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

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

[tex] \dot{x}=\frac{dx}{dt}[/tex]

You know the chain rule?

[tex] \frac{df}{dx}=\frac{dt}{dx}\frac{df}{dt}[/tex]

Let:

[tex] f=\frac{dy}{dt}[/tex]

and you have your answer.
 
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Okay,

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

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

Thanks
I think I see ;)
 
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
[tex]\frac{d^2y}{dx^2}+ 3x\frac{dy}{dx}+ y= cos(ln(x))[/tex]
by letting [itex]y= x^r[/itex]?