3.5.3.21 Example 21

$$\frac{y^{\prime }y}{1+\frac{1}{2}\sqrt{1+{\left(y^{\prime }\right)}^2}}=-x$$

Let $y^{\prime }=p$ and rearranging gives

$$\begin{array}{rl}py& =-x(1+\frac{1}{2}\sqrt{1+p^2})\\ y& =-x(\frac{1}{p}+\frac{1}{2p}\sqrt{1+p^2})\\ & =-x(\frac{2}{2p}+\frac{1}{2p}\sqrt{1+p^2})\\ & =-x\left(\frac{2+\sqrt{1+p^2}}{2p}\right)\\ \text{(1)}& & =xf+g\end{array}$$

Hence

$$\begin{array}{rl}f& =-\frac{2+\sqrt{1+p^2}}{2p}\\ g& =0\end{array}$$

Since $f\left(p\right)\ne p$ then this is d’Almbert ode. Taking derivative of (1) w.r.t. $x$ gives

$$\begin{array}{rl}p& =\frac{d}{dx}(xf\left(p\right)+g\left(p\right))\\ & =f\left(p\right)+x{f}^{\prime }\left(p\right)\frac{dp}{dx}+g^{\prime }\left(p\right)\frac{dp}{dx}\end{array}$$

But $f\left(p\right)=-\frac{2+\sqrt{1+p^2}}{2p},f^{\prime }\left(p\right)=\frac{-1}{p^2},g=0,g^{\prime }=0$ and the above becomes

$$\begin{array}{rl}p& =-\frac{2+\sqrt{1+p^2}}{2p}+x(-\frac{1}{2\sqrt{1+p^2}}-\frac{-2-\sqrt{1+p^2}}{2p^2})\frac{dp}{dx}\\ \text{(2)}& p+\frac{2+\sqrt{1+p^2}}{2p}& =x(-\frac{1}{2\sqrt{1+p^2}}-\frac{-2-\sqrt{1+p^2}}{2p^2})\frac{dp}{dx}\end{array}$$

The singular solution is found by setting $\frac{dp}{dx}=0$ which results in $p+\frac{2+\sqrt{1+p^2}}{2p}=0$. Hence $p=\pm i$ or $y^{\prime }=\pm i$ $$ or $y=\pm ix$. But these do not satisfy the ode, hence no singular solutions exist.

The general solution is when $\frac{dp}{dx}\ne 0$ in (2). This gives the ode

$$\begin{array}{rl}\frac{dp}{dx}& =\frac{1}{x}\frac{(p+\frac{2+\sqrt{1+p^2}}{2p})}{(-\frac{1}{2\sqrt{1+p^2}}-\frac{-2-\sqrt{1+p^2}}{2p^2})}\\ & =\frac{1}{x}(p^3+p)\end{array}$$

But this is non-linear in $p$. Hence inversion is needed. This becomes

$$\begin{array}{rl}\frac{dx}{dp}& =x\frac{(-\frac{1}{2\sqrt{1+p^2}}-\frac{-2-\sqrt{1+p^2}}{2p^2})}{(p+\frac{2+\sqrt{1+p^2}}{2p})}\\ \frac{dx}{dp}& =\frac{x}{p^3+p}\\ \frac{dx}{dp}-\frac{1}{p+p^3}x& =0\end{array}$$

Which is now linear in $x\left(p\right)$.  The solution is

$$\begin{array}{}\text{(3)}& x=\frac{p}{\sqrt{1+p^2}}{c}_1\end{array}$$

We now need to eliminate $p$. We have two equations to do that, (1) and (3). Here they are side by side

$$\begin{array}{}\text{(1)}& y& =-x\left(\frac{2+\sqrt{1+p^2}}{2p}\right)\text{(3)}& x& =\frac{p}{\sqrt{1+p^2}}{c}_1\end{array}$$

We can either solve for $p$ from (1) and plugin in the value found into (3). Or we can solve for $p$ from (3) and plugin the value found in (1). Using CAS we can just use the solve command. For an example, using Maple it gives

eq1:=y=-x*(  (2+sqrt(1+p^2))/(2*p)); 
eq2:=x=p/sqrt(1+p^2)*_C1 
sol:=solve([eq1,eq2],[p,y],'allsolutions'); 
[[p = x*RootOf((c__1^2 - x^2)*_Z^2 - 1), y = -(RootOf((c__1^2 - x^2)*_Z^2 - 1)*c__1 + 2)/(2*RootOf((c__1^2 - x^2)*_Z^2 - 1))]]
 

Now we can use allvalues

map(X->allvalues(X),sol) 
[[p = x*sqrt(1/(c__1^2 - x^2)), y = -(sqrt(1/(c__1^2 - x^2))*c__1 + 2)/(2*sqrt(1/(c__1^2 - x^2)))], 
[p = -x*sqrt(1/(c__1^2 - x^2)), y = (-sqrt(1/(c__1^2 - x^2))*c__1 + 2)/(2*sqrt(1/(c__1^2 - x^2)))]]
 

Hence the solutions are

$$\begin{array}{rl}{y}_1& =-\frac{\sqrt{\frac{1}{c_1^2-x^2}}{c}_1+2}{2\sqrt{\frac{1}{c_1^2-x^2}}}\\ y_2& =-\frac{-\sqrt{\frac{1}{c_1^2-x^2}}{c}_1+2}{2\sqrt{\frac{1}{c_1^2-x^2}}}\end{array}$$

These are verified valid solutions to the ode (had to use assuming positive)