Is there a way to solve this O.D.E without using Lagrange equation method? $$x(y')^2 -yy'=-1$$
As the title suggests can I use some other method like exact equations or separation of variables? I was never taught this method but the assignment suggests to use this method by looking it up in the book, however, the explanation in the book is difficult to understand...
So is there any other way to solve it using the methods I suggested?
 A: You can reverse the dependency from $y(x)$ to $x(y)$, on segments of a solution where this is possible due to monotonicity, to get
$$
x=yx'(y)-[x'(y)]^2
$$
This now is a standard example of a Clairaut equation. It has a linear solution family
$$
x=Cy-C^2
$$
and a singular solution
$$
x=\frac{y^2}4.
$$
And all combinations from changing the solution due to the non-uniqueness of the singular solution.
A: Use the ansatz that $y$ is a polynomial: $y(x) = \sum_{i=0}^{N}a_i x^i$. Then $y'(x) = \sum_{i = 0}^{N}ia_i x^{i-1}$ and $(yy')(x) = \sum_{m=0}^{2N-1}(\sum_{i+j = m+1, i,j \leq N} ia_ia_j)x^m$ and $x(y')^2(x) = x\sum_{m=0}^{2N-1}(\sum_{i+j = m+1, i,j \leq N} ija_ia_j)x^m$. Put it into the differential equation.
Now compare coefficients: for $m=0$ we get $a_1a_0 -1 = 0$, so $a_0 = C$ and $a_1 = \frac{1}{C}$ for some $C \neq 0$. For $m = 1$ we get $2a_2a_0 = 0$, so $a_2 = 0$. For $m = 3$ we get $0 = 3a_3a_0 + a_1a_2 = 3a_3a_0$, so $a_3 = 0$. Inductively $a_i = 0$ for $i \geq 2$.
Now we got for $y$: $y(x) = \frac{1}{C}x+C$. These are solutions.
