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

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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
 
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so can I say c2=0 making:

y=e^x cosx ?
 
Sneakatone said:
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?
 
Sneakatone said:
so can I say c2=0 making:

y=e^x cosx ?

No, you've already established that the coefficient of [itex]e^x \cos(x)[/itex] must be zero. It's the coefficient of [itex]e^x \sin(x)[/itex] which is arbitrary, since [itex]c_2e^0 \sin(0) = 0 = c_2e^{\pi} \sin(\pi)[/itex] for any [itex]c_2 \in \mathbb{R}[/itex].
 
can c2 be -cot(x) ? or can I put any number and it will be correct?
 
Sneakatone said:
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.