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

[hunt_mat]

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.

 

[thomas49th]

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 ;)

 

[HallsofIvy]

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?

 

[hunt_mat]

For this particular problem why not do as I suggested? It's a valid teachnique, it's a known solution.