ODE review question for PDE class. (word question)

Nick Bruno
Messages
98
Reaction score
0
1. This problem concerns a nonhomogeneous second order linear ODE L[y] = g(t).
Suppose that:
y1(t) satisfies the ODE with the initial conditions y(0)=1, y'(0) = 0,
y2(t) satisfies the ODE with the initial conditions y(0)=0, y'(0) = 1, and
y3(t) satisfies the ODE with the initial conditions y(0)=0, y'(0) = 0.
Find the solution to the ODE with the initial conditions y(0) = alpha, y'(0) = beta. Justify your answer.




2. Homework Equations ... this is my problem



3. I haven't started this class yet, but this should be a review from my ODE class. (Its been a long while) This homework question is actually from a partial differential equation class starting in the fall.

from what I remember, second order ODE means there is a second derivative involved. The linear term implies an intremental decrease (or increase) in exponents and the nonhomogeneous term implies that Q(x) does not equal 0.

It seems the question is giving an equation L[y] = g(t) but I don't see any primes or derivatives in this equation, no exponents, and I think the notation may be screwing me up? Any advice would help. Regards,
 
Physics news on Phys.org
L[y] is an operator that represents a 2nd order, linear, nonhomogeneous differential equation. For your problem, I think you can assume that L[y] = a(t)y'' + b(t)y' + c(t)y.
 
thank you
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top