HallsofIvy said:You really have to think a little. You know, I hope, that the general solution to y"+ y= 0 is y(x)= Ccos(x)+ Dsin(x) so that y'(x)= -Csin(x)+ Dcos(x). Now put those functions into your boundary conditions:
y(1)+ y(-1)= Ccos(1)+ Dsin(1)+ Ccos(-1)+ Dsin(-1)= 0
y'(1)+ y'(-1)= -Csin(1)+ Dcos(1)- Csin(-1)+ Dcos(-1)= 2.
Using the fact that cosine is an even function and sine is an odd function will simplify these a lot.
Didn't we just do this in another thread?
The differential equation "y''(x) + y(x) = 0" is a second-order ordinary differential equation. It is also known as a homogeneous linear differential equation, as the dependent variable y and its derivatives appear only in a linear form.
The initial conditions y(1)+y(-1)=0 and y'(1)+y'(-1)=2 represent boundary conditions for the differential equation. They indicate the value of the dependent variable y and its derivative at specific points in the domain of the function.
To solve "y''(x) + y(x) = 0" with the given boundary conditions, we can use the method of undetermined coefficients. This involves assuming a solution of the form y(x) = Ae^(mx), where A and m are constants. We can then substitute this solution into the differential equation and the boundary conditions to determine the values of A and m.
Yes, for example, if we assume the solution y(x) = Ae^(mx), we can substitute it into the differential equation to get m^2Ae^(mx) + Ae^(mx) = 0. This simplifies to (m^2 + 1)Ae^(mx) = 0. Since e^(mx) is never equal to 0, we can conclude that m^2 + 1 = 0. Solving for m, we get m = ±i. Therefore, the general solution to the differential equation is y(x) = Ae^(ix) + Be^(-ix). Using the boundary conditions, we can determine the values of A and B, giving us the specific solution y(x) = sin(x) - cos(x).
Boundary conditions are important in solving differential equations because they help determine the specific solution to the equation. Without boundary conditions, the solution to a differential equation would only be a general solution, which could have an infinite number of possible solutions. Boundary conditions give us specific values to work with, allowing us to find the unique solution that satisfies both the differential equation and the given conditions.