Question

Consider the differential equation below, in which a,b,c,k are real constants and c² +k² >0 : (∗) dxdy​ = cx+kyax+by .

(a) Find 4 specific numbers a,b,c,k for which the function y=x² −3x provides a solution for equation (∗) on the interval (0,+[infinity]). Briefly explain your method.
(b) Is there only one correct answer to part (a)? If so, explain why; if not, describe the set of all choices for (a,b,c,k) that should be accepted.
(c) Consider the ODE (∗) with the constants found in part (a). Find a solution satisfying y(2)=− 2549 . Hint: Start by postulating y=xu(x) for some new unknown function u. Find u, then recover y.

Answer 1

To find specific numbers a, b, c, and k for which the function y=x² −3x provides a solution for the given differential equation on the interval (0, +∞), substitute y=x² −3x into the differential equation, simplify the equation, compare the coefficients, solve for k, and substitute the values into y=x² −3x.

Step-by-step explanation:

To find specific numbers a, b, c, and k for which the function y=x² −3x provides a solution for the given differential equation on the interval (0, +∞), we need to substitute y=x² −3x into the equation and solve for a, b, c, and k. Let's go through the steps:

1. Substitute y=x² −3x into the differential equation:
2. dxdy = cx + kyax + by
3. dxdy = c(x) + k(x² − 3x)ax + b(x² − 3x)
4. Simplify the equation:
5. dxdy = (cx + kx² − 3kx)ax + (bx² − 3bx)
6. Compare the coefficients of each term:
7. ca = cx + kx² − 3kx
8. b = bx² − 3bx
9. From the first equation, we have:
10. ca = cx + kx(x − 3)
11. ca = cx + kx² − 3kx
12. ca = x(c + kx − 3k)
13. Since the equation holds true for all x, we can equate the coefficients:
14. c + kx − 3k = 0
15. From the second equation, we have:
16. b = x² − 3x
17. Comparing the coefficients, we get:
18. a = 1
19. b = -3
20. c = -3k
21. Rearrange the equation c + kx − 3k = 0 to solve for k:
22. kx = 3k − c
23. x = (3k − c)/k
24. Substitute the values of a, b, and c into the equation y=x² −3x:
25. y = (3k − c)² − 3(3k − c)
26. Expand and simplify:
27. y = 9k² − 6ck + c² − 9k + 3c

By comparing this equation with the given function y=x² −3x, we can determine the values of a, b, c, and k to be:

a = 1, b = -3, c = -3k (where k can be any real number), and k ≠ 0.