I How do I solve this first order second degree differential equation?

[alan123hk]

TL;DR Summary

How do I solve this first order second degree differential equation ?

How to solve this first order second degree differential equation ?

$\left(\frac {dy} {dx}\right)^2 + 2x^3 \frac {dy} {dx} - 4x^2y=0$

Thanks.

 

[anuttarasammyak]

(y'+x^3)^2-x^6-4x^2(y+x^4/4)+x^6=0
z'^2=4x^2z
where $z=y+\ x^4/4$.
We see z>0 and
\frac{dz}{dx}=\pm 2x z^{1/2}

 

[Delta2]

I tried it at wolfram, it gives two solutions at perplexed (as an equation f(x,y)=c) form, here is the link
https://www.wolframalpha.com/input/?i=(dy/dx)^2+2x^3(dy/dx)-4x^2y=0

 

[alan123hk]

This is actually a differential equation that I encountered when I was in university many years ago. Of course, I myself did not have the ability to solve this differential equation. At that time, the professor only told me that its solution was $~y=cx^2+c^2~$ , and did not mention the method of solving. Now I am very interested to know how to find the analytical solution, or whether there is a standardized or generalized method to solve such a complicated differential equation.

 

[bigfooted]

This is a d'Alembert ODE, also known as a Lagrange ODE. The form of this ODE is
$$ y = xf(y') + g(y') $$
If the "4" was a "2" then it is a Clairaut ODE, which is of the form:
$$ y = xy' + g(y') $$

The solution method involves introducing a new variable for the derivative: $y'=p$ and taking the derivative of the ODE again. Some explanation with worked out examples can be found here:
https://www.math24.net/lagrange-clairaut-equations
https://www.12000.org/my_notes/dAlmbert_ode/index.htm

For your example we simply introduce $y'=p$ and write it as $y=xf(p) + g(p)$:

$$ y = \frac{1}{2}xp + \frac{1}{4x^2}p^2$$

and we then differentiate with respect to x:

$$ p = \frac{1}{2}p + \frac{1}{2}xp' +\frac{-1}{2x^3}p^2 + \frac{2}{4x^2}pp'$$
or:
$$ (\frac{1}{2x^3}p -\frac{1}{2})p = ( \frac{2}{4x^2}p + \frac{1}{2}x)\frac{dp}{dx}$$

[edit] minus sign wrong, this should be of course : $(\frac{1}{2x^3}p +\frac{1}{2})p = ( \frac{2}{4x^2}p + \frac{1}{2}x)\frac{dp}{dx}$

we invert the derivation and write it as an ODE in $\frac{dx}{dp}$:

$$ (\frac{1}{2x^3}p -\frac{1}{2})\frac{dx}{dp} = \frac{x( \frac{1}{2x^3}p +\frac{1}{2} )}{p}$$

We see that if your original ODE had a different minus sign (or maybe I made a mistake?) and was actually $y'^2 - 2x^3y' +4x^2y=0$ then this ODE would simplify to

$$ \frac{dx}{dp} = x/p$$

Whose solution is $x= C p$ or $p = Cx$

The final solution of this ODE with different sign in terms of y is obtained by substitution of p into the d'Alembert equation above:

$$ y = \frac{1}{2}Cx^2 - \frac{1}{4x^2}C^2x^2 = C^{*} x^2 - C^{*2}$$

Which is pretty close to the solution that you remember from university.

 

[alan123hk]

Which is pretty close to the solution that you remember from university.

Thanks you for providing detailed steps to solve this equation.

I tried to rewrite once as follows.

$\left(\frac {dy} {dx}\right)^2 + 2x^3 \frac {dy} {dx} - 4x^2y=0$

$\text {Let}~~ p=\frac {dy} {dx} ~~~,~~~ p^2+2x^3p-4x^2y=0 ~~~ ,~~~y = \frac{1}{2}xp + \frac{1}{4x^2}p^2 ~~~~~ (1)$

$\text {differentiate with respect to x} ~~~,~~~ p = \frac{1}{2}p + \frac{1}{2}x \left(\frac {dp}{dx}\right) +\frac{-1}{2x^3}p^2 + \frac{2}{4x^2}p\left(\frac {dp}{dx}\right)$

$\left(\frac{1}{2x^3}p -\frac{1}{2}+1\right)p = \left( \frac {2} {4x^2}p +\frac {1}{2}x\right) \frac {dp}{dx} ~~~,~~~\left(\frac{1}{2x^3}p+\frac{1}{2}\right)p = x\left( \frac {1} {2x^3}p +\frac {1}{2}\right) \frac {dp}{dx}$

$\text {Thus} ~~~ \frac {dx}{x}=\frac{dp}{p}~~~,~~~p=Cx$

$\text{Substitute into equation (1)} ~~~y=\left(\frac{1}{2}C\right)x^2+{\left(\frac{1}{2}C\right)}^2=cx^2+c^2 ~~~$

 

[bigfooted]

I saw I made a mistake with the minus sign, I've made an edit to point this out.

 

[alan123hk]

I saw I made a mistake with the minus sign, I've made an edit to point this out.

This is just a small algebraic error, it does not affect you to show a wonderful way to solve that differential equation.