1}{n(n-1)\cdot\cdot\cdot3}\cdot\frac{1}{2}=\frac{1}{n(n-1)}\ (n \ge 3)，\\\\ &\ \ \ \ \ \ \ \ 所以y=x+\frac{1}{2}x^2+\frac{1}{6}x^3+\cdot\cdot\cdot+\frac{1}{n(n-1)}x^n+\cdot\cdot\cdot. & \end{aligned} ​  (1) 因为y∣x=0​=21​，所以设方程特解为y=21​+n=1∑∞​an​xn，有y′=n=1∑∞​nan​xn−1=a1​+n=1∑∞​(n+1)an+1​xn，        代入方程，有a1​+n=1∑∞​(n+1)an+1​xn=x3+(21​+n=1∑∞​an​xn)2=x3+41​+n=1∑∞​an​xn+(n=1∑∞​an​xn)2=        x3+41​+n=1∑∞​an​xn+[a12​x2+2a1​a2​x3+(a22​+2a1​a3​)x4+⋅⋅⋅+(i+j=n∑​ai​aj​)xn+⋅⋅⋅]，即        a1​+(2a2​−a1​)x+(3a3​−a2​−a12​)x2+(4a4​−a3​−2a1​a2​)x3+⋅⋅⋅+[(n+1)an+1​−an​−i+j=n∑​ai​aj​]xn+⋅⋅⋅=        41​+x3，比较系数，得a1​=41​，2a2​−a1​=0，3a3​−a2​−a12​=0，4a4​−a3​−2a1​a2​=1，⋅⋅⋅，        (n+1)an+1​−an​−i+j=n∑​ai​aj​=0 (n≥4)，解得a1​=41​，a2​=81​，a3​=161​，a4​=329​，⋅⋅⋅，        所以y=21​+41​x+81​x2+161​x3+329​x4+⋅⋅⋅.  (2) 因为y∣x=0​=0，所以设y=n=1∑∞​an​xn是方程的特解，则y′=n=1∑∞​nan​xn−1，代入方程，有        (1−x)n=1∑∞​nan​xn−1+n=1∑∞​an​xn=1+x，即n=1∑∞​nan​xn−1−n=1∑∞​nan​xn+n=1∑∞​an​xn=1+x，        或a1​+n=1∑∞​[(n+1)an+1​+(1−n)an​]xn=1+x，比较系数，得a1​=1，a2​=21​，an+1​=n+1n−1​an​ (n≥2)，        或an​=nn−2​an−1​=n(n−1)⋅⋅⋅3(n−2)(n−3)⋅⋅⋅1​⋅21​=n(n−1)1​ (n≥3)，        所以y=x+21​x2+61​x3+⋅⋅⋅+n(n−1)1​xn+⋅⋅⋅.​​

5.  验证函数 y ( x ) = 1 + x 3 3 ! + x 6 6 ! + ⋅ ⋅ ⋅ + x 3 n ( 3 n ) ! + ⋅ ⋅ ⋅ ( − ∞ < x < + ∞ ) 满足微分方程 y ′ ′ + y ′ + y = e x ，并     利用此结果求幂级数 ∑ n = 0 ∞ x 3 n ( 3 n ) ! 的和函数 . \begin{aligned}&5. \ 验证函数y(x)=1+\frac{x^3}{3!}+\frac{x^6}{6!}+\cdot\cdot\cdot+\frac{x^{3n}}{(3n)!}+\cdot\cdot\cdot(-\infty \lt x \lt +\infty)满足微分方程y''+y'+y=e^x，并\\\\&\ \ \ \ 利用此结果求幂级数\sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!}的和函数.&\end{aligned} ​5. 验证函数y(x)=1+3!x3​+6!x6​+⋅⋅⋅+(3n)!x3n​+⋅⋅⋅(−∞<x<+∞)满足微分方程y′′+y′+y=ex，并    利用此结果求幂级数n=0∑∞​(3n)!x3n​的和函数.​​

解：

  因为 y ( x ) = 1 + x 3 3 ! + x 6 6 ! + ⋅ ⋅ ⋅ + x 3 n ( 3 n ) ! + ⋅ ⋅ ⋅ ，    y ′ ( x ) = x 2 2 ! + x 5 5 ! + ⋅ ⋅ ⋅ + x 3 n − 1 ( 3 n − 1 ) ! + ⋅ ⋅ ⋅ ，    y ′ ′ ( x ) = x + x 4 4 ! + ⋅ ⋅ ⋅ + x 3 n − 2 ( 3 n − 2 ) ! + ⋅ ⋅ ⋅ ，则 y ′ ′ ( x ) + y ′ ( x ) + y ( x ) = ∑ n = 0 ∞ x n n ! = e x ，   所以函数 y ( x ) 满足微分方程 y ′ ′ + y ′ + y = e x ，    y ′ ′ + y ′ + y = e x 对应的齐次方程 y ′ ′ + y ′ + y = 0 的特征方程为 r 2 + r + 1 = 0 ，根为 r 1 ， r 2 = − 1 2 ± 3 2 i ，   因此齐次方程的通解为 Y = e − x 2 ( C 1 c o s 3 2 x + C 2 s i n 3 2 x ) ，设非齐次微分方程的特解为 y ∗ = A e x ，   代入方程 y ′ ′ + y ′ + y = e x ，得 A = 1 3 ，则 y ∗ = 1 3 e x ，非齐次微分方程的通解为    y = Y + y ∗ = e − x 2 ( C 1 c o s 3 2 x + C 2 s i n 3 2 x ) + 1 3 e x ，幂级数的和函数 y ( x ) 满足， y ( 0 ) = 1 ， y ′ ( 0 ) = 0 ，   则 y ( 0 ) = 1 = C 1 + 1 3 ， y ′ ( 0 ) = 0 = − 1 2 C 1 + 3 2 C 2 + 1 3 ，得 C 1 = 2 3 ， C 2 = 0 ，根据微分方程初值问题解的唯一性，   可得幂级数的和函数为 y ( x ) = 2 3 e − x 2 c o s 3 2 x + 1 3 e x   ( − ∞ < x < + ∞ ) . \begin{aligned} &\ \ 因为y(x)=1+\frac{x^3}{3!}+\frac{x^6}{6!}+\cdot\cdot\cdot+\frac{x^{3n}}{(3n)!}+\cdot\cdot\cdot，\\\\ &\ \ y'(x)=\frac{x^2}{2!}+\frac{x^5}{5!}+\cdot\cdot\cdot+\frac{x^{3n-1}}{(3n-1)!}+\cdot\cdot\cdot，\\\\ &\ \ y''(x)=x+\frac{x^4}{4!}+\cdot\cdot\cdot+\frac{x^{3n-2}}{(3n-2)!}+\cdot\cdot\cdot，则y''(x)+y'(x)+y(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}=e^x，\\\\ &\ \ 所以函数y(x)满足微分方程y''+y'+y=e^x，\\\\ &\ \ y''+y'+y=e^x对应的齐次方程y''+y'+y=0的特征方程为r^2+r+1=0，根为r_1，r_2=-\frac{1}{2}\pm \frac{\sqrt{3}}{2}i，\\\\ &\ \ 因此齐次方程的通解为Y=e^{-\frac{x}{2}}\left(C_1cos\frac{\sqrt{3}}{2}x+C_2sin\frac{\sqrt{3}}{2}x\right)，设非齐次微分方程的特解为y^*=Ae^x，\\\\ &\ \ 代入方程y''+y'+y=e^x，得A=\frac{1}{3}，则y^*=\frac{1}{3}e^x，非齐次微分方程的通解为\\\\ &\ \ y=Y+y^*=e^{-\frac{x}{2}}\left(C_1cos\frac{\sqrt{3}}{2}x+C_2sin\frac{\sqrt{3}}{2}x\right)+\frac{1}{3}e^x，幂级数的和函数y(x)满足，y(0)=1，y'(0)=0，\\\\ &\ \ 则y(0)=1=C_1+\frac{1}{3}，y'(0)=0=-\frac{1}{2}C_1+\frac{\sqrt{3}}{2}C_2+\frac{1}{3}，得C_1=\frac{2}{3}，C_2=0，根据微分方程初值问题解的唯一性，\\\\ &\ \ 可得幂级数的和函数为y(x)=\frac{2}{3}e^{-\frac{x}{2}}cos\frac{\sqrt{3}}{2}x+\frac{1}{3}e^x\ (-\infty \lt x \lt +\infty). & \end{aligned} ​  因为y(x)=1+3!x3​+6!x6​+⋅⋅⋅+(3n)!x3n​+⋅⋅⋅，  y′(x)=2!x2​+5!x5​+⋅⋅⋅+(3n−1)!x3n−1​+⋅⋅⋅，  y′′(x)=x+4!x4​+⋅⋅⋅+(3n−2)!x3n−2​+⋅⋅⋅，则y′′(x)+y′(x)+y(x)=n=0∑∞​n!xn​=ex，  所以函数y(x)满足微分方程y′′+y′+y=ex，  y′′+y′+y=ex对应的齐次方程y′′+y′+y=0的特征方程为r2+r+1=0，根为r1​，r2​=−21​±23 ​​i，  因此齐次方程的通解为Y=e−2x​(C1​cos23 ​​x+C2​sin23 ​​x)，设非齐次微分方程的特解为y∗=Aex，  代入方程y′′+y′+y=ex，得A=31​，则y∗=31​ex，非齐次微分方程的通解为  y=Y+y∗=e−2x​(C1​cos23 ​​x+C2​sin23 ​​x)+31​ex，幂级数的和函数y(x)满足，y(0)=1，y′(0)=0，  则y(0)=1=C1​+31​，y′(0)=0=−21​C1​+23 ​​C2​+31​，得C1​=32​，C2​=0，根据微分方程初值问题解的唯一性，  可得幂级数的和函数为y(x)=32​e−2x​cos23 ​​x+31​ex (−∞<x<+∞).​​

6.  利用欧拉公式将函数 e x c o s   x 展开成 x 的幂级数 . \begin{aligned}&6. \ 利用欧拉公式将函数e^xcos\ x展开成x的幂级数.&\end{aligned} ​6. 利用欧拉公式将函数excos x展开成x的幂级数.​​

解：

  根据欧拉公式 e i x = c o s   x + i   s i n   x 可知， c o s   x = R e ( e i x ) ，所以 e x c o s   x = e x ⋅ R e ( e i x ) = R e ( e x ⋅ e i x ) = R e [ e ( 1 + i ) x ] ，   因为 e ( 1 + i ) x = ∑ n = 0 ∞ 1 n ! ( 1 + i ) n x n = ∑ n = 0 ∞ [ 2 ( c o s   π 4 + i   s i n   π 4 ) ] n x n n ! = ∑ n = 0 ∞ ( c o s n π 4 + i   s i n n π 4 ) 2 n 2 ⋅ x n n ! ，    x ∈ ( − ∞ ,   + ∞ ) ，所以 e x c o s   x = R e [ e ( 1 + i ) x ] = ∑ n = 0 ∞ c o s n π 4 ⋅ 2 n 2 ⋅ x n n ! ， x ∈ ( − ∞ ,   + ∞ ) . \begin{aligned} &\ \ 根据欧拉公式e^{ix}=cos\ x+i\ sin\ x可知，cos\ x=Re(e^{ix})，所以e^xcos\ x=e^x\cdot Re(e^{ix})=Re(e^x\cdot e^{ix})=Re[e^{(1+i)x}]，\\\\ &\ \ 因为e^{(1+i)x}=\sum_{n=0}^{\infty}\frac{1}{n!}(1+i)^nx^n=\sum_{n=0}^{\infty}\left[\sqrt{2}\left(cos\ \frac{\pi}{4}+i\ sin\ \frac{\pi}{4}\right)\right]^n\frac{x^n}{n!}=\sum_{n=0}^{\infty}\left(cos\frac{n\pi}{4}+i\ sin\frac{n\pi}{4}\right)2^{\frac{n}{2}}\cdot\frac{x^n}{n!}，\\\\ &\ \ x \in (-\infty, \ +\infty)，所以e^xcos\ x=Re[e^{(1+i)x}]=\sum_{n=0}^{\infty}cos\frac{n\pi}{4}\cdot2^{\frac{n}{2}}\cdot\frac{x^n}{n!}，x \in (-\infty, \ +\infty). & \end{aligned} ​  根据欧拉公式eix=cos x+i sin x可知，cos x=Re(eix)，所以excos x=ex⋅Re(eix)=Re(ex⋅eix)=Re[e(1+i)x]，  因为e(1+i)x=n=0∑∞​n!1​(1+i)nxn=n=0∑∞​[2 ​(cos 4π​+i sin 4π​)]nn!xn​=n=0∑∞​(cos4nπ​+i sin4nπ​)22n​⋅n!xn​，  x∈(−∞, +∞)，所以excos x=Re[e(1+i)x]=n=0∑∞​cos4nπ​⋅22n​⋅n!xn​，x∈(−∞, +∞).​​

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