Is There a Solution to the Boundary Condition Problem for y-2y'+2y=0?

[Lancelot59]

I'm asked to determine if for the solution
$$y=c_{1}e^{x}cos(x)+c_{2}e^{x}sin(x)$$
for:
$$y"-2y'+2y=0$$

whether a member of the family can be found that satisfies the boundary conditions:
$$y(0)=1$$, $$y'(\pi)=0$$

Not quite sure what to do here. The examples in my book give boundary conditions for the same function, not derivatives.

When I put the first condition into y, I got c₁=1, then substituting that result into the derivative condition I found c₂=-1. So I found the constants, does this mean that there is a member of the family that can satisfy the boundary condition? For some reason I think there should be a Wronskian involved.

 

[HallsofIvy]

Yes, there is a member of the family that satisfies the boundary condition and you found it:
$$y(x)= e^x cos(x)- e^x sin(x)$$

 

[Lancelot59]

Alright, thanks for the help!