Find the values of lambda for which the given problem has a nontrivial solution

Question

Find the values of lambda for which the given problem has a nontrivial solution also determine the corresponding eigenfunctions y"-2y'+(lambda)y=0 0

Explanation / Answer

There are three possible ranges that lambda can take on, negative, 0, or positive. Case 1: If it's 0, y" = 0. y' = c. y = cx + d. Since y(0) = 0, d = 0. y = cx. y'(pi) = 0, c = 0. This trivial. Case 2: If it's negative, you have two real repeated roots for your R equation (r^2 - constant = 0). So you will have an answer in the form of C1e^(R1x) + C2e^(-R1x). y(0) = 0 C1+C2 = 0. The derivative of that is C1R1e^(R1x) - C2R1e^(-R1x). C1R1e^(R1pi) - C2R1e^(-R1pi) = 0. C1R1e^(R1pi) = C2R1e^(-R1pi) ---> C1/C2 = e^(-2R1pi). Substitute. C2e^(-2R1pi) + C2 = 0. C2(1+e^(-2R1pi) = 0, C2 = 0 because e^x is never 0. C1 = 0. Trivial again. Case 3: If it's positive, you have imaginary roots and trigonometric answer. Your answer is in the form of C1cos(R1x) + C2sin(R1x). y(0) = 0 = C1 + 0 = C1 = 0. Your answer becomes C2sin(R1x). Derivative of that is C2R1cos(R1x). At y'(pi) = 0. C2R1cos(R1pi) = 0. We see that C2 is forced to be 0 if cos(R1pi) is not 0. So, cos(R1pi) has to be 0. cos is 0 at 1/2pi and 3/2pi. R1*pi has to be in the form of 1/2 pi + k*pi. R1 = 1/2 + k, where k is any integer. Lambda has to be in the form of (1/2 + k)^2. Eigenfunctions = C2sin(R1x).

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.