this is the review sheet I do not know how professor got y=ce^(2t) and Yp=At +B

Question

this is the review sheet I do not know how professor got y=ce^(2t) and Yp=At +B

can you show me the steps in detail how my professor got those answer

(15%, 15 min) Write down the pseudo-code for the improved Euler method with h = 05 and final value t = 2. Perform the first 2 iterations by hand and compare those values with the exact solution. y' = .5 - t + 2y, y (0) = 1 Solve the homogeneous equation to get: y = ce2t Now use the method of undetermined coefficients: yp = At + B A = .5 - t + 2At + B A = .5, B = 0 Hence y = ce2t + .5t and the initial conditions c = 1 Thus y = e 2t + .5t

Explanation / Answer

y ' = .5 - t + 2y

You want the diff eq in the form: ay ' + by = g(t)

y ' - 2y = .5 - t

Then you want to solve the characteristic equation where ar + b = 0.

r - 2 = 0

r = 2

If your r is a real root, then y = ce^(rt) is a solution. This is how your teacher got

y = ce^(2t).

Since g(t) = .5 - t or -t + .5, this is a polynomial so you want to choose

Yp = At + B.

Yp ' = A

Plugging Yp and Yp ' back into the diff eq y ' - 2y = .5 - t gives:

A - 2(At + B) = .5 - t

A - 2At + 2B = .5 - t

-2At + (A + 2B) = -t + .5

matching up coefficients:

-2At = -t

A + 2B = .5

-2At = -t gives A = 1/2 then A + 2B = .5 becomes 1/2 + 2B = 1/2.

2B = 0

B = 0

Then Yp = At + B is Yp = (1/2)t

final solution equation is y = y1 + Yp

y = ce^(2t) + (1/2)t

y(0) = 1 so:

1 = ce^(0) + (1/2)(0)

c = 1

final equation: y = e^(2t) + (1/2)t

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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