Many general and singular solutions from Clairaut DE

Question

The ODE $$ (y-xy')^2-y'^2=1$$ I think since it is first order second degree ODE , it can be solved by writing it in this form y=F(y',x) and then differentiating wrt x , or writing it in the form x=F(y,y') and then differentiating wrt y . (please tell me if this is wrong).

Instead of doing this, I solved it in the following way by transforming it to two Clairaut DEs:

I obtained two general solutions and four singular solutions, I feel that some solutions are refused but which solutions and why :

First solution :
$$y=xy'+\sqrt{1+y'^2}$$ Differentiate wrt x $$y''(x+\frac{y'}{\sqrt{1+y'^2}})=0$$ so the general solution is $$y=xc+\sqrt{1+c^2}$$ The singular solution is when $$x=-\frac{y'}{\sqrt{1+y'^2}}$$ or we can write this as $$y'=\pm \frac{x}{\sqrt{1-x^2}}$$ Eliminate y' from the DE $$y=xy'+\sqrt{1+y'^2}$$ So we have 2 singular solutions since y' is positive or negative , the singular solutions are: $$y=\frac{x^2+1}{\sqrt{1-x^2}}$$ $$y=\frac{-x^2+1}{\sqrt{1-x^2}}=\sqrt{1-x^2}$$

The second solution , we will take the negative square root for y $$y=xy'-\sqrt{1+y'^2}$$ Apply same steps ,so differentiate wrt x $$y''(x-\frac{y'}{\sqrt{1+y'^2}})=0$$ so the second general solution is $$y=xc-\sqrt{1+c^2}$$ The singular solution is when $$x=\frac{y'}{\sqrt{1+y'^2}}$$ or we can write this as $$y'=\pm \frac{x}{\sqrt{1-x^2}}$$ Eliminate y' from the DE $$y=xy'-\sqrt{1+y'^2}$$ So the two other singular solutions will be $$y=\frac{x^2-1}{\sqrt{1-x^2}}=-\sqrt{1-x^2}$$ $$y=\frac{-x^2-1}{\sqrt{1-x^2}}$$

Answer 1

You found the linear solutions
$$y=cx+s\sqrt{1+c^2}\tag{LINSOL}$$ in both cases $s=\pm 1$ for the constant sign of the square root in $$y=y'x+s\sqrt{1+y'^2}\tag{ODE}$$ correctly. However, for the singular solutions one has to observe that in the equation $$ x+s\frac{y'}{\sqrt{1+y'^2}}=0\tag1 $$ the signs of $x$ and $y'$ are coupled. Thus while $x^2=\frac{y'^2}{1+y'^2}$ implies indeed $$ |y'|=\frac{|x|}{\sqrt{1-x^2}},\tag2 $$ there is only one solution for each value of $s$, $$ y'=-s\frac{x}{\sqrt{1-x^2}},\tag3 $$ Inserting back into the original equation (ODE) gives $$ y=-s\frac{x^2}{\sqrt{1-x^2}}+s\frac1{\sqrt{1-x^2}}=s\sqrt{1-x^2}\tag{SINGSOL} $$

Answer 2

The given differential equation is $$(y-xy')^2-y'^2=1$$

$$\implies y=xy'\pm\sqrt{1+y'^2}$$ which is of Clairaut form.

Differentiating both side with respect to $x$, we have

$$y'=y'+xy''\pm\frac{y'y''}{\sqrt{1+y'^2}}\implies y''(x\pm\frac{y'}{\sqrt{1+y'^2}})=0$$which gives $y''=0\implies y'=c,\quad$ where $c$ is an arbitrary constant

Hence the general solution of the given differential solution is $$y=cx\pm\sqrt{1+c^2}$$ where $c$ is an arbitrary constant.

Also $$x\pm\frac{y'}{\sqrt{1+y'^2}}=0\implies y'=\pm\frac{x}{\sqrt{1-x^2}}$$Integrating, $$y=\pm\sqrt{1-x^2}$$These are the singular solutions.

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Conclusion:

• When we consider the given differential equation $$(y-xy')^2-y'^2=1$$ is of the form $$ y=xy'+\sqrt{1+y'^2}$$then the general solution of the differential equation is $$y=cx+\sqrt{1+c^2}$$ where $c$ is an arbitrary constant $\qquad$ and the singular solution be $$y=\sqrt{1-x^2}.$$

• Again when we consider the given differential equation $$(y-xy')^2-y'^2=1$$ is of the form $$ y=xy'-\sqrt{1+y'^2}$$then the general solution of the differential equation is $$y=cx-\sqrt{1+c^2}$$ where $c$ is an arbitrary constant $\qquad$ and the singular solution be $$y=-\sqrt{1-x^2}.$$