How to solve $yy'' + (y')^2 = (y')^3$?

Question

I want to solve the following differential equation

$$yy'' + (y')^2 = (y')^3.$$

But I don't understand how should I do it. At first, I thought that I should let $$u=y'$$

But the solution is

$$Cy^2 − x + y = D$$

And I can't understand how to solve it. Thanks.

Answer 1

let $$u(y)=\frac{dy(x)}{dx}$$ then we have $$u(y)^2+yu'(y)u(y)=u(y)^3$$ factorizing $$u(y)\left(-y\frac{du(y)}{dy}+u(y)^2-u(y)\right)=0$$ you must solve $$u(y)=0$$ or $$-y\frac{du(y)}{dy}+u(y)^2-u(y)=0$$ this Can be written as $$\frac{\frac{du(y)}{dy}}{u(y)^2-u(y)}=\frac{1}{y}$$ Can you finish?

Answer 2

First, rewrite your ODE as

$$y'' = \dfrac{(y')^3-(y')^2}{y}=f(y,y').$$

This type of ODE can be solved by the following substitution

$$y'(x)=v(y(x)) \implies y''(x)=\dfrac{dv(y)}{dy}\dfrac{dy(x)}{dx}=v'(y)y'(x)=v'(y)v(y)$$

By applying this substitution for the ODE we obtain

$$v'v=\dfrac{v^3-v^2}{y} \implies v'(y)=\dfrac{v^2(y)-v(y)}{y},$$

which is separable.

You will also need to exclude the cases $y\equiv 0$ and $y' \equiv 0$ separately, which are trivial solutions to the ODE.

Answer 3

(You may see this as another form of Dr. Sonnhard Grauber's answer.)

Define $u$ as: $$u:=y'=\frac{\mathrm{d}y}{\mathrm{d}x}$$

You can see: $$y''=\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}u}{\mathrm{d}y}\cdot\frac{\mathrm{d}y}{\mathrm{d}x}=\color{red}{u\frac{\mathrm{d}u}{\mathrm{d}y}}$$

Substituting these into the equation gets: $$yu\frac{\mathrm{d}u}{\mathrm{d}y}+u^2=u^3$$ or equally: $$y\frac{\mathrm{d}u}{\mathrm{d}y}+u=u^2\\\implies \frac{\mathrm{d}y}{y}=\frac{\mathrm{d}u}{u^2-u}$$

You can now integrate bothsides and find $u$ as a function of $y$. After that, by substituting $u$ back with $\frac{\mathrm{d}y}{\mathrm{d}x}$, you'll end up with a simple (separable) first order equation involving $x%$ and $y$.

Hope it helps.

Answer 4

To obtain a direct solution, mimicing the other answers, transform the equation to be composed of integrable terms as $$ \frac{y''}{y'^2-y'}=\frac{y'}{y} $$ where we need to exclude the points in the state space were $y=0$ or $y'=0$ or $y'=1$. All three cases provide solution classes, $y=c$, $y=x+c$.

Integrating gives $$ \ln|y'-1|-\ln|y'|=\ln|y|+c $$ or with $C=\pm e^c$ $$ 1-\frac1{y'}=Cy\implies x=y-\frac C2 y^2+D $$ etc.

Answer 5

$$yy'' + (y')^2 = (y')^3$$ $$y''y=(y')^3-(y')^2$$ $$ydy'=((y')^2-(y'))\frac {dy}{dx}dx$$ $$\frac {dy'}{((y')^2-(y'))}=\frac {dy}{y}$$ Simply integrate now $$\int \frac {dy'}{y'(y'-1)}=\int \frac {dy}{y}=\ln|y|+K$$ $$......$$

Another simple way is to consider the derivative of $y'y$ $$yy'' + (y')^2 = (y')^3$$ $$(y'y)' = \frac {(y'y)^3}{y^3}$$ $$-\frac {(y'y)'}{(y'y)^2} = -\frac {y'}{y^2}$$ Integrate to get $$\frac {1}{(y'y)} =\frac {1}{y}+K$$ $$\frac {1}{(y')} =1+Ky$$ Which is easy to solve... $$\boxed{x' =1+Ky \to x=y+K_1y^2+K_2}$$