fix generation errors for chp07

Author: longfei-chen

chp07/subsections/7.1.html

@@ -721,20 +721,19 @@ <h4>习题1 <span class="difficulty medium">中等难度</span></h4>
 
                 <div class="practice-problem">
                     <h4>习题2 <span class="difficulty hard">较难</span></h4>
-                    <p>验证 $y = \frac{1}{2}x^2 + Cx$ 是微分方程 $x\frac{d^2y}{dx^2} - \frac{dy}{dx} = -1$ 的通解，并求满足初始条件 $y(1) = 2$，$y'(1) = 3$ 的特解。</p>
+                    <p>验证 $y = C_1 x^2 + x + C_2$ 是微分方程 $x\frac{d^2y}{dx^2} - \frac{dy}{dx} = -1$ 的通解，并求满足初始条件 $y(1) = 2$，$y'(1) = 3$ 的特解。</p>
 
                     <div class="solution">
                         <div class="solution-header">解：</div>
                         <p><strong>验证过程：</strong></p>
-                        <p>$y = \frac{1}{2}x^2 + Cx$</p>
-                        <p>$\frac{dy}{dx} = x + C$</p>
-                        <p>$\frac{d^2y}{dx^2} = 1$</p>
-                        <p>代入微分方程：$x \cdot 1 - (x + C) = x - x - C = -C$</p>
-                        <p>但这不等于$-1$，所以需要重新检查...</p>
-                        <p><strong>正确的验证：</strong>设$y = \frac{1}{2}x^2 + C$</p>
-                        <p>$\frac{dy}{dx} = x$，$\frac{d^2y}{dx^2} = 1$</p>
-                        <p>代入：$x \cdot 1 - x = 0 \neq -1$</p>
-                        <p><strong>实际通解：</strong>$y = \frac{x^2}{2} - x + Cx$</p>
+                        <p>$y = C_1 x^2 + x + C_2$</p>
+                        <p>$\frac{dy}{dx} = 2C_1 x + 1$</p>
+                        <p>$\frac{d^2y}{dx^2} = 2C_1$</p>
+                        <p>代入微分方程：$x \cdot (2C_1) - (2C_1 x + 1) = -1$</p>
+                        <p>代入初值，当 $y(1)=2$ 时，得 $C_1 + 1 + C_2 = 2$</p>
+                        <p>当 $y'(1)=3$ 时，得 $2C_1 + 1 = 3$</p>
+                        <p>得，$C_1 = 1, C_2 = 0$</p>
+                        <p>即，$y = x^2 + x$</p>
                     </div>
                 </div>
 

chp07/subsections/7.10.html

@@ -598,7 +598,7 @@ <h4>一阶常系数线性微分方程组</h4>
                     \vdots \\
                     \frac{dx_n}{dt} = a_{n1}x_1 + a_{n2}x_2 + \cdots + a_{nn}x_n + f_n(t)
                     \end{cases}$$
-                    <p>的方程组称为<strong>$n$ 阶一阶常系数线性微分方程组</strong>，其中 $a_{ij}$ 是常数。</p>
+                    <p>的方程组称为<strong>一阶常系数线性微分方程组</strong>，其中 $a_{ij}$ 是常数。</p>
                 </div>
 
                 <div class="matrix-form">
@@ -1063,7 +1063,7 @@ <h4>习题3：二阶微分方程组求解 <span class="difficulty very-hard">很
                         $$y = C_1 e^{\alpha t} + C_2 e^{-\alpha t} + C_3 \cos \beta t + C_4 \sin \beta t + e^t$$
 
                         <p>进而解得：</p>
-                        $$x = -\alpha^3 C_1 e^{\alpha t} + \alpha^3 C_2 e^{-\alpha t} + \beta^3 C_3 \cos \beta t -\beta^3 C_4 \sin \beta t - 2e^t$$
+                        $$x = -\frac{\alpha^2+1}{\alpha} C_1 e^{\alpha t} + \frac{\alpha^2+1}{\alpha} C_2 e^{-\alpha t} - \frac{1-\beta^2}{\beta} C_3 \sin \beta t + \frac{1-\beta^2}{\beta} C_4 \cos \beta t - 2e^t$$
 
                         <div style="background: linear-gradient(135deg, #e8f5e8, #d4edda); padding: 15px; border-radius: 8px; margin: 15px 0;">
                             <p><strong>解题要点：</strong></p>

chp07/subsections/7.3.html

@@ -791,7 +791,7 @@ <h4>例4：比例系数相等的情况</h4>
                         <div class="solution-step">
                             <div class="step-title">步骤1：检查比例关系</div>
                             <p>$\frac{2}{4} = \frac{3}{6} = \frac{1}{2}$，所以 $a_1 : a_2 = b_1 : b_2$</p>
-                            <p>方程可写成：$(2x + 3y + 1)dx + 2(2x + 3y) + 5)dy = 0$</p>
+                            <p>方程可写成：$(2x + 3y + 1)dx + (2(2x + 3y) + 5)dy = 0$</p>
                         </div>
 
                         <div class="solution-step">
@@ -817,9 +817,11 @@ <h4>例4：比例系数相等的情况</h4>
 
                         <div class="solution-step">
                             <div class="step-title">步骤5：求解并回代</div>
-                            <p>这仍然不是直接可分离的。重新考虑，用 $y$ 作为独立变量：</p>
-                            <p>从 $u = 2x + 3y$ 得 $x = \frac{u - 3y}{2}$</p>
-                            <p>原方程最终可化简求解得到隐式解。</p>
+                            <p>上式可进一步分离变量：$\frac{u+1}{u+7}du = -dy$</p>
+                            <p>对左边进行积分得：$\int \frac{u+1}{u+7}du = u - 6 \ln|u+7| + C_1$</p>
+                            <p>对右边进行积分得：$-\int dy = -y + C_2$</p>
+                            <p>所以，$u - 6 \ln|u+7| = -y + C$</p>
+                            <p>代入 $u = 2x + 3y$ 得：x + 2y - 3 \ln|2x + 3y +7| = C</p>
                         </div>
                     </div>
                 </div>
@@ -856,8 +858,7 @@ <h4>习题2 <span class="difficulty hard">较难</span></h4>
                         <p><strong>步骤3：</strong>这是齐次方程，用 $v = \frac{Y}{X}$ 求解</p>
                         <p>$\frac{dY}{dX} = \frac{Y - X}{X + Y} = \frac{v - 1}{1 + v}$</p>
 
-                        <p><strong>步骤4：</strong>分离变量并积分得到解</p>
-                        <p><strong>最终解：</strong>$(x-1)^2 + (y-2)^2 = C(x-1)$</p>
+                        <p><strong>步骤4：</strong>剩下的步骤可以参考上面的例3</p>
                     </div>
                 </div>
 
@@ -884,7 +885,7 @@ <h4>习题3 <span class="difficulty very-hard">很难</span></h4>
                         </div>
 
                         <div class="solution-step">
-                            <div class="step-title">步骤3：建立关于v的微分方程</div>
+                            <div class="step-title">步骤3：建立关于 $v$ 的微分方程</div>
                             <p>代入得：$v + x\frac{dv}{dx} = \frac{1}{v} + 2v$</p>
                             <p>简化：$x\frac{dv}{dx} = \frac{1}{v} + 2v - v = \frac{1}{v} + v$</p>
                             <p>进一步简化：$x\frac{dv}{dx} = \frac{1 + v^2}{v}$</p>

chp07/subsections/7.5.html

@@ -865,7 +865,7 @@ <h4>习题2 <span class="difficulty hard">较难</span></h4>
                         $$2\sqrt{y} = C_1x + C_2$$
                         $$y = \frac{(C_1x + C_2)^2}{4}$$
 
-                        <p><strong>答案：</strong>$y = \frac{(C_1x + C_2)^2}{4}$ 及 $y = 0$</p>
+                        <p><strong>答案：</strong>$y = \frac{(C_1x + C_2)^2}{4}$ 及 $y = C$</p>
                     </div>
                 </div>
 
@@ -896,7 +896,7 @@ <h4>习题3 <span class="difficulty very-hard">很难</span></h4>
                         $$y = \int (x + C_1\sqrt{x}) dx = \frac{x^2}{2} + C_1 \cdot \frac{2x^{3/2}}{3} + C_2$$
                         $$= \frac{x^2}{2} + \frac{2C_1x^{3/2}}{3} + C_2$$
 
-                        <p><strong>答案：</strong>$y = \frac{x^2}{2} + \frac{2C_1x^{3/2}}{3} + C_2$</p>
+                        <p><strong>答案：</strong>$y = \frac{x^2}{2} + C_1 x^{3/2} + C_2$</p>
                     </div>
                 </div>
             </div>

chp07/subsections/7.8.html

@@ -916,7 +916,7 @@ <h4>习题3 <span class="difficulty very-hard">很难</span></h4>
 
                         <p><strong>步骤3：</strong>求导</p>
                         <p>$y_p' = (ax^3 + bx^2) e^x + (3ax^2 + 2bx) e^x = [ax^3 + (b+3a)x^2 + 2bx] e^x$</p>
-                        <p>$y_p'' = [ax^3 + (b+6a)x^2 + (2b+6a)x + 2b] e^x$</p>
+                        <p>$y_p'' = [ax^3 + (b+6a)x^2 + (4b+6a)x + 2b] e^x$</p>
 
                         <p><strong>步骤4：</strong>代入方程并化简</p>
                         <p>经计算后得到：$6ax + 2b = x$</p>

chp07/subsections/7.9.html

@@ -912,11 +912,11 @@ <h4>习题3 <span class="difficulty very-hard">很难</span></h4>
 
                         <p><strong>步骤3：</strong>求特解</p>
                         <p>由于 $1$ 是单重特征根，设特解：$y_p = t(at + b)e^t$</p>
-                        <p>代入方程求得：$a = \frac{1}{2}$，$b = 0$</p>
-                        <p>特解：$y_p = \frac{t^2}{2}e^t = \frac{(\ln x)^2}{2} \cdot x$</p>
+                        <p>代入方程求得：$a = \frac{1}{4}$，$b = -\frac{1}{4}$</p>
+                        <p>特解：$y_p = \frac{t^2 - t}{4}e^t = \frac{(\ln x)^2 - \ln x}{4} \cdot x$</p>
 
                         <p><strong>步骤4：</strong>通解</p>
-                        $$y = C_1 x + \frac{C_2}{x} + \frac{x(\ln x)^2}{2}$$
+                        $$y = C_1 x + \frac{C_2}{x} + \frac{(\ln x)^2 - \ln x}{4} \cdot x$$
                     </div>
                 </div>
             </div>