Is there a way to solve this O.D.E without using Lagrange equation method?
$x {(y^{'})}^{2} - y y^{'} = - 1$

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Is there a way to solve this O.D.E without using Lagrange equation method?x(y′)2−yy′=−1

Marin Lowe · Accepted Answer

Step 1You can reverse the dependency from y(x) to x(y), on segments of a solution where this is possible due to monotonicity, to getx=yx′(y)−[x′(y)]2This now is a standard example of a Clairaut equation. It has a linear solution familyx=Cy−C2and a singular solutionx=y24.And all combinations from changing the solution due to the non-uniqueness of the singular solution.

ncruuk7ikt · Accepted Answer

Step 1Use the ansatz that y is a polynomial: y(x)=∑i=0Naixi. Then y′(x)=∑i=0Niaixi−1and (yy′)(x)=∑m=02N−1(∑i+j=m+1,i,j≤Niaiaj)xmandx(y′)2(x)=x∑m=02N−1(∑i+j=m+1,i,j≤Nijaiaj)xm.Put it into the differential equation.Now compare coefficients: for m=0 we get a1a0−1=0, so a0=C and a1=1C for some C≠0. For m=1 we get 2a2a0=0, so a2=0For m=3 we get 0=3a3a0+a1a2=3a3a0, so a3=0. Inductively ai=0 for i≥2Now we got for y: y(x)=1Cx+C. These are solutions.

Is there a way to solve this O.D.E

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