# TT'(t) - cT(t) = 0

1. Oct 15, 2006

### Dragonfall

What sort of ODE is

$$tT'(t)-cT(t)=0$$ for a real positive c?

2. Oct 15, 2006

### quasar987

linear homogeneous of first order?

3. Oct 16, 2006

### Dragonfall

I tried solving it using the substitution T=tv like in the book for homogeneous equations, but it doesn't work.

4. Oct 16, 2006

### quasar987

$$y'+P(t)y=Q(t)$$, the solution can be found by multiplying the equation by an integrating factor

$$\mu=e^{\int P(t) dt}$$

Last edited: Oct 16, 2006
5. Oct 16, 2006

### Dragonfall

Got it. Thanks a lot.

6. Oct 17, 2006

### d_leet

In the original posters case, the equation is even simpler to solve since the ODE is separable.

7. Oct 17, 2006

### HallsofIvy

Staff Emeritus
Also that is an "Euler-type" equation or "equipotential" equation since the coefficient of each derivative (here only one) is t to a power equal to the order of the derivative and so T= tr, for some r, is a solution.

But, as d leet said, it is separable and easily integrable.