Solve the initial value problem d^2 y / dx^2 - 5 dy/dx + 6 y = 0 with y(0) = 2 and y'(0) = 5. What is the value of y(ln 2)?

Answer Key:

6

### Solution: Characteristic equation: r^2 - 5r + 6 = 0 ==> (r - 2)(r - 3) = 0 ==> r = 2, 3. General solution: y(x) = C_1 e^{2x} + C_2 e^{3x}. y(0) = C_1 + C_2 = 2. y'(x) = 2 C_1 e^{2x} + 3 C_2 e^{3x} ==> y'(0) = 2 C_1 + 3 C_2 = 5. Solving: C_1 = 1, C_2 = 1. y(x) = e^{2x} + e^{3x}. At x = ln 2: y(ln 2) = e^{2 ln 2} + e^{3 ln 2} = 2^2 + 2^3 = 4 + 8 = 12. Let's make y(ln 2) = 12.

← Prev Question Next Question →