Can Differential Equations with Boundary Conditions Have Non-Unique Solutions?

[Sneakatone]

Homework Statement

The given two-parameter family is a solution of the indicated differential equation on the interval
(−infinity, infinity).
Determine whether a member of the family can be found that satisfies the boundary conditions. (If yes, enter the solution. If an answer does not exist, enter DNE.)

y = c1e^x cos x + c2e^x sin x; y'' − 2y' + 2y = 0

I completed the 1st 3 but I don't know this one
(d) y(0) = 0, y(π) = 0

Homework Equations

The Attempt at a Solution

when y(0) = 0 :
0=c1 e^(0)cos(0)+c2 e^(0)sin(0)
c1=0

when y(π) = 0:
0=c1e^π cosπ+c2 e^π sinπ
0=c1e^π+0
c1=0

 

[pasmith]

That tells you that c_1 = 0. But c_2 can be anything! You're not asked to find a unique solution, but to find at least one solution.

 

[Sneakatone]

so can I say c2=0 making:

y=e^x cosx ?

 

[LCKurtz]

so can I say c2=0 making:

y=e^x cosx ?

Your original problem should have stated you were looking for a solution that isn't identically zero that satisfies the boundary conditions. Your answer I have quoted doesn't satisfy $y(\pi)=0$. The point is, can you find a $c_2$ that isn't zero so you have a nontrivial solution?

 

[pasmith]

so can I say c2=0 making:

y=e^x cosx ?

No, you've already established that the coefficient of e^x \cos(x) must be zero. It's the coefficient of e^x \sin(x) which is arbitrary, since c_2e^0 \sin(0) = 0 = c_2e^{\pi} \sin(\pi) for any c_2 \in \mathbb{R}.

 

[HallsofIvy]

so can I say c2=0 making:

y=e^x cosx ?

That is a valid answer but you are taking c2= 1, not 0.

 

[LCKurtz]

That is a valid answer but you are taking c2= 1, not 0.

No it isn't. See post #4.

 

[Sneakatone]

can c2 be -cot(x) ? or can I put any number and it will be correct?

 

[LCKurtz]

can c2 be -cot(x) ? or can I put any number and it will be correct?

$c_1$ and $c_2$ are constants. Post #2 pointed out to you $c_1=0$ and $c_2$ can be anything. Try something. Then check if the solution you get satisfies the two boundary conditions.