Understanding the Type of ODE for tT'(t)-cT(t)=0 with a Real Positive Constant

[Dragonfall]

What sort of ODE is

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

 

[quasar987]

linear homogeneous of first order?

 

[Dragonfall]

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

 

[quasar987]

I don't know about this trick. But when you have an ode of the form

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

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

 

[Dragonfall]

Got it. Thanks a lot.

 

[d_leet]

I don't know about this trick. But when you have an ode of the form

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

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

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

 

[HallsofIvy]

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= t^r, for some r, is a solution.

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