Why is the solution of Clairaut's Differential Equation valid?

Question

Clairaut's Differential Equation is:

$$y=xy^\prime+f(y^\prime)$$

where $f$ is supposed to be continuously differentiable.

Every proof for the solution of this equation that I have seen starts by differentiating both sides of this equation; however, this makes $y^{\prime\prime}$ appear.

Why can we assume that $y$ has a second derivative?

Answer 1

When approaching this differential equation most people begin by restricting themselves to finding a smooth solution $y$.

So the logic begins as follows: Suppose $y$ is a smooth solution to \begin{align} y=xy'+f(y') \end{align} then $y$ must also satisfy the equation \begin{align} y'=y'+xy''+f'(y')y''. \end{align} There are many advantages to this approach, one begins that the complexity of $f$ lowers. Hence with a sufficient number of derivatives, we might be able to reduce the problem to a linear equation which we know have a solution.

Example: For instance, consider a smooth solution to \begin{align} y= xy'+(y')^2 \end{align} then we see that \begin{align} 0=2y'y''+xy''= (2y'+x)y''. \end{align}

Hence $y$ must be $y = -\frac{1}{4}x^2+C$. Here we found a family of solutions, but we did not answer whether there are rougher solutions since we implicitly restricted ourselves to the class of smooth functions.

Additional Example: Here is a better example. Consider \begin{align} y=xy'+y'=(1+x)y' \end{align} which can be solved directly to yield the rough solution \begin{align} y= |x+1|. \end{align} However, if we restrict ourselves to smoother solution, then we see \begin{align} y''(1+x)=0 \ \ \implies y''=0 \end{align} which yields $y(x) = a(x+1)$.

Answer 2

The equation $y_0=px_0+f(p)$ need not have a solution $p$ for all initial values $(x_0,y_0)$. In these points there is no solution of the Clairaut differential equation.

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Assume now that the initial point $(x_0,y_0)$ is well-behaved, that is a $p_0$ with $y_0=x_0p_0+f(p_0)$ exists and $x_0+f'(p_0)\ne 0$. Then by the implicit function theorem for the function $$ F(x,y,p)=px+f(p)-y $$ at $(x,y)\approx(x_0,y_0)$ the equation $$F=0\iff y=xp+f(p)$$ is solvable for $p\approx p_0$, and the local solution $p=g(x,y)$ is as smooth as $f$. That is, as $f\in C^k$, with $k\ge 1$, giving $F\in C^k$, so is $g\in C^k$ (locally).

The solution of $y'=g(x,y)$ now in consequence is $C^{k+1}$, so at least $C^2$, as long as it stays in the neighborhood of $(x_0,y_0)$.

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Note that where the linear solutions $y=cx+f(c)$ meet the singular solution of $x+f'(y')=0$, which can be parametrized as $$x(p)=-f'(p), ~~ y(p)=f(p)-f'(p)p,$$ one can switch the branches of solutions and at that point the second derivative jumps.