求解满足法线y截距等于该点到原点距离的曲线微分方程
Let's walk through how to solve the differential equation $y + \frac{x}{y'} = \sqrt{x^2 + y^2}$ step by step, using standard techniques for first-order ODEs.
Step 1: Rewrite the equation with a substitution for homogeneity
First, let's use the common substitution for homogeneous equations: let $u = \frac{y}{x}$, so $y = ux$. This means $y' = u + x\frac{du}{dx}$ (using the product rule). We also know $\sqrt{x^2 + y^2} = x\sqrt{1 + u^2}$ (assuming $x \neq 0$; we can handle $x=0$ separately later).
Substitute these into the original equation:
$$ux + \frac{x}{u + x\frac{du}{dx}} = x\sqrt{1 + u^2}$$
Divide both sides by $x$ to simplify:
$$u + \frac{1}{u + x\frac{du}{dx}} = \sqrt{1 + u^2}$$
Step 2: Isolate the derivative term
Rearrange to get the term with $\frac{du}{dx}$ by itself:
$$\frac{1}{u + x\frac{du}{dx}} = \sqrt{1 + u^2} - u$$
Take the reciprocal of both sides:
$$u + x\frac{du}{dx} = \frac{1}{\sqrt{1 + u^2} - u}$$
Rationalize the right-hand side (multiply numerator and denominator by $\sqrt{1 + u^2} + u$):
$$\frac{1}{\sqrt{1 + u^2} - u} \times \frac{\sqrt{1 + u^2} + u}{\sqrt{1 + u^2} + u} = \sqrt{1 + u^2} + u$$
(The denominator simplifies to $(1+u^2) - u^2 = 1$.)
Now our equation becomes:
$$u + x\frac{du}{dx} = \sqrt{1 + u^2} + u$$
Subtract $u$ from both sides to get a separable equation:
$$x\frac{du}{dx} = \sqrt{1 + u^2}$$
Step 3: Separate variables and integrate
Separate the variables:
$$\frac{du}{\sqrt{1 + u^2}} = \frac{dx}{x}$$
Integrate both sides. The left integral is a standard form:
$$\ln\left(u + \sqrt{1 + u^2}\right) = \ln|x| + C$$
where $C$ is the constant of integration.
Exponentiate both sides to eliminate the logs:
$$u + \sqrt{1 + u^2} = K|x|$$
where $K = e^C$ is a positive constant. We can rewrite this as $u + \sqrt{1 + u^2} = Cx$ (absorbing the absolute value into $C$, which can be any non-zero constant).
Step 4: Solve for $u$ and substitute back to $y$
Let $A = Cx$, so:
$$\sqrt{1 + u^2} = A - u$$
Square both sides to eliminate the square root:
$$1 + u^2 = A^2 - 2Au + u^2$$
Cancel $u^2$ from both sides:
$$1 = A^2 - 2Au$$
Solve for $u$:
$$u = \frac{A^2 - 1}{2A}$$
Substitute back $A = Cx$ and $u = \frac{y}{x}$:
$$\frac{y}{x} = \frac{(Cx)^2 - 1}{2Cx}$$
Multiply both sides by $x$:
$$y = \frac{C2x2 - 1}{2C}$$
Rearrange terms to get the final form:
$$2Cy = C2x2 - 1$$
$$C2x2 = 2Cy + 1$$
Divide both sides by $C^2$ (let $C' = \frac{1}{C}$, a new constant):
$$x^2 = 2C'y + (C')^2$$
Renaming $C'$ to $C$ (since constants are arbitrary), we get:
$$x^2 = C(2y + C)$$
Quick verification
To confirm this works, take $C=1$: $x^2 = 2y +1$ → $y = \frac{x^2 -1}{2}$. The slope of the tangent is $y'=x$, so the normal slope is $-\frac{1}{x}$. The y-intercept of the normal line at $(x,y)$ is $y +1 = \frac{x^2 +1}{2}$. The distance from $(x,y)$ to the origin is $\sqrt{x^2 + y^2} = \sqrt{x^2 + \left(\frac{x2-1}{2}\right)2} = \frac{x^2 +1}{2}$, which matches the y-intercept. Perfect!
备注：内容来源于stack exchange，提问作者math forever
