# Equation of Paths for Second order ODE: d^y/dx^2 - x - 2x^3 = 0

#### Belief7

##### New member
Find the equation of the paths of 𝑑 2𝑦 𝑑𝑥 2 − 𝑥 + 2𝑥 3 = 0 sketch the paths in the plane. Locate their critical points and determine its nature.

Can someone please provide a solution with an easy to understand explanation?

#### Belief7

##### New member
Hi,

The question is as follows:
Find the equation of the paths of 𝑑²y/𝑑𝑥² − 𝑥 + 2𝑥³ = 0. Sketch the paths in the plane.
Locate their critical points and determine its nature.

I have calculated the system of LDE to be:
dy/dx = t + 0*y
dt/dx d= x - 2*x³

I am stuck in the same step.
Please help me proceed further,.

#### HallsofIvy

##### Elite Member
Your differential equation is $$\displaystyle \frac{d^2y}{dx^2}- x- 2x^3= 0$$ and you are asked to "sketch the paths in the xy-plane"?

I would just write the equation as $$\displaystyle \frac{d^2y}{dx^2}= 2x^3+ x$$ and integrate twice:
$$\displaystyle \frac{dy}{dx}= \frac{1}{2}x^4+ \frac{1}{2}x^2+ C$$
$$\displaystyle y(x)= \frac{1}{10}x^5+\frac{1}{6}x^3+ Cx+ D$$

Each choice of C and D gives a different curve.

#### Belief7

##### New member
Your differential equation is $$\displaystyle \frac{d^2y}{dx^2}- x- 2x^3= 0$$ and you are asked to "sketch the paths in the xy-plane"?

I would just write the equation as $$\displaystyle \frac{d^2y}{dx^2}= 2x^3+ x$$ and integrate twice:
$$\displaystyle \frac{dy}{dx}= \frac{1}{2}x^4+ \frac{1}{2}x^2+ C$$
$$\displaystyle y(x)= \frac{1}{10}x^5+\frac{1}{6}x^3+ Cx+ D$$

Each choice of C and D gives a different curve.
thank you! that helped! Although, I was wondering whether we could we first formulate a system of linear DE from the given second-order DE rather than just separating the variables and integrating. I would like to know how you would do that.

#### HallsofIvy

##### Elite Member
The standard method would be to introduce the new variable, $$\displaystyle u= \frac{dy}{dx}$$. Then $$\displaystyle \frac{d^2y}{dx^2}= \frac{du}{dx}= 2x^3+ x$$.

Now you have the system of equations
$$\displaystyle \frac{dy}{dx}= u$$
$$\displaystyle \frac{du}{dx}= 2x^3+ x$$.

Since the second equation does not involve y those equations are "uncoupled" and the simplest way to handle this is to integrate the second equation directly.