5 Change of variable in a differential equation | 微分方程中的变量代换

📚 5 Change of variable in a differential equation | 微分方程中的变量代换

Many first-order and second-order differential equations encountered in IB Mathematics: Analysis and Approaches Higher Level cannot be solved directly using separation of variables or integrating factors. A carefully chosen substitution can transform a complicated differential equation into a recognisable form – typically separable, linear, or exactly integrable. This technique, known as change of variable, is one of the most powerful analytical tools in a student’s toolbox. By introducing a new dependent or independent variable, we can simplify nonlinear terms, eliminate constants inside composite functions, or reduce the order of an equation. Mastering change of variable not only solves the immediate problem but also deepens your understanding of the geometric and algebraic structure of differential equations.

许多 IB 数学分析与方法高级水平中的一阶和二阶微分方程无法直接用分离变量法或积分因子法求解。精心选择的代换可以将复杂的微分方程转化为可辨识的形式——通常是可分离的、线性的或可直接积分的。这种技巧叫作变量代换，是学生工具箱中最有力的分析方法之一。通过引入新的因变量或自变量，我们可以简化非线性项，消除复合函数内部的常数，或降低方程的阶数。掌握变量代换不仅能解决眼前的问题，还能加深你对微分方程的几何和代数结构的理解。

1. The Core Idea of Substitution | 代换的核心思想

At its heart, a change of variable rewrites the relationship between the function and its derivatives in a more convenient coordinate system. Suppose we have an original dependent variable y and independent variable x. We define a new variable v that is a clever function of y and x, often suggested by the structure of the equation. The chain rule then relates dy/dx to dv/dx, transforming the original ODE into an equation for v. Once solved for v, we back‑substitute to recover y. Common choices in the IB syllabus include v = y/x for homogeneous functions, v = ax + by + c for linear combinations, and v = y^(1−n) for Bernoulli equations. The key skill is spotting which part of the equation looks like a derivative or which grouping repeats itself.

变量代换的核心在于将函数及其导数之间的关系用一个更方便的坐标系统重新表达。假设我们有一个因变量 y 和自变量 x。我们定义一个新变量 v，它是 y 和 x 的一个巧妙函数，通常由方程的结构暗示。然后利用链式法则将 dy/dx 与 dv/dx 联系起来，将原常微分方程转化为关于 v 的方程。一旦解出 v，就回代求出 y。IB 教学大纲中常见的选择包括：齐次函数中用 v = y/x，线性组合中用 v = ax + by + c，伯努利方程中用 v = y^(1−n)。关键技巧在于识别方程中哪一部分看起来像导数，或哪一组项反复出现。

To perform the substitution systematically, follow these steps: (i) Identify the promising substitution v = g(x,y). (ii) Express dy/dx in terms of x, v, and dv/dx by differentiating your substitution with respect to x. (iii) Replace all occurrences of y and dy/dx in the original equation. (iv) Simplify the new equation until it becomes separable, linear, or can be solved by another known method. (v) Integrate to find v, then substitute back to get y(x).

要系统地进行代换，请遵循以下步骤：(i) 确定有希望的代换关系 v = g(x,y)。(ii) 将代换式两边对 x 求导，用 x、v 和 dv/dx 表出 dy/dx。(iii) 将原方程中所有出现的 y 和 dy/dx 替换掉。(iv) 简化新方程，直至它变成可分离的、线性的或可以用其他已知方法求解的形式。(v) 积分求出 v，然后回代得出 y(x)。

The introduction of a new variable can feel artificial, but it is backed by a simple principle: if an equation contains a block like (y/x) or (x+y) repeatedly, treating that block as a single variable often collapses the complexity. As you practise, you will develop an intuition for “good” substitutions.

引入新变量可能显得有些人为，但其背后有一个简单的原理：如果方程中反复出现像 (y/x) 或 (x+y) 这样的块，把这一块当作单一变量往往能消解复杂性。随着练习，你会培养出对”好”代换的直觉。

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2. Homogeneous Equations: v = y/x | 齐次方程：代换 v = y/x

A first-order differential equation is called homogeneous if it can be written in the form dy/dx = F(y/x). The function on the right depends only on the ratio y/x. The standard substitution is v = y/x, or equivalently y = v x. Differentiating y = v x with respect to x using the product rule gives dy/dx = v + x dv/dx. Substituting into the homogeneous equation yields v + x dv/dx = F(v). This rearranges to a separable equation: x dv/dx = F(v) − v, provided F(v) ≠ v. The resulting equation can be solved by separating variables: dv / (F(v) − v) = dx / x. After integration, we replace v by y/x and simplify to obtain the general solution, often expressed implicitly.

若一阶微分方程可以写成 dy/dx = F(y/x) 的形式，则称其为齐次方程。等号右边的函数仅依赖于比值 y/x。标准的代换是 v = y/x，等价于 y = v x。对 y = v x 关于 x 求导，利用乘法法则得 dy/dx = v + x dv/dx。代入齐次方程得到 v + x dv/dx = F(v)。若 F(v) ≠ v，可将其整理为可分离变量方程：x dv/dx = F(v) − v。由此可用分离变量法求解：dv / (F(v) − v) = dx / x。积分后，将 v 替换为 y/x 并化简，即得通解，通常为隐式表达。

Worked example: Solve dy/dx = (x² + y²) / (2xy). First rewrite the right-hand side by dividing numerator and denominator by x²: dy/dx = (1 + (y/x)²) / (2(y/x)). Here F(v) = (1 + v²) / (2v). Let v = y/x, then dy/dx = v + x dv/dx. Substituting gives v + x dv/dx = (1+v²)/(2v). Subtract v: x dv/dx = (1+v²)/(2v) − v = (1+v²−2v²)/(2v) = (1−v²)/(2v). So we have (2v/(1−v²)) dv = dx/x. Integrate both sides: ∫ 2v/(1−v²) dv = ∫ dx/x. The left integral is −ln|1−v²| (by noticing derivative of denominator). Thus −ln|1−v²| = ln|x| + C, leading to 1/(1−v²) = k|x|. Replace v = y/x: 1/(1−(y/x)²) = k|x|. This simplifies to x²/(x²−y²) = k|x|, or x² = k|x|(x²−y²). Solving for y gives the implicit family of curves.

例题：求解 dy/dx = (x² + y²) / (2xy)。先将右边分子分母同除以 x² 改写：dy/dx = (1 + (y/x)²) / (2(y/x))。这里 F(v) = (1 + v²) / (2v)。令 v = y/x，则 dy/dx = v + x dv/dx。代入得 v + x dv/dx = (1+v²)/(2v)。减去 v：x dv/dx = (1+v²)/(2v) − v = (1+v²−2v²)/(2v) = (1−v²)/(2v)。于是有 (2v/(1−v²)) dv = dx/x。两边积分：∫ 2v/(1−v²) dv = ∫ dx/x。左边积分是 −ln|1−v²|（注意到分母的导数）。故 −ln|1−v²| = ln|x| + C，得出 1/(1−v²) = k|x|。回代 v = y/x：1/(1−(y/x)²) = k|x|，化简为 x²/(x²−y²) = k|x|，即 x² = k|x|(x²−y²)。对 y 求解可得隐式曲线族。

Tip: Always check if the equation is homogeneous by inspecting whether replacing x with tx and y with ty leaves the differential equation unchanged. If dy/dx remains the same after this scaling, the substitution v = y/x will work. Remember to consider the special case F(v) = v, which leads directly to dy/dx = y/x, a separable equation with solution y = Cx.

提示：检验齐次性的一种方法是检查将 x 换成 tx，y 换成 ty 后，dy/dx 是否保持不变。若经过这种尺度变换后 dy/dx 不变，代换 v = y/x 便有效。还需注意特殊情况 F(v) = v，它将直接导致 dy/dx = y/x，是可分离的，其解为 y = Cx。

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3. Equations of the form dy/dx = f(ax + by + c) | 形如 dy/dx = f(ax + by + c) 的方程

When the right-hand side depends on a linear combination of x and y, such as dy/dx = sin(x + 2y) or dy/dx = (2x + y + 1)², a direct substitution v = ax + by + c can turn the equation into a separable one. Let v = ax + by + c. Differentiating with respect to x yields dv/dx = a + b (dy/dx). Hence dy/dx can be expressed as (1/b)(dv/dx − a), provided b ≠ 0. The original equation becomes (1/b)(dv/dx − a) = f(v), which is separable: dv/dx = a + b f(v). Then we can write dv / (a + b f(v)) = dx, integrate both sides, and finally back‑substitute v = ax + by + c to obtain the solution in terms of x and y.

当方程右边依赖于 x 和 y 的线性组合时，例如 dy/dx = sin(x + 2y) 或 dy/dx = (2x + y + 1)²，直接代换 v = ax + by + c 可以将方程化为可分离变量的类型。令 v = ax + by + c。两边对 x 求导得 dv/dx = a + b (dy/dx)。因此 dy/dx 可表为 (1/b)(dv/dx − a)，假设 b ≠ 0。原方程变为 (1/b)(dv/dx − a) = f(v)，它是可分离的：dv/dx = a + b f(v)。然后我们可以写成 dv / (a + b f(v)) = dx，两边积分，最后回代 v = ax + by + c 即得用 x 和 y 表示的解。

Worked example: Solve dy/dx = (x + y)². In this case a = 1, b = 1, c = 0, so set v = x + y. Then dv/dx = 1 + dy/dx, giving dy/dx = dv/dx − 1. The equation becomes dv/dx − 1 = v², or dv/dx = v² + 1. Separate variables: dv/(v²+1) = dx. Integration gives arctan(v) = x + C. Replace v by x + y: arctan(x + y) = x + C. The general solution can be left implicitly as x + y = tan(x + C). This is a simple yet elegant result obtained by recognising the linear block inside the function.

例题：求解 dy/dx = (x + y)²。此时 a = 1，b = 1，c = 0，故设 v = x + y。则 dv/dx = 1 + dy/dx，得 dy/dx = dv/dx − 1。方程变为 dv/dx − 1 = v²，即 dv/dx = v² + 1。分离变量：dv/(v²+1) = dx。积分得 arctan(v) = x + C。将 v 替换为 x + y：arctan(x + y) = x + C。通解可保留为隐式形式 x + y = tan(x + C)。通过识别函数内部的线性块，我们得到了一个简洁而优美的结果。

If b = 0, the equation is simply dy/dx = f(ax + c), which does not contain y on the right – it is already directly integrable: y = ∫ f(ax + c) dx. Thus the substitution is needed only when both a and b are non‑zero, or when the combination genuinely mixes x and y. In cases where the expression inside f is linear in both variables but with different coefficients, the method still applies by redefining v appropriately; sometimes a preliminary shift of variables eliminates constants and simplifies the block further.

若 b = 0，方程只是 dy/dx = f(ax + c)，右边不含 y，它已经可以直接积分：y = ∫ f(ax + c) dx。因此只有在 a 和 b 均不为零或组合确实混合了 x 和 y 时才需要代换。如果 f 内部的表达式对两个变量都是线性的但系数不同，该方法依然适用，只需适当重新定义 v；有时预先对变量进行平移可以消去常数，进一步简化块。

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4. The Bernoulli Equation: Reducing Nonlinearity with z = y^(1−n) | 伯努利方程：用 z = y^(1−n) 降低非线性

A Bernoulli equation has the standard form dy/dx + P(x) y = Q(x) yⁿ, where n is a real number, typically n ≠ 0, n ≠ 1. For n = 0 the equation is linear; for n = 1 it is separable. For other n, the presence of yⁿ makes the equation nonlinear, but a clever substitution z = y^(1−n) transforms it into a linear first‑order equation in z. To see this, differentiate z = y^(1−n) with respect to x: dz/dx = (1−n) y^(−n) dy/dx. Multiply the original Bernoulli equation by (1−n) y^(−n): (1−n) y^(−n) dy/dx + (1−n) P(x) y^(1−n) = (1−n) Q(x). Notice that the first term is exactly dz/dx, and y^(1−n) = z. Thus we obtain the linear equation dz/dx + (1−n) P(x) z = (1−n) Q(x). This can be solved using an integrating factor μ(x) = exp(∫ (1−n) P(x) dx). After finding z, we back‑substitute y = z^(1/(1−n)) to recover y.

伯努利方程的标准形式为 dy/dx + P(x) y = Q(x) yⁿ，其中 n 是实数，通常 n ≠ 0 且 n ≠ 1。当 n = 0 时方程为线性；当 n = 1 时可分离。对于其他 n，yⁿ 的存在使方程非线性，但巧妙的代换 z = y^(1−n) 可以将其转化为关于 z 的一阶线性方程。为看清这一点，将 z = y^(1−n) 对 x 求导：dz/dx = (1−n) y^(−n) dy/dx。将原伯努利方程乘以 (1−n) y^(−n) 得：(1−n) y^(−n) dy/dx + (1−n) P(x) y^(1−n) = (1−n) Q(x)。注意到第一项恰好是 dz/dx，而 y^(1−n) = z。于是我们得到线性方程 dz/dx + (1−n) P(x) z = (1−n) Q(x)。可用积分因子 μ(x) = exp(∫ (1−n) P(x) dx) 求解。求出 z 后，回代 y = z^(1/(1−n)) 即得 y。

Worked example: Solve dy/dx + (1/x) y = x y². Here P(x) = 1/x, Q(x) = x, and n = 2. Let z = y^(1−2) = y^(−1). Then dz/dx = −y^(−2) dy/dx. Multiply the original equation by −y^(−2): −y^(−2) dy/dx − (1/x) y^(−1) = −x. This is dz/dx − (1/x) z = −x. The integrating factor is μ(x) = exp(∫ −(1/x) dx) = exp(−ln|x|) = 1/x. Multiply through: (1/x) dz/dx − (1/x²) z = −1. The left side is d/dx (z/x). So d/dx (z/x) = −1. Integrate: z/x = −x + C, so z = −x² + Cx. Recall z = 1/y, thus 1/y = Cx − x², giving y = 1 / (Cx − x²).

例题：求解 dy/dx + (1/x) y = x y²。此处 P(x) = 1/x，Q(x) = x，n = 2。令 z = y^(1−2) = y^(−1)。则 dz/dx = −y^(−2) dy/dx。将原方程乘以 −y^(−2)：−y^(−2) dy/dx − (1/x) y^(−1) = −x。即 dz/dx − (1/x) z = −x。积分因子为 μ(x) = exp(∫ −(1/x) dx) = exp(−ln|x|) = 1/x。两边乘以积分因子：(1/x) dz/dx − (1/x²) z = −1。左边为 d/dx (z/x)。故 d/dx (z/x) = −1。积分：z/x = −x + C，于是 z = −x² + Cx。因为 z = 1/y，所以 1/y = Cx − x²，得 y = 1 / (Cx − x²)。

A common mistake is forgetting to handle n = 0 or 1 separately, or misapplying the power when back‑substituting. Remember that the substitution z = y^(1−n) is chosen precisely to make the derivative term match the coefficient after multiplication by (1−n). Always check your final solution by differentiating and plugging into the original equation, especially when n is negative or fractional.

一个常见错误是忘记单独处理 n = 0 或 1 的情形，或者在回代时错误地处理幂次。务必记住，代换 z = y^(1−n) 的选取正是为了使导数项在乘以 (1−n) 后与系数匹配。最终解总是应该通过求导并代入原方程进行验证，尤其在 n 为负数或分数时更应如此。

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5. Equations Reducible to Homogeneous Form | 可化为齐次形式的方程

Some differential equations have the form dy/dx = f ( (a₁x + b₁y + c₁) / (a₂x + b₂y + c₂) ). When c₁ and c₂ are both zero, the equation is homogeneous and we can use v = y/x. If the constants are non‑zero but the lines a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0 intersect at a point (h, k), we can eliminate the constant terms by shifting the origin. Specifically, let X = x − h, Y = y − k. Then the linear expressions become a₁X + b₁Y and a₂X + b₂Y, and the equation transforms into a homogeneous one in X and Y. The substitution V = Y / X then reduces it to a separable equation. Finally, we translate back to the original variables.

有些微分方程具有 dy/dx = f ( (a₁x + b₁y + c₁) / (a₂x + b₂y + c₂) ) 的形式。当 c₁ 和 c₂ 均为零时，该方程是齐次的，可用 v = y/x。若常数非零，但直线 a₁x + b₁y + c₁ = 0 和 a₂x + b₂y + c₂ = 0 相交于一点 (h, k)，我们可以通过平移原点来消去常数项。具体做法是令 X = x − h，Y = y − k。此时线性表达式变为 a₁X + b₁Y 和 a₂X + b₂Y，方程转化为关于 X 和 Y 的齐次方程。接着用代换 V = Y / X 将其化为可分离变量的方程。最后再转换回原始变量。

Worked example: Solve dy/dx = (x + y − 1) / (x − y + 3). First solve the system x + y − 1 = 0 and x − y + 3 = 0. Adding gives 2x + 2 = 0 → x = −1, then y = 2. So h = −1, k = 2. Set X = x + 1, Y = y − 2. Then x = X − 1, y = Y + 2. Compute numerator: (X − 1) + (Y + 2) − 1 = X + Y. Denominator: (X − 1) − (Y + 2) + 3 = X − Y. The ODE becomes dY/dX = (X + Y) / (X − Y). This is homogeneous. Let V = Y/X, so Y = VX, dY/dX = V + X dV/dX. Substitute: V + X dV/dX = (1 + V)/(1 − V). Rearranging: X dV/dX = (1+V)/(1−V) − V = (1+V−V+V²)/(1−V) = (1+V²)/(1−V). Separate variables: (1−V)/(1+V²) dV = dX/X. Integrate: ∫ (1/(1+V²) − V/(1+V²)) dV = ln|X| + C. This gives arctan(V) − ½ ln(1+V²) = ln|X| + C. Back‑substitute V = Y/X, then Y = y−2, X = x+1. The solution can be simplified to an implicit relationship involving arctan((y−2)/(x+1)) and a logarithmic term. Such transformations are powerful whenever the intersection point exists.

例题：求解 dy/dx = (x + y − 1) / (x − y + 3)。首先解方程组 x + y − 1 = 0 和 x − y + 3 = 0。相加得 2x + 2 = 0 → x = −1，进而 y = 2。故 h = −1, k = 2。令 X = x + 1, Y = y − 2。则 x = X − 1, y = Y + 2。计算分子：(X − 1) + (Y + 2) − 1 = X + Y。分母：(X − 1) − (Y + 2) + 3 = X − Y。原方程化为 dY/dX = (X + Y) / (X − Y)。这是齐次方程。令 V = Y/X，于是 Y = VX, dY/dX = V + X dV/dX。代入得 V + X dV/dX = (1 + V)/(1 − V)。整理：X dV/dX = (1+V)/(1−V) − V = (1+V²)/(1−V)。分离变量：(1−V)/(1+V²) dV = dX/X。积分：∫ (1/(1+V²) − V/(1+V²)) dV = ln|X| + C。得到 arctan(V) − ½ ln(1+V²) = ln|X| + C。回代 V = Y/X，Y = y−2, X = x+1。解可简化为包含 arctan((y−2)/(x+1)) 与对数项的隐式关系。只要存在交点，这种变换便非常有效。

If the lines are parallel (a₁/a₂ = b₁/b₂ ≠ c₁/c₂), the intersection point does not exist. In that situation, the substitution reduces to the method of Section 3: set v = a₁x + b₁y (or a suitable linear combination) because the ratio is essentially a function of that linear combination. Recognising this subtlety can save time and prevent algebraic dead ends.

如果两条直线平行（a₁/a₂ = b₁/b₂ ≠ c₁/c₂），则不存在交点。此时代换简化为第 3 节中的方法：令 v = a₁x + b₁y（或适当的线性组合），因为比值本质上是该线性组合的函数。认识到这一细微差别可以节省时间，并避免代数上的死胡同。

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6. Second‑order ODEs with y Missing: p = dy/dx | 不显含 y 的二阶方程：令 p = dy/dx

For a second‑order differential equation of the form F(x, y’, y”) = 0 where the dependent variable y does not

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