Solve the differential equation $\left(\frac{dy}{dx}\right)^2 [1-2\frac{dy}{dx}y]=c$ where $c\in\Bbb{R}$.

This is a non linear first order differential equation. Although I have no idea to solve the problem. I tried couple of substitutions. One of them is taking-

$z=\left(\frac{dy}{dx}\right)^2$. But this substitution is working because there is a $y$ term in the equation.

Can anyone give any hint or idea regarding the problem? Thanks for help in advance.

Solution 1:

Notice that $2y'y=(y^2)',$ hence you have $(y')^2[1-(y^2)']=c,$ and on that note, $(y')^2=\left(\frac{2yy'}{2y}\right)^2=\left[\frac{(y^2)'}{2y}\right]^2=\frac{[(y^2)']^2}{4y^2},$ so $\frac{[(y^2)']^2}{4y^2}[1-(y^2)']=c.$ Let $z=y^2,$ thus $\frac{(z')^2}{4z}(1-z')=c.$ This can be rewritten as $(z')^2(1-z')=4cz,$ and differentiating results in $2z'z''(1-z')-(z')^2z''=4cz'.$ Let $f=z',$ so $2f(1-f)f'-f^2f'=[2(1-f)-f]ff'=(2-3f)ff'=4cf.$ This is a separable equation. Notice that $f=0$ trivially solves this equation, but for other solutions, you get that $(2-3f)f'=4c,$ which simplifies to $(2f-\frac32f^2)'=4c.$ This brings us to $2f(x)-\frac32f(x)^2=4cx+k,$ equivalent to $12f(x)-9f(x)^2=24cx+6k=24cx+k_0,$ equivalent to $[3f(x)-2]^2=k_1-24cx.$ This translates to $f(x)=\frac{2\pm\sqrt{k_1-24cx}}{3}.$ Accompanied by $f(x)=0,$ and remembering that $f=z',$ this means $z'(x)=\frac{2\pm\sqrt{k_1-24cx}}{3}=\frac23(1\pm\sqrt{k_2-6cx}),\,z'(x)=0.$ This implies $z(x)=\frac23x\mp\frac2{27c}(k_2-6cx)\sqrt{k_2-6cx}+C,\,z(x)=C.$

Here, we should remember that we differentiated the equation $(z')^2(1-z')=4cz,$ so we must check our solutions. $$\left[\frac23(1\pm\sqrt{k_2-6cx})\right]^2\left(1-\left[\frac23(1\pm\sqrt{k_2-6cx})\right]\right)=\left[\frac49(1+k_2-6cx\pm2\sqrt{k_2-6cx})\right]\left[\frac{1\mp2\sqrt{k_2-6cx}}{3}\right]=\frac4{27}(1+k_2-6cx\pm2\sqrt{k_2-6cx})\mp\frac8{27}(1+k_2-6cx\pm2\sqrt{k_2-6cx})\sqrt{k_2-6cx}=\frac4{27}+\frac4{27}(k_2-6cx)\pm\frac8{27}\sqrt{k_2-6cx}\mp\frac8{27}\sqrt{k_2-6cx}\mp\frac8{27}(k_2-6cx)\sqrt{k_2-6cx}-\frac{16}{27}(k_2-6cx)=\frac4{27}-\frac{12}{27}(k_2-6cx)\mp\frac8{27}(k_2-6cx)\sqrt{k_2-6cx}=\frac4{27}-\frac{12}{27}k_2+\frac83cx\mp4c\frac2{27c}(k_2-6cx)\sqrt{k_2-6cx}=4c\left[\frac1{27c}-\frac{3}{27c}k_2+\frac23x\mp\frac2{27c}(k_2-6cx)\sqrt{k_2-6cx}\right]=4c\left[\frac23x\mp\frac2{27c}(k_2-6cx)\sqrt{k_2-6cx}+C\right]$$ if and only if $C=\frac1{27c}-\frac{k_2}{9c}.$ This means that the non-trivial solutions found solve the equation precisely with the condition that $C=\frac1{27c}-\frac{k_2}{9c}.$ Meanwhile, for $z(x)=C,$ we have that $0^2(1-0)=0=4cC,$ forcing $C=0.$ So with that, we have confirmed that $z(x)=\frac1{27c}-\frac{k_2}{9c}+\frac23x\mp\frac2{27c}(k_2-6cx)\sqrt{k_2-6cx}=\frac1{27c}-\frac{k_2}{9c}+\frac6{9c}cx\mp\frac2{27c}(k_2-6cx)\sqrt{k_2-6cx}=\frac1{27c}-\frac1{9c}(k_2-6cx)\mp\frac2{27c}(k_2-6cx)\sqrt{k_2-6cx}$ and $z(x)=0$ are solutions. All that remains is notice that $z=y^2,$ and you have the solutions to your equation. In other words, $$y(x)=0,$$ $$y(x)=\sqrt{\frac1{27c}-\frac1{9c}(k_2-6cx)\mp\frac2{27c}(k_2-6cx)\sqrt{k_2-6cx}},$$ $$y(x)=-\sqrt{\frac1{27c}-\frac1{9c}(k_2-6cx)\mp\frac2{27c}(k_2-6cx)\sqrt{k_2-6cx}}.$$ You can also combine this into a family of piecewise functions by aportioning each solution to a different subset of the domain, and this creates more solutions under suitable conditions.