From a7ad825faed55d16721dccdbd296bda0fcbca032 Mon Sep 17 00:00:00 2001 From: Cym10x Date: Wed, 29 Apr 2026 11:57:24 +0800 Subject: [PATCH] =?UTF-8?q?M=20=E9=AB=98=E6=95=B0=E4=B8=8B/=E4=BA=94?= =?UTF-8?q?=E4=B8=80=E5=A4=8D=E4=B9=A0/DAY5.md=20=E9=AB=98=E6=95=B0?= =?UTF-8?q?=E4=B8=8B/=E4=BA=94=E4=B8=80=E5=A4=8D=E4=B9=A0/=E9=AB=98?= =?UTF-8?q?=E7=AD=89=E6=95=B0=E5=AD=A6=E6=9C=9F=E4=B8=AD=E8=AF=95=E9=A2=98?= =?UTF-8?q?=E5=88=86=E7=B1=BB=E6=B1=87=E7=BC=96=EF=BC=88=E4=B8=80=EF=BC=89?= =?UTF-8?q?=E5=8E=9F=E5=8D=B7=E7=89=88.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 高数下/五一复习/DAY5.md | 7 ------- ...期中试题分类汇编（一）原卷版.md | 14 +++++++++++++- 2 files changed, 13 insertions(+), 8 deletions(-) diff --git a/高数下/五一复习/DAY5.md b/高数下/五一复习/DAY5.md index 0e9c115..7467656 100644 --- a/高数下/五一复习/DAY5.md +++ b/高数下/五一复习/DAY5.md @@ -106,10 +106,3 @@ - **2022年 证明题18(2)** 设 $y=y(x)$ 的反函数满足 $\dfrac{\mathrm{d}^{2}x}{\mathrm{d}y^{2}} + (y+2xe^{x})\left(\dfrac{\mathrm{d}x}{\mathrm{d}y}\right)^{3} = 0$。 (2) 求满足初始条件 $y(0)=0,\ y'(0)=\frac{3}{2}$ 的解。 - - **解析**： - (1) 已证 $y'' - y = 2xe^{x}$。 - 特征方程 $r^{2}-1=0$，$r=\pm1$。设特解 $y^{*} = xe^{x}(ax+b) = e^{x}(ax^{2}+bx)$，代入得 $a=\frac{1}{2}, b=-\frac{1}{2}$，特解为 $y^{*}=e^{x}(\frac{1}{2}x^{2}-\frac{1}{2}x)$。通解 - $$y = C_{1}e^{x} + C_{2}e^{-x} + e^{x}\left(\frac{1}{2}x^{2} - \frac{1}{2}x\right).$$ - 由 $y(0)=0,\ y'(0)=\frac{3}{2}$ 得 $C_{1}=1,\ C_{2}=-1$，故特解为 - $$y = e^{x} - e^{-x} + e^{x}\left(\frac{1}{2}x^{2} - \frac{1}{2}x\right).$$ \ No newline at end of file diff --git a/高数下/五一复习/高等数学期中试题分类汇编（一）原卷版.md b/高数下/五一复习/高等数学期中试题分类汇编（一）原卷版.md index 99845d6..7fe36b3 100644 --- a/高数下/五一复习/高等数学期中试题分类汇编（一）原卷版.md +++ b/高数下/五一复习/高等数学期中试题分类汇编（一）原卷版.md @@ -180,4 +180,16 @@ $$\cos\theta = \frac{\vec{\tau_{1}}\cdot\vec{\tau_{2}}}{|\vec{\tau_{1}}||\vec{\tau_{2}}|} = \frac{x(x-y)+y(x+y)+z^{2}}{\sqrt{x^{2}+y^{2}+z^{2}}\sqrt{(x-y)^{2}+(x+y)^{2}+z^{2}}} = \frac{2z^{2}}{\sqrt{2z^{2}}\sqrt{3z^{2}}} = \frac{2}{\sqrt{6}},$$ 故夹角恒定。 ---- \ No newline at end of file +--- + +### 微分 +- **2022年 证明题18(2)** + 设 $y=y(x)$ 的反函数满足 $\dfrac{\mathrm{d}^{2}x}{\mathrm{d}y^{2}} + (y+2xe^{x})\left(\dfrac{\mathrm{d}x}{\mathrm{d}y}\right)^{3} = 0$。 + (2) 求满足初始条件 $y(0)=0,\ y'(0)=\frac{3}{2}$ 的解。 + + **解析**： + (1) 已证 $y'' - y = 2xe^{x}$。 + 特征方程 $r^{2}-1=0$，$r=\pm1$。设特解 $y^{*} = xe^{x}(ax+b) = e^{x}(ax^{2}+bx)$，代入得 $a=\frac{1}{2}, b=-\frac{1}{2}$，特解为 $y^{*}=e^{x}(\frac{1}{2}x^{2}-\frac{1}{2}x)$。通解 + $$y = C_{1}e^{x} + C_{2}e^{-x} + e^{x}\left(\frac{1}{2}x^{2} - \frac{1}{2}x\right).$$ + 由 $y(0)=0,\ y'(0)=\frac{3}{2}$ 得 $C_{1}=1,\ C_{2}=-1$，故特解为 + $$y = e^{x} - e^{-x} + e^{x}\left(\frac{1}{2}x^{2} - \frac{1}{2}x\right).$$ \ No newline at end of file