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

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

• • alan123hk and Delta2
Delta2
Homework Helper
Gold Member
• 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.

• Delta2
bigfooted
Gold Member
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:
$$(\frac{1}{2x^3}p -\frac{1}{2})p = ( \frac{2}{4x^2}p + \frac{1}{2}x)\frac{dp}{dx}$$

 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.

Last edited:
• Delta2 and 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 ~~~ ## • Delta2
bigfooted
Gold Member
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.