Second ODE - Using x = e^t show that the equation

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Homework Help Overview

The discussion revolves around transforming a second-order ordinary differential equation (ODE) using the substitution \( x = e^t \). The original equation is given as \( ax^{2}\frac{d^{2}y}{dx^{2}} + bx\frac{dy}{dx} + cy = 0 \), where \( a, b, c \) are coefficients. Participants are exploring the implications of this substitution on the derivatives involved.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of the chain rule to express derivatives with respect to \( t \) in terms of derivatives with respect to \( x \). There are attempts to differentiate \( y \) as a function of \( x(t) \) and to apply product rules in the context of second derivatives. Some participants express confusion about the correct application of these rules and the implications of the substitution on the derivatives.

Discussion Status

There is ongoing exploration of the transformation process, with various participants attempting to clarify the relationships between the derivatives. Some have made progress in understanding the differentiation process, while others are still grappling with specific steps and notation. No consensus has been reached, but several productive lines of reasoning are being pursued.

Contextual Notes

Participants are navigating the complexities of differentiating functions that depend on both \( x \) and \( t \), with some expressing uncertainty about the implications of implicit differentiation in this context. The discussion reflects a mix of understanding and confusion regarding the application of the chain and product rules in the transformation of the ODE.

  • #31
Standard terminology really, sorry if i didn't explain.

<br /> \dot{x}=\frac{dx}{dt}<br />

You know the chain rule?

<br /> \frac{df}{dx}=\frac{dt}{dx}\frac{df}{dt}<br />

Let:

<br /> f=\frac{dy}{dt}<br />

and you have your answer.
 
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  • #32
Okay,

<br /> <br /> \frac{df}{dx}=\frac{dt}{dx}\frac{df}{dt}<br /> <br />

You were using f but in our case we set y = dy/dt?

Thanks
I think I see ;)
 
  • #33
hunt_mat said:
In order to avoid all this nonsense about tranformations, just look for solutions of the form x^n of your original equation
In other words just don't do the problem you are given?

There are many good reasons for knowing that this substitution will change an Euler-type (or "equipotential") equation to an equation with constant coefficients having the same characteristic equation.

For example, how would you solve
\frac{d^2y}{dx^2}+ 3x\frac{dy}{dx}+ y= cos(ln(x))
by letting y= x^r?
 
  • #34
For this particular problem why not do as I suggested? It's a valid teachnique, it's a known solution.
 

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