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

• I
• alan123hk
In summary: Thank you again!In summary, we discussed how to solve a first order second degree differential equation given by ##(dy/dx)^2 + 2x^3(dy/dx) - 4x^2y = 0.## By introducing a new variable for the derivative, we can rewrite the equation in the form ##y = xf(p) + g(p)## and take the derivative again to obtain an ODE in terms of ##dx/dp.## By solving this ODE, we can substitute back to find the solution in terms of y, which is ##y = Cx^2 - C^2.##
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.

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

alan123hk and Delta2
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
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
bigfooted said:
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
I saw I made a mistake with the minus sign, I've made an edit to point this out.

alan123hk
bigfooted said:
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.

## 1. How do I identify a first order second degree differential equation?

A first order second degree differential equation is an equation that contains a first derivative and a second derivative of a dependent variable with respect to an independent variable. It can be written in the form: y'' = f(x,y,y').

## 2. What is the general solution to a first order second degree differential equation?

The general solution to a first order second degree differential equation is a function that satisfies the equation for all possible values of the independent variable. It is typically expressed in terms of the independent variable and one or more arbitrary constants.

## 3. How do I solve a first order second degree differential equation using separation of variables?

To solve a first order second degree differential equation using separation of variables, you need to rearrange the equation so that all terms containing the dependent variable are on one side and all terms containing the independent variable are on the other side. Then, integrate both sides with respect to their respective variables and solve for the constant of integration.

## 4. What is the difference between an initial value problem and a boundary value problem for a first order second degree differential equation?

An initial value problem for a first order second degree differential equation involves finding a solution that satisfies the equation and a given set of initial conditions, usually in the form of a specific value for the dependent variable at a certain value of the independent variable. A boundary value problem, on the other hand, involves finding a solution that satisfies the equation and a given set of boundary conditions, usually in the form of specific values for the dependent variable at two different values of the independent variable.

## 5. Can I use software to solve a first order second degree differential equation?

Yes, there are many software programs and online tools available that can solve first order second degree differential equations. These tools use numerical methods to approximate the solution, which can be helpful for more complex or non-linear equations. However, it is still important to understand the concepts and methods used to solve these equations by hand.

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