# Determine whether a member of the family can be found that satisfies the boundary conditions.

## 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 dont know this one
(d) y(0) = 0, y(π) = 0

## 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
Homework Helper
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.

so can I say c2=0 making:

y=e^x cosx ?

LCKurtz
Homework Helper
Gold Member
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
Homework Helper
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
Homework Helper
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
Homework Helper
Gold Member
That is a valid answer but you are taking c2= 1, not 0.
No it isn't. See post #4.

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

LCKurtz