Solve ODE $2yy''+(y')^2+(y')^4=0.$

Question

Find the general solution of the ODE: $$ 2yy''+(y')^2+(y')^4=0.$$

Attempt. I tried substitution $y=e^z$, but it got me nowhere.

Thanks in advance.

Answer 1

Here is an attempt. Note that $$y''=y'\frac{dy'}{dy}.$$ So, the DE becomes $$2yy'\frac{dy'}{dy}+(y')^2+(y')^4=0.$$ That is, $y'=0$ (which gives $y(x)=k$ for some constant $k$), or $$2y\frac{dy'}{dy}+y'\big(1+(y')^2\big)=0.$$ So, $$-\frac{2\ dy'}{y'\big(1+(y')^2\big)}=\frac{dy}{y}.$$ By integrating this equation, we obtain $$\ln\left(\frac{1+(y')^2}{(y')^2}\right)=\ln(cy)$$ for some constant $c$. So, $$cy-1=\frac{1}{(y')^2}.$$ That is, $$y'=\pm\frac{1}{\sqrt{cy-1}}.$$ Thus, $$\sqrt{cy-1}\ dy=\pm dx.$$ By integrating the equation above, we obtain $$\frac{2}{3c}(cy-1)^{\frac32}=\pm(x-a)$$ for some constant $a$. That is, $$y(x)=\frac{1+\left(\frac{3c}{2}(x-a)\right)^{\frac23}}{c}$$ is another form of solutions (apart from the constant functions).

Answer 2

You can transform the equation (assuming $y'\ne 0$, $y=C=const.$ also being solutions) to $$ -\frac{y'}{y}=\frac{2y'y''}{y'^2(1+y'^2)}=\frac{2y''}{y'}-\frac{2y'y''}{1+y'^2} $$ which integrates to $$ -\ln|y|+c=2\ln|y'|-\ln(1+y'^2)\\~\\ \implies 1+y'^{-2}=Cy,~~\sqrt{Cy-1}\,y'=\pm 1,~~\sqrt{Cy-1}^3=\pm\frac32Cx+D $$

Answer 3

$$ 2yy''+(y')^2+(y')^4=0.$$ The usual change of function to solve an ODE of autonomous kind is : $$\frac{dy}{dx}=f(y)$$ $\frac{d^2y}{dx^2}=\frac{df}{dy}\frac{dy}{dx}=f'(y)f(y)$ $$2yf'f+f^2+f^4=0 $$ This is a separable ODE. $$\frac{dy}{y}=-2\frac{df}{f+f^3}$$ $$y=c_1\left(1+\frac{1}{f^2}\right)$$ $f=\sqrt{\frac{c_1}{y-c_1}}\quad;\quad\frac{dy}{dx}=\sqrt{\frac{c_1}{y-c_1}}.\quad$ Again a separable ODE. $$x=\int \sqrt{\frac{y-c_1}{c_1}}dy$$ After elementary calculus : $$y(x)=c_1+c_1\left(\frac{3}{2c_1}(x+c_2)\right)^{2/3}$$