# Constant or variable coefficient?

1. Mar 14, 2005

### loli12

i was asked to identify the following, linear? homogeneous? variable/constant coefficient?

note: x' = dx/dt

3(t^2)x'' = tx' + 4x -In(t)

I thought this is a linear, non-homogenous equation with variable coefficient, but the answer said this eqt has constant coefficient.. can someone please explain to me why?
Thanks

2. Mar 14, 2005

### saltydog

The equation has variable coefficients: the t's in front of the derivatives. I bet a dollar your answer is right.

3. Mar 14, 2005

### HallsofIvy

Staff Emeritus
It is an "equi-potential" or "Euler type" equation since the coefficient of each derivative has degree equal to the order of the derivative. It can be converted to an equation with constant coefficients by the change of variable, u= ln(t). Then
x'= (dx/du)(du/dt)= (1/t)(dx/du) etc.

By the way, it is "ln(t)" not "In(t)". I see that over and over again. Does it not occur to any one that "I" would be a very peculiar initial for a "l"ogarithm?

4. Mar 17, 2005

### saltydog

Thanks Hall. Glad it was only a dollar. I took the time to work through that. So for the benefit of me and loli12:

$$ax^2\frac{d^2y}{dx^2}+bx\frac{dy}{dx}+cy=0$$

Letting $t=ln(x)$ or $x=e^t$ and using the chain rule we have:

$$\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}$$

$$\frac{d^2y}{dx^2}=\frac{dy}{dx}\frac{d^2x}{dt^2}+\frac{dx}{dt}[\frac{d^2y}{dx^2}\frac{dx}{dt}]$$

And so:

$$\frac{dy}{dx}=e^{-t}\frac{dy}{dt}$$

$$\frac{d^2y}{dx^2}=e^{-2t}[\frac{d^2y}{dt^2}-\frac{dy}{dt}]$$

Substituting this into the equation leads to:

$$ay^{''}+(b-a)y^{'}+cy=0$$

Oh yea, when you solve for y(t) in the converted equation, remember to convert it back to x using the substitution: $t=ln(x)$. Ok, I'm done.

Last edited: Mar 17, 2005
5. Mar 17, 2005

### HallsofIvy

Staff Emeritus
But your original post just said you were supposed to identify it as "constant coefficients" or "variable coefficients"??