学科知识点表达

光华启迪英语组-薛老师

学科知识点表达

光华启迪英语组-薛老师

类别一：开启对话与厘清问题

在拿到问题后，切忌直接计算。这前30秒是展示你严谨思维的关键。

1. 请求重复或澄清：

Could you please rephrase the question? I want to make sure I'm tackling the right problem.
您能否重新表述一下问题？我想确保我在解决正确的问题。
I'm not sure I caught the last part. Did you say...?
我不确定我是否听清了最后一部分。您是说...？
Just to clarify, when you mention [某个术语]， are we assuming it's under ideal conditions?
只是为了澄清一下，当您提到[某个术语]时，我们是否假设它处于理想条件下？

2. 复述问题以确认理解：

So, if I understand correctly, the task is to derive/explain/find [核心目标] given that [已知条件].
所以，如果我理解正确的话，任务是在给定[已知条件]的情况下推导/解释/找到[核心目标]。
In essence, this is a problem about [核心概念]， and I need to establish a relationship between A and B.
本质上，这是一个关于[核心概念]的问题，我需要建立A和B之间的关系。
Let me restate the problem in my own words: We have a [描述系统]， and we're observing [某种现象]， and the goal is to find out why/how.
让我用自己的话重新表述问题：我们有一个[描述系统]，正在观察[某种现象]，目标是找出原因/方式。

3. 界定范围与假设：

I will assume that [例如： air resistance is negligible / the object is a particle / the system is in equilibrium] unless stated otherwise.
除非另有说明，否则我将假设[例如：空气阻力可忽略不计/物体为质点/系统处于平衡状态]。
To simplify the problem initially, I'm going to consider a simplified model where...
为了简化初始问题，我将考虑一个简化模型，其中...
This problem seems to bridge the concepts of [概念A] and [概念B]. I'll need to draw from both areas.
这个问题似乎连接了[概念A]和[概念B]。我需要从这两个领域汲取知识。

类别二：构建思考框架

这是你展示逻辑架构能力的地方，告诉面试官你的 解题路线图 。

1. 提出初步思路：

My first instinct is to apply [某个定律/定理/方法]， because the problem involves [该定律的适用条件].
我的第一直觉是应用[某个定律/定理/方法]，因为这个问题涉及[该定律的适用条件]。
Looking at the structure of this equation, it reminds me of the form for [某种标准形式]， which suggests using [相应技巧].
观察这个方程的结构，它让我想起了[某种标准形式]，这提示我使用[相应技巧]。
I recall that a key principle in this area is [原理]， so I will start by writing that down.
我记得这个领域的一个关键原理是[原理]，所以我将从写下这个原理开始。

2. 规划步骤：

I propose to tackle this in three stages: first, I'll identify all the relevant forces; second, I'll resolve them into components; third, I'll apply Newton's second law.
我建议分三个阶段解决这个问题：首先，我会识别所有相关的力；其次，我会将它们分解为分量；第三，我会应用牛顿第二定律。
A systematic approach would be to first consider the general case, and then apply the boundary conditions.
一个系统的方法是首先考虑一般情况，然后应用边界条件。
Before I dive into calculus, I want to see if this can be solved using a purely geometric argument.
在深入微积分之前，我想看看是否可以使用纯几何论证来解决这个问题。

3. 考虑不同方法：

There are a couple of ways to approach this. I could use [方法A]， which is more direct, or [方法B]， which might offer more insight. I'll start with [方法A] for now.
有几种方法可以解决这个问题。我可以使用[方法A]，这更直接，或者[方法B]，这可能提供更多的见解。我现在先从[方法A]开始。
I wonder if this can be solved by considering energy instead of forces. That might be more efficient here.
我想知道是否可以通过考虑能量而不是力来解决这个问题。这在这里可能更有效。

类别三：推导与解释过程中的核心句式

这是面试的主体，需要将无声的思维变成有声的解说。

1. 陈述步骤与理由：

I am now going to differentiate this expression. Since it's a product of two functions, I must use the product rule.
我现在要对这个表达式求导。由于它是两个函数的乘积，我必须使用乘积法则。
The next logical step is to integrate both sides. Notice that the left side is with respect to y and the right with respect to x, but because I have separated the variables, this is valid.
下一步逻辑是对两边积分。注意左边是关于y的，右边是关于x的，但因为我已经分离了变量，所以这是有效的。
Here, I'm applying conservation of mechanical energy because there are no non-conservative forces doing work.
在这里，我应用机械能守恒定律，因为没有非保守力做功。

2. 解释 为什么 ：

The reason for taking the square root here is to solve for the velocity, v, which is currently squared.
在这里取平方根的原因是为了解出速度v，它目前是平方的。
I'm substituting this value back into the first equation. This will allow me to eliminate one variable and solve for the other.
我将这个值代回到第一个方程中。这将允许我消去一个变量并解出另一个变量。
This step is crucial because it decouples the two equations, making them solvable independently.
这一步至关重要，因为它解耦了两个方程，使它们可以独立求解。

3. 描述数学操作：

I'll rearrange this equation to make y the subject.
我将重新排列这个方程，使y成为主语。
Let me expand these brackets and collect like terms.
让我展开这些括号并合并同类项。
I can factorise this quadratic expression to find the roots.
我可以分解这个二次表达式来找到根。

4. 描述物理过程：

As the capacitor charges up, the potential difference across it increases, thereby opposing the flow of current.
随着电容器充电，其两端的电势差增加，从而阻碍电流的流动。
The wave superposition at this point is constructive because the path difference is an integer multiple of the wavelength.
此时的波叠加是相长的，因为路径差是波长的整数倍。
When the string is plucked, it vibrates at its fundamental frequency and a series of harmonics.
当弦被拨动时，它以其基频和一系列谐波振动。

类别四：应对困难与调整思路

卡住并不可怕，可怕的是沉默地卡住。如何应对困难恰恰是面试官最看重的素质之一。

1. 承认困境：

This is an interesting twist. My initial approach seems to have led me to a dead end.
这是一个有趣的转折。我的初始方法似乎让我走进了死胡同。
I'm a bit stuck at this point. The expression is becoming quite messy.
我在这一点上有点卡住了。表达式变得相当混乱。
That can't be right. The units on both sides don't match. I must have made a mistake earlier.
这不可能是对的。两边的单位不匹配。我之前一定犯了一个错误。

2. 自我检查与修正：

Let me trace back my steps. Ah, I see the error. I forgot the chain rule when I differentiated that term.
让我追溯我的步骤。啊，我看到错误了。我在对那个项求导时忘记了链式法则。
Perhaps my assumption that [某个假设] was incorrect. What if I relax that assumption?
也许我关于[某个假设]的假设是不正确的。如果我放宽这个假设会怎样？
Let me verify the dimensions of this equation to see if it makes physical sense.
让我验证这个方程的量纲，看看它是否有物理意义。

3. 请求提示或引导：

Could you give me a small nudge in the right direction?
您能在正确的方向上给我一个小小的提示吗？
Is there a particular concept you think I should be focusing on here?
您认为我应该在这里关注某个特定的概念吗？
I feel like I'm close. Am I right in thinking that [某个想法] is relevant?
我觉得我很接近了。我认为[某个想法]是相关的，对吗？

4. 转换思路：

Let me abandon that method and try a more brute-force approach.
让我放弃那种方法，尝试一种更直接的方法。
What if I consider the limiting cases? For example, as t approaches infinity, what happens?
如果我考虑极限情况会怎样？例如，当t趋近于无穷大时，会发生什么？
A different perspective might be to consider the symmetry of the system. That often simplifies things.
一个不同的视角可能是考虑系统的对称性。这通常会简化问题。

类别五：得出结论与深入反思

得到一个答案不是终点，展示你对其意义的理解才是。

1. 陈述结论：

Therefore, putting it all together, the solution is...
因此，综合所有因素，解决方案是...
So, the final expression for the velocity as a function of position is...
因此，速度作为位置函数的最终表达式是...
In conclusion, the proof is complete. We have shown that P(k+1) is true assuming P(k), and since the base case holds, the statement is true for all n.
总之，证明完成。我们已经证明了假设P(k)为真时，P(k+1)也为真，并且由于基础情况成立，所以该陈述对所有n都为真。

2. 解释结果的意义：

This result makes sense intuitively because as mass increases, acceleration decreases for a given force.
这个结果在直觉上是合理的，因为对于给定的力，随着质量增加，加速度减小。
The negative sign in the answer indicates that the direction of the force is opposite to the displacement, which is consistent with a restoring force.
答案中的负号表示力的方向与位移相反，这与恢复力一致。
It's interesting to note that the final answer is independent of [某个参数]， which was not obvious at the beginning.
有趣的是，最终答案与[某个参数]无关，这在开始时并不明显。

3. 总结核心洞察：

The key insight here was to recognise the hidden quadratic form.
这里的关键洞察是识别隐藏的二次形式。
The crux of this problem was applying conservation laws at the correct instants in time.
这个问题的关键是在正确的时间点应用守恒定律。
This demonstrates the power of [某个方法] in solving problems that appear very complicated at first glance.
这展示了[某个方法]在解决乍看之下非常复杂的问题时的威力。

4. 展望与延伸：

A natural extension of this problem would be to consider what happens if we include friction.
这个问题的自然延伸是考虑如果我们包括摩擦会发生什么。
This model has limitations; in a real-world scenario, we would have to account for [真实世界因素].
这个模型有局限性；在现实世界的场景中，我们必须考虑[真实世界因素]。
I wonder if this result can be generalised for a broader class of functions.
我想知道这个结果是否可以推广到更广泛的函数类。

数学 (Mathematics) AS/A2

知识点： 微积分 (Calculus) - 链式法则 (Chain Rule)

概念解释：

The chain rule is used when we have a function composed within another function, which we call a composite function. It allows us to find the rate of change of the outer function with respect to the inner one, and then multiply it by the rate of change of the inner function with respect to the variable.

链式法则用于当我们有一个嵌套在另一个函数内的函数时，我们称之为复合函数。它允许我们找到外部函数相对于内部函数的变化率，然后将其乘以内部函数相对于变量的变化率。

案例演示：

问题： Explain how you would find the derivative of y = sin(x²).

问题：解释你将如何找到y = sin(x²)的导数。

学生回答：

Okay, so I need to find dy/dx for y = sin(x²). I recognise that this is a composite function. The outer function is sin(u), and the inner function is u = x².

The chain rule states that dy/dx = (dy/du) * (du/dx).

So, first, I find the derivative of the outer function with respect to u: dy/du = cos(u).

Then, I find the derivative of the inner function with respect to x: du/dx = 2x.

Now, I combine them: dy/dx = cos(u) * 2x.

Finally, I substitute back u = x² to express everything in terms of x. So, the final answer is dy/dx = 2x * cos(x²).

The crucial part was identifying the inner and outer functions. If I had just taken the derivative of sin(x) and gotten cos(x), it would have been incorrect because the argument of the sine function is x², not x.

好的，所以我需要找到y = sin(x²)的dy/dx。我认识到这是一个复合函数。外部函数是sin(u)，而内部函数是u = x²。

链式法则指出dy/dx = (dy/du) * (du/dx)。

所以，首先，我找到外部函数相对于u的导数：dy/du = cos(u)。

然后，我找到内部函数相对于x的导数：du/dx = 2x。

现在，我将它们结合起来：dy/dx = cos(u) * 2x。

最后，我代回u = x²，用x表示所有内容。所以，最终答案是dy/dx = 2x * cos(x²)。

关键部分是识别内部和外部函数。如果我只是对sin(x)求导并得到cos(x)，那将是不正确的，因为正弦函数的自变量是x²，而不是x。

知识点： 代数 (Algebra) - 数学归纳法 (Proof by Induction)

概念解释：

Proof by induction is a method to prove a statement is true for all natural numbers. It works in two core steps. First, the base case: we show the statement holds for the first number, usually n=1. Second, the inductive step: we assume the statement is true for some n=k (this is the inductive hypothesis), and then, using that assumption, we prove it must also be true for n=k+1. This creates a domino effect; if it's true for 1, it's true for 2, and so on ad infinitum.

数学归纳法是一种证明一个陈述对所有自然数都为真的方法。它通过两个核心步骤工作。首先，基础情况：我们证明该陈述对第一个数成立，通常是n=1。其次，归纳步骤：我们假设该陈述对某个n=k成立（这是归纳假设），然后，使用该假设，我们证明它对n=k+1也必须为真。这会产生多米诺骨牌效应；如果它对1成立，它对2也成立，依此类推，无限延伸。

案例演示：

问题： Prove that the sum of the first n positive integers is n(n+1)/2.

问题：证明前n个正整数的和为n(n+1)/2。

学生回答：

I'll prove this by induction.

Let P(n) be the statement: 1 + 2 + ... + n = n(n+1)/2.

Step 1: Base Case. For n=1, the left-hand side is 1. The right-hand side is 1(1+1)/2 = 1. So P(1) is true.

Step 2: Inductive Step. I now assume P(k) is true for some arbitrary k. That is, I assume 1+2+...+k = k(k+1)/2. This is my inductive hypothesis.

My goal is to show that P(k+1) is true. That is, I need to show 1+2+...+k+(k+1) = (k+1)(k+2)/2.

Starting with the left-hand side of P(k+1): (1+2+...+k) + (k+1). By my inductive hypothesis, I can replace the sum in the bracket with k(k+1)/2.

So now I have: k(k+1)/2 + (k+1).

I can factor out (k+1) and write this as: (k+1)(k/2 + 1) = (k+1)( (k+2)/2 ) = (k+1)(k+2)/2.

This is exactly the right-hand side of P(k+1). Therefore, I have shown that if P(k) is true, then P(k+1) is also true.

Combining the base case and the inductive step, by the principle of mathematical induction, P(n) is true for all natural numbers n.

我将用归纳法证明这一点。

设P(n)为陈述：1 + 2 + ... + n = n(n+1)/2。

步骤1：基础情况。对于n=1，左边是1。右边是1(1+1)/2 = 1。所以P(1)为真。

步骤2：归纳步骤。我现在假设P(k)对某个任意k为真。也就是说，我假设1+2+...+k = k(k+1)/2。这是我的归纳假设。

我的目标是证明P(k+1)为真。也就是说，我需要证明1+2+...+k+(k+1) = (k+1)(k+2)/2。

从P(k+1)的左边开始：(1+2+...+k) + (k+1)。根据我的归纳假设，我可以用k(k+1)/2替换括号中的和。

所以现在我有：k(k+1)/2 + (k+1)。

我可以提出(k+1)并将其写成：(k+1)(k/2 + 1) = (k+1)( (k+2)/2 ) = (k+1)(k+2)/2。

这正是P(k+1)的右边。因此，我已经证明了如果P(k)为真，那么P(k+1)也为真。

结合基础情况和归纳步骤，根据数学归纳法原理，P(n)对所有自然数n都为真。

物理 (Physics) AS/A2

知识点： 力学 (Mechanics) - 牛顿第二定律 (Newton's Second Law)

概念解释：

Newton's second law fundamentally states that the vector sum of the forces (the net force) acting on a body is equal to the rate of change of its momentum. For a constant mass, this simplifies to the famous F_net = m*a. The acceleration is in the same direction as the net force. It's crucial to consider all forces acting on the object and resolve them into components if necessary.

牛顿第二定律从根本上指出，作用在物体上的力的矢量和（净力）等于其动量的变化率。对于恒定质量，这简化为著名的F_net = m*a。加速度与净力方向相同。考虑作用在物体上的所有力并在必要时将它们分解为分量是至关重要的。

案例演示：

问题： A block slides down a frictionless inclined plane. Explain the forces acting on it and its motion.

问题：一个物块沿无摩擦的斜面下滑。解释作用在它上面的力及其运动。

学生回答：

Let me sketch this mentally. We have a block on a slope at an angle θ. I'll first identify all the forces acting on the block.

  1. Weight (mg): This acts vertically downwards.
  2. Normal Contact Force (R): This acts perpendicularly away from the surface of the incline.

Since the plane is frictionless, there is no friction force.

The weight force is not parallel to the plane, so it's useful to resolve it into components.

I'll resolve parallel and perpendicular to the incline.

The component parallel to the incline is mg sinθ. This is the component that pulls the block down the slope.

The component perpendicular to the incline is mg cosθ. This is balanced by the normal force R, because there is no acceleration in that direction. So, R = mg cosθ.

Now, applying Newton's second law parallel to the incline: The net force F_net = mg sinθ.

So, mg sinθ = m * a.

The mass m cancels out, giving us a = g sinθ.

The key insight is that the acceleration is constant and depends only on the gravity and the angle of the slope. It is independent of the mass of the block.

让我在脑海中勾勒这个场景。我们有一个物块在角度为θ的斜面上。我将首先识别作用在物块上的所有力。

  1. 重力(mg)：垂直向下作用。
  2. 法向接触力(R)：垂直远离斜面表面作用。

由于平面是无摩擦的，所以没有摩擦力。

重力不平行于平面，因此将其分解为分量是有用的。

我将分解为平行和垂直于斜面的分量。

平行于斜面的分量是mg sinθ。这是拉动物块沿斜面向下的分量。

垂直于斜面的分量是mg cosθ。这由法向力R平衡，因为在该方向上没有加速度。所以，R = mg cosθ。

现在，沿斜面方向应用牛顿第二定律：净力F_net = mg sinθ。

所以，mg sinθ = m * a。

质量m抵消，得到a = g sinθ。

关键洞察是加速度是恒定的，仅取决于重力和斜面的角度。它与物块的质量无关。

知识点： 电学 (Electricity) - 电容器充放电 (Capacitor Charge and Discharge)

概念解释：

A capacitor stores charge. The time it takes to charge or discharge is characterized by the time constant, τ, which is the product of resistance and capacitance (R x C). Physically, the time constant is the time for the charge to decrease to about 37% of its initial value during discharge, or to reach about 63% of its maximum during charging. The exponential nature comes from the fact that the rate of flow of charge (current) is proportional to the remaining charge yet to flow.

电容器储存电荷。充电或放电所需的时间由时间常数τ表征，它是电阻和电容的乘积(R x C)。从物理上讲，时间常数是放电期间电荷减少到其初始值的约37%的时间，或充电期间达到其最大值的约63%的时间。指数特性源于电荷流动速率（电流）与尚未流动的剩余电荷成正比的事实。

案例演示：

问题： Why does the charge on a capacitor not instantly become maximum when you connect it to a battery in a simple RC circuit?

问题：为什么在简单的RC电路中，当您将电容器连接到电池时，电容器上的电荷不会立即达到最大值？

学生回答：

That's because of the fundamental way a capacitor works. When the circuit is first connected, the voltage across the capacitor is zero. The battery 'sees' a potential difference and pushes charge onto the capacitor plates. This initial current is high, limited only by the resistor (I = V/R).

However, as charge builds up on the plates, a voltage V_c = Q/C develops across the capacitor, which opposes the voltage of the battery.

This means the net voltage driving the current through the resistor is now (V_battery - V_c). As V_c increases, this net voltage decreases.

Since the current is proportional to this net voltage, the current decreases over time.

The rate at which charge builds up is therefore high at first and slows down asymptotically. It's an exponential process. It never quite reaches the maximum theoretical charge, but gets arbitrarily close to it. The time constant RC tells us how quickly this process happens.

这是因为电容器的基本工作方式。当电路首次连接时，电容器两端的电压为零。电池"看到"电势差并将电荷推到电容器极板上。初始电流很高，仅受电阻限制(I = V/R)。

然而，随着电荷在极板上积累，电容器两端会产生电压V_c = Q/C，这与电池的电压相反。

这意味着驱动电流通过电阻的净电压现在是(V_battery - V_c)。随着V_c增加，这个净电压减小。

由于电流与这个净电压成正比，电流随时间减小。

因此，电荷积累的速率最初很高，然后渐近地减慢。这是一个指数过程。它永远不会完全达到理论上的最大电荷，但会任意接近它。时间常数RC告诉我们这个过程发生的速度有多快。

高数 (Further Mathematics) AS/A2

知识点： 复数 (Complex Numbers) - 欧拉公式 (Euler's Formula) & 极坐标形式 (Polar Form)

概念解释：

Euler's formula, e^(iθ) = cosθ + i sinθ, provides a deep connection between exponential and trigonometric functions in the complex plane. It allows us to represent a complex number not just in Cartesian form (a + bi), but also in a powerful polar form: r * e^(iθ), where r is the modulus (distance from origin) and θ is the argument (angle from the positive real axis). This form makes multiplication and division of complex numbers much easier, as you simply multiply moduli and add arguments.

欧拉公式e^(iθ) = cosθ + i sinθ在复平面上提供了指数函数和三角函数之间的深刻联系。它允许我们不仅以笛卡尔形式(a + bi)表示复数，还可以以强大的极坐标形式r * e^(iθ)表示，其中r是模（到原点的距离），θ是辐角（与正实轴的夹角）。这种形式使复数的乘法和除法变得容易得多，因为您只需将模相乘并将辐角相加。

案例演示：

问题： Multiply the complex numbers (√3 + i) and (1 + i). First do it the standard way, then explain a more elegant method and interpret the result geometrically.

问题：将复数(√3 + i)和(1 + i)相乘。首先用标准方法做，然后解释一个更优雅的方法并从几何上解释结果。

学生回答：

Sure. The standard way is to expand the brackets: (√3 + i)(1 + i) = √3*1 + √3*i + i*1 + i*i = √3 + (√3 + 1)i - 1 = (√3 - 1) + (√3 + 1)i.

Now, the more elegant way is to use the polar form. Let me find the modulus and argument of each.

For z1 = √3 + i. Its modulus r1 = √( (√3)² + 1² ) = 2. Its argument θ1 = arctan(1/√3) = π/6.

For z2 = 1 + i. Its modulus r2 = √(1² + 1²) = √2. Its argument θ2 = arctan(1/1) = π/4.

So, z1 = 2 e^(i π/6) and z2 = √2 e^(i π/4).

When I multiply them, z1 * z2 = (2 * √2) * e^( i*(π/6 + π/4) ) = 2√2 * e^(i 5π/12).

So, the product has a modulus of 2√2 and an argument of 5π/12.

Geometrically, this means that multiplying these two numbers corresponds to scaling the first number by a factor of √2 (the modulus of the second) and rotating it counter-clockwise by an angle of π/4 (the argument of the second). This geometric interpretation is a direct consequence of the polar form and Euler's formula.

当然。标准方法是展开括号：(√3 + i)(1 + i) = √3*1 + √3*i + i*1 + i*i = √3 + (√3 + 1)i - 1 = (√3 - 1) + (√3 + 1)i。

现在，更优雅的方法是使用极坐标形式。让我找到每个数的模和辐角。

对于z1 = √3 + i。它的模r1 = √( (√3)² + 1² ) = 2。它的辐角θ1 = arctan(1/√3) = π/6。

对于z2 = 1 + i。它的模r2 = √(1² + 1²) = √2。它的辐角θ2 = arctan(1/1) = π/4。

所以，z1 = 2 e^(i π/6)，z2 = √2 e^(i π/4)。

当我将它们相乘时，z1 * z2 = (2 * √2) * e^( i*(π/6 + π/4) ) = 2√2 * e^(i 5π/12)。

所以，乘积的模为2√2，辐角为5π/12。

从几何上讲，这意味着将这两个数相乘相当于将第一个数缩放√2倍（第二个数的模）并将其逆时针旋转π/4角（第二个数的辐角）。这种几何解释是极坐标形式和欧拉公式的直接结果。

知识点： 微分方程 (Differential Equations) - 分离变量法 (Separable Equations)

概念解释：

A first-order differential equation is separable if we can algebraically manipulate it to have all terms involving one variable (say, y) on one side, and all terms involving the other variable (x) on the other. The strategy is then to integrate both sides independently. This works because we are effectively integrating with respect to different variables on each side, which is justified by the chain rule in reverse.

一阶微分方程是可分离的，如果我们可以通过代数操作使其一侧包含所有涉及一个变量（例如y）的项，另一侧包含所有涉及另一个变量（x）的项。然后策略是独立地对两边积分。这之所以有效，是因为我们实际上是对每一侧的不同变量进行积分，这由反向的链式法则证明是合理的。

案例演示：

问题： Solve the differential equation dy/dx = x / y², given that y=1 when x=0.

问题：求解微分方程dy/dx = x / y²，已知当x=0时y=1。

学生回答：

I can see this equation is separable because I can rearrange it to group y's and x's.

My first step is to write it as y² dy = x dx. I've effectively 'multiplied' both sides by y² and by dx.

Now, I integrate both sides.

The left side, ∫ y² dy, becomes (1/3)y³ + C1.

The right side, ∫ x dx, becomes (1/2)x² + C2.

I can combine the constants of integration into one: (1/3)y³ = (1/2)x² + C.

Now, I use the initial condition to find C. When x=0, y=1.

Substituting: (1/3)(1)³ = (1/2)(0)² + C => 1/3 = C.

So, the particular solution is (1/3)y³ = (1/2)x² + 1/3.

I can multiply through by 6 to simplify: 2y³ = 3x² + 2.

So, the solution is y³ = (3x² + 2)/2. The key was recognizing the equation as separable and being careful with the integration and initial condition.

我可以看到这个方程是可分离的，因为我可以重新排列它来分组y和x的项。

我的第一步是将其写成y² dy = x dx。我实际上是将两边都"乘以"了y²和dx。

现在，我对两边积分。

左边，∫ y² dy，变为(1/3)y³ + C1。

右边，∫ x dx，变为(1/2)x² + C2。

我可以将积分常数合并为一个：(1/3)y³ = (1/2)x² + C。

现在，我使用初始条件来找到C。当x=0时，y=1。

代入：(1/3)(1)³