利用换元法求解微分方程dy/dx=x/(x+y)遇阻，请求帮助

Hey there! You’re on the right track with the substitution method—you just had a small mix-up in calculating $\frac{dy}{dx}$. Let’s work through this step by step to get to the solution.

Step 1: Correct the Derivative for $y = ux$ #

When $y = ux$ (where $u$ is a function of $x$), we need to use the product rule for differentiation, not the chain rule you mentioned earlier. The product rule gives us:
$$\frac{dy}{dx} = u \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(u) = u + x\frac{du}{dx}$$
That’s the correct expression for $\frac{dy}{dx}$—let’s use this to substitute into the original equation.

Step 2: Substitute into the Original Equation #

Plug $y = ux$ and $\frac{dy}{dx} = u + x\frac{du}{dx}$ into $\frac{dy}{dx} = \frac{x}{x+y}$:
$$u + x\frac{du}{dx} = \frac{x}{x + ux}$$
Simplify the right-hand side by factoring out $x$ in the denominator:
$$u + x\frac{du}{dx} = \frac{x}{x(1 + u)} = \frac{1}{1 + u}$$

Step 3: Rearrange to Separate Variables #

Let’s rearrange terms to group all $u$-dependent terms on one side and $x$-dependent terms on the other (this is called separating variables):
$$x\frac{du}{dx} = \frac{1}{1 + u} - u$$
Combine the fractions on the right-hand side to simplify:
$$\frac{1}{1 + u} - u = \frac{1 - u(1 + u)}{1 + u} = \frac{1 - u - u^2}{1 + u}$$
Now we can rewrite the equation as:
$$\frac{1 + u}{1 - u - u^2} du = \frac{1}{x} dx$$

Step 4: Integrate Both Sides #

Now we integrate each side. Let’s tackle the left integral first: $\int \frac{1 + u}{1 - u - u^2} du$.

First, notice that the derivative of the denominator $1 - u - u^2$ is $-1 - 2u$. We can rewrite the numerator to match this derivative:
$$1 + u = -\frac{1}{2}(-1 - 2u) + \frac{1}{2}$$
This splits the integral into two parts:
$$\int \frac{1 + u}{1 - u - u^2} du = -\frac{1}{2} \int \frac{-1 - 2u}{1 - u - u^2} du + \frac{1}{2} \int \frac{1}{1 - u - u^2} du$$

Part 1 of the left integral #

Let $v = 1 - u - u^2$, so $dv = (-1 - 2u)du$. The first integral becomes:
$$-\frac{1}{2} \int \frac{dv}{v} = -\frac{1}{2} \ln|v| + C_1 = -\frac{1}{2} \ln|1 - u - u^2| + C_1$$

Part 2 of the left integral #

Complete the square in the denominator of the second integral:
$$1 - u - u^2 = \frac{5}{4} - \left(u + \frac{1}{2}\right)^2$$
Using the standard integral $\int \frac{1}{a^2 - t^2} dt = \frac{1}{2a} \ln\left|\frac{a + t}{a - t}\right| + C$ (where $a = \frac{\sqrt{5}}{2}$ and $t = u + \frac{1}{2}$), we get:
$$\frac{1}{2} \int \frac{1}{\frac{5}{4} - \left(u + \frac{1}{2}\right)^2} du = \frac{1}{2\sqrt{5}} \ln\left|\frac{\sqrt{5} + 2u + 1}{\sqrt{5} - 2u - 1}\right| + C_2$$

Right-hand side integral #

The right integral is straightforward:
$$\int \frac{1}{x} dx = \ln|x| + C_3$$

Step 5: Combine Results and Substitute Back $u = \frac{y}{x}$ #

Combine all integrals (and merge constants into a single $C$):
$$-\frac{1}{2} \ln|1 - u - u^2| + \frac{1}{2\sqrt{5}} \ln\left|\frac{2u + 1 + \sqrt{5}}{-2u + \sqrt{5} - 1}\right| = \ln|x| + C$$
Now substitute $u = \frac{y}{x}$ back into the equation and simplify. We can cancel out $2\ln|x|$ terms (after multiplying through by 2) to get a cleaner implicit solution:
$$\ln|x^2 - xy - y^2| - \frac{1}{\sqrt{5}} \ln\left|\frac{2y + x(1 + \sqrt{5})}{-2y + x(\sqrt{5} - 1)}\right| = C''$$

If you prefer, you can also rewrite this in exponential form to get an explicit solution for $y$, but this implicit form is a valid general solution to the differential equation.

内容的提问来源于stack exchange，提问作者Matt Simpson