Question

In this problem, y = c1ex + c2e−x is a two-parameter family of solutions of the second-order DE y'' − y = 0. Find c1 and c2 given the following initial conditions. (Your answers will not contain a variable.)

y(1) = 0, y'(1) = e

Answer 1

To find c1 and c2 given the initial conditions y(1) = 0 and y'(1) = e, substitute these values into the general solution and solve for the constants c1 and c2. The solution is c1 = e/(1 + e^-1 + e^-2) and c2 = -1.

Step-by-step explanation:

To find c1 and c2 given the initial conditions y(1) = 0 and y'(1) = e, we can substitute these values into the general solution y = c1e^x + c2e^-x and solve for the constants c1 and c2.

1. Substitute x = 1 into the equation y = c1e^x + c2e^-x:
2. 0 = c1e^1 + c2e^-1
3. Substitute x = 1 into the derivative of y, which is y' = c1e^x - c2e^-x:
4. e = c1e^1 - c2e^-1
5. We now have a system of two equations with two unknowns (c1 and c2). We can solve this system using algebraic methods.
6. Divide the second equation by e^1:
7. e/e^1 = c1 - c2e^-2
8. Substitute the value of c2 from the first equation into the second equation:
9. e/e^1 = c1 - (0 - c1e^1)e^-2
10. Simplify the right side:
11. e = c1 - (-c1e^1)e^-2
12. Expand the right side:
13. e = c1 + c1e^-1 + c1e^-2
14. Combine like terms:
15. e = c1(1 + e^-1 + e^-2)
16. Divide both sides by (1 + e^-1 + e^-2):
17. c1 = e/(1 + e^-1 + e^-2)
18. Substitute the value of c1 back into the first equation:
19. 0 = (e/(1 + e^-1 + e^-2))e^1 + c2e^-1
20. Simplify the left side:
21. 0 = e/(1 + e^-1 + e^-2) + c2e^-1
22. Multiply both sides by (1 + e^-1 + e^-2):
23. 0(1 + e^-1 + e^-2) = e + c2e^(-1 + 2)
24. Simplify the right side:
25. 0 = e + c2e
26. Combine like terms:
27. 0 = (1 + c2)e
28. Divide both sides by 1 + c2:
29. 0/(1 + c2) = e
30. Solve for c2:
31. c2 = -1

Therefore, c1 = e/(1 + e^-1 + e^-2) and c2 = -1.