- #1
Lancelot59
- 640
- 1
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.
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.
