# Linear ODE Non-constant coefficient

1. Jan 19, 2010

### jschmid2

Hi. I'm having difficulty remembering how to solve for u(r).
The equation is r*u''+u'=0 with BC u(2)=20; u(1)=540.

Any help would be appreciated. I really need help setting up how to solve. Thanks.

2. Jan 19, 2010

### elibj123

This is an Euler equation (a standard form would be with u'' coefficient being r^2, so just multiply the equation by r), and is solved with guessing a solution $$u(r)=r^{\lambda}$$
Substituting into the equation gives:

$$\lambda (\lambda-1)r^{\lambda -1}+\lambda r^{\lambda-1}$$

Then divding by the power of r gives you the characteristic polynomial:

$$P(\lambda)= \lambda ^ {2} =0$$

So you have one solution $$u(r)=r^{0}=1$$, the second solution will e $$u(r)=ln(r) r^{0}=ln(r)$$ (which is a result of a manipulation analogous to constant coefficient theory)

And a general solution is $$u(r)=A+B ln(r)$$, and then just use initial conditions

3. Jan 19, 2010

### jschmid2

Thanks so much. It reminded me of Cauchy-Euler, but I did not know how to approach it with lambda=0.

4. Jan 20, 2010

### HallsofIvy

Another way to do this problem is to note that u does not appear explicitely in the problem.

Let v= y' and the equation becomes rv'+ v= 0, a simple, separable, first order equation.

rv'= -v so dv/v= -dr/r and, integrating, ln v= ln -r+ C or v= C'/r.

Now we have u'= C'/r so integrating again, u(t)= C'/2 ln r+ C", exactly what elibj123 got.