Finding Nonzero Solutions for a Differential Equation with Boundary Conditions

[LMKIYHAQ]

Homework Statement

Using the given differential eq. M'' + bM=0 where b is a real-valued parameter, find all nonzero values for b that adhere to the following boundary conditions:
M'(0)=0
M'(Pi)=0

Homework Equations

When I apply the boundary conditions, after I set up the equation for f(x) using the gen. sol for the characteristic equation M^2+bM=0, do I need to figure out when f(x) has nonzero solutions or when f ' (x) has nonzero solutions?

The Attempt at a Solution

ex: when b < 0 taking b= -a where a>0 we get the gen. sol:
f(x)=c_1exp(ax) + c_2exp(-ax).
applying the bound. conds we find that c_1= c_2 from f '(0)=0 and using f '(Pi)=0 we find that 0=c_1a [exp(aPi)-exp(-aPi)]. If c_1 and c_2 are not equal to 0 we see the exponential part can never be zero, implying c_1 and c_2 are equal to 0.
Does this mean there are no nonzero sols for f '(x) which is what we want? and or does this say there are no non-zero sols. for f(x) if that is what we want?

I see I need 2 other cases for the value of b. I am just wondering if I am on the right track.

Thanks.

 

[HallsofIvy]

Acutually, when b< 0, you want to take b= -a² to get e^ax and e^-ax as solutions. However, it is still true that for b< 0, this problem has only the "trivial" solution, M(x)= 0.

If b= 0, then M"= 0 and you can get the general solution just by integrating twice.

If b> 0, then let b= a² and two independent solutions are cos(ax) and sin(ax).

 

[LMKIYHAQ]

Acutually, when b< 0, you want to take b= -a² to get e^ax and e^-ax as solutions. However, it is still true that for b< 0, this problem has only the "trivial" solution, M(x)= 0.

If b= 0, then M"= 0 and you can get the general solution just by integrating twice.

If b> 0, then let b= a² and two independent solutions are cos(ax) and sin(ax).

I meant take b= -a^2 for b<0 like you said. I have that part.
For b=0 solving I get M(x)= c_1 + c_2 x. I got M'(x)=c_2. When applying the boundary conditions M'(0)=0 I found that c_2= 0. M'(Pi)=0 gives me the same info that c_2=0. Does this mean a nonzero solution is M(x)=c_1 when c_1 >0?

Also, for b>0 I found that M(x)=c_1 cos(ax)+c_2 sin(ax). applying the boundary conditions M'(x)=0 I got 0=c_2 a. Since a>0 by assumption, c_2=0. For M'(Pi)=0 I got 0= -c_1asin(aPi). taking c_1 not equal to 0 and a>0 (assumed) sin(aPi)=0 when a=n n=1,2,3...

Does this mean the nonzero solutions are M_n(x)=sin(a_nx)=sin(nx)?

Thanks.