求解微分方程cosy sin2x dx +(cos²y - cos²x)dy = 0的方法问询

Let’s walk through solving this equation step by step— I totally get why online solutions might feel confusing at first, so we’ll break every part down clearly.

First, let’s rewrite the equation in the standard $M(x,y)dx + N(x,y)dy = 0$ form to analyze its properties:

• $M(x,y) = \cos y \sin2x$
• $N(x,y) = \cos^2y - \cos^2x$

Step 1: Check if it’s an Exact Differential Equation #

A differential equation is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. Let’s compute both partial derivatives:

• $\frac{\partial M}{\partial y} = -\sin y \sin2x$ (differentiate $M$ with respect to $y$, treating $x$ as constant)
• $\frac{\partial N}{\partial x} = 2\cos x \sin x = \sin2x$ (use the double-angle identity $\sin2x=2\sin x\cos x$)

Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, this isn’t an exact equation. We’ll need an integrating factor to make it exact.

Step 2: Find an Integrating Factor #

Let’s test if the integrating factor depends only on $y$. We use the formula:
$$\frac{1}{M}\left(\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}\right) = \text{function of } y \text{ only}$$

Plug in our values:
$$\frac{1}{\cos y \sin2x}\left(\sin2x - (-\sin y \sin2x)\right) = \frac{\sin2x(1 + \sin y)}{\cos y \sin2x} = \frac{1 + \sin y}{\cos y}$$

Great, this is a function of $y$ alone! The integrating factor $\mu(y)$ is:
$$\mu(y) = e^{\int \frac{1 + \sin y}{\cos y} dy}$$

Split the integral into simpler terms: $\frac{1}{\cos y} + \frac{\sin y}{\cos y} = \sec y + \tan y$. We know:
$$\int (\sec y + \tan y) dy = \ln|\sec y + \tan y| + \ln|\sec y| + C$$

Exponentiating to cancel the log gives:
$$\mu(y) = \sec y(\sec y + \tan y) = \frac{1 + \sin y}{\cos^2 y}$$

Step 3: Multiply the Original Equation by the Integrating Factor #

Multiply every term by $\mu(y)$ to get an exact equation:
$$\frac{(1 + \sin y)\sin2x}{\cos y} dx + \frac{(1 + \sin y)(\cos^2y - \cos^2x)}{\cos2 y} dy = 0$$

Let’s call the new terms $M'(x,y)$ and $N'(x,y)$. You can verify $\frac{\partial M'}{\partial y} = \frac{\partial N'}{\partial x}$ now— this confirms the equation is exact.

Step 4: Find the Potential Function $u(x,y)$ #

We need a function $u(x,y)$ such that $\frac{\partial u}{\partial x}=M'$ and $\frac{\partial u}{\partial y}=N'$.

First integrate $M'$ with respect to $x$ (treating $y$ as constant):
$$u(x,y) = \int \frac{(1+\sin y)\sin2x}{\cos y} dx = -\frac{1+\sin y}{2\cos y} \cos2x + C(y)$$

Next, differentiate $u$ with respect to $y$ and set it equal to $N'$ to solve for $C'(y)$. After simplifying (using trig identities like $\cos2x=2\cos²x-1$), we find:
$$C'(y) = 1 + \sin y - \frac{1}{2}\sec²y - \frac{1}{2}\tan y \sec y$$

Integrate term by term to get $C(y)$:
$$C(y) = y - \cos y - \frac{1}{2}\tan y - \frac{1}{2}\sec y + K$$

Step 5: Write the General Solution #

The general solution is $u(x,y) = C$ (a constant). Combining all parts and simplifying (multiplying through by 2 to eliminate fractions), we get:
$$-\frac{1+\sin y}{\cos y} \cos2x + 2y - 2\cos y - \tan y - \sec y = C$$

You can further simplify $\frac{1+\sin y}{\cos y}$ to $\frac{\cos y}{1-\sin y}$ if preferred, but the above form is perfectly valid.

内容的提问来源于stack exchange，提问作者fitzmerl duron