- Thread starter Belief7
- Start date

The question is as follows:

Find the equation of the paths of πΒ²y/ππ₯Β² β π₯ + 2π₯Β³ = 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*xΒ³

I am stuck in the same step.

Please help me proceed further,.

- Joined
- Jan 27, 2012

- Messages
- 4,938

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.

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.

- Joined
- Jan 27, 2012

- Messages
- 4,938

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.