University:
Whitman College
Course:
MATH-244 | Differential Equations
Academic year:

2022
Views:

460

Pages:

4
Author:

supergurl87675m9

Solutions: Section 2.4 1. In Problems 1, 3 and 5 we want to find a single connected interval in which both p(t) and g(t) are continuous. (a) Problem 1: In standard form, y0 + ln(t) 2t y= t−3 t−3 The restriction on the log makes t > 0, and the t − 3 makes t 6= 3. Together, this breaks the number line into two possibilities: Either 0 < t < 3 or t > 3. The initial time is t0 = 1, which makes us choose the first interval, 0 < t < 3. (b) Problem 3: Similarly, y 0 = tan(t)y = sin(t) The tangent function has vertical asymptotes, which separates the possible intervals that could be used. Some possibilities: − π π 3π 3π 5π π < t < , or < t < , or 2. Since the initial time is t0 = 1, we choose the middle interval, −2 < t < 2. y0 + 2. Problem 7: In this case, f (t, y) = t−y 2t + 5y We see that f is continuous away from the line 2t + 5y = 0, which splits the plane into two pieces (above/below the line). Check the partial derivative with respect to y: ∂f −7t = ∂y (2t + 5y)2 so we still have the same restriction. 3. Problem 9: f (t, y) = ln |ty| 1 − t2 + y 2 This is continuous away from ty = 0 (away from the t− and y−axis, and where 1 − t2 + y 2 6= 0. The points where this is zero forms a hyperbola in the plane. Compute the partial derivative: ∂f = ∂y t (1 ty − t2 + y 2 ) − ln |ty|(2y) (1 − t2 + y 2 )2 This does not add any new restrictions. 4. Problem 13: Solve y 0 = −4t/y, y(0) = y0 . This is separable, so Z y dy = −4 Z 1 t dt ⇒ y 2 = −2t2 + C 2 With the initial condition, 1 1 1 2 y0 = 0 + C ⇒ y 2 = −2t2 + y02 2 2 2 We can simplify a bit: y 2 = −4t2 + y02 q y(t) = ± y02 − 4t2 ⇒ We would choose the positive root if y0 > 0, or the negative root if y0 < 0. Furthermore, for the square root to be defined, y02 − 4t2 > 0 ⇒ (y0 − 2t)(y0 + 2t) > 0 This separates the plane into three regions. We choose the middle since t0 = 0. Assuming y0 > 0, we have: y0 y0 − ≤t≤ 2 2 (If y0 < 0, flip the inequalities). 5. Problem 14: y 0 = 2ty 2 , y(0) = y0 . Separable: Z y −2 dy = Z 2t dt ⇒ 1 − = t2 + C y With the initial condition, C = −1/y0 , so: y(t) = 1 y0 = (1/y0 ) − t2 1 − y0 t2 This is valid as long as y0 6= 0. What if it is? Then we see that y(t) = 0 is the unique solution. If y0 6= 0, then we continue by looking at where 1 − y0 t2 = 0 1 t = ±√ y0 ⇒ This is valid only if y0 > 0. If y0 < 0, then the denominator, 1 − y0 t2 is never zero (for any t). Thus, if y0 < 0, the solution that we previously obtained is valid for all t. The last case is where y0 > 0. Since the initial time is t0 = 0, then the solution y(t) is only valid for: 1 1 −√ < t < √ y0 y0 Summary: In this homework problem, we saw that the time interval on which the solution is valid depended greatly on the initial value of y, • If y0 < 0, y(t) is valid for all time. • If y0 = 0, y(t) = 0 is the solution, valid for all time. √ • If y0 > 0, y(t) is valid for a short segment of time, between ±1/ y0 . 6. Problem 15: y 0 + y 3 = 0, y(0) = y0 . This is separable: y 0 = −y 3 − y −3 dy = dt 1 −2 y =t+C 2 This technique is only valid if y 6= 0. From the original differential equation, we see that y(t) = 0 would be the (unique) solution if y0 = 0. Otherwise, we can solve for C: 1 =0+C 2y02 We now want to solve for y explicitly: y −2 = 2t + The main issue here: 2t + 1 y02 1 y02 y2 = 1 2t + 1 y02 > 0 in order for us to solve for y. In that case, t>− 1 2y02 Also, we need to decide on whether to take the positive or negative square root when solving for y(t): This depends on the sign of y0 ; if y0 > 0, then take the positive root, otherwise take the negative root. In any event, we still have the restriction on time. 7. Problem 22: Parts (a) and (b) were done in class. The point of part (c) is to show that the second solution, y2 (t) cannot be found as a constant multiple of the first (why? This will become clearer in Chapter 3). Clearly, no value of a constant in ct + c2 can give you the function − 41 t2 . 8. Problem 23: (a) Show that e2t and ce2t (c any constant) are both solutions to the ODE: y 0 −2y = 0. You can show this directly (by substitution), or by actually solving the DE. You should see that the general solution is y(t) = Ae2t (b) Show that 1t is a solution to y 0 + y 2 = 0, but Ct is not. You can again show this directly (by substitution), or by actually solving the DE. If you solve it, you should get: y(t) = 1 t−C (or y(t) = 0). 9. Problem 24: To show this, first note that, if y(t) = φ(t) is a solution to y 0 + p(t)y = 0, then: φ0 + p(t)φ = 0 Now substitute y(t) = cφ: y 0 = cφ0 , and: cφ0 + p(t)cφ = c (φ0 + p(t)φ) = c · 0 = 0 10. Problem 25: Same idea as 24. Substitute the expression in to see what you get. Assume that y1 solves y 0 + p(t)y = 0. This means that y10 + p(t)y10 = 0. Assume that y2 solves y 0 + p(t)y = g(t). This means that y20 + p(t)y2 = g(t). Now, substitute y = y1 + y2 , y 0 = y10 + y20 into the DE: (y10 + y20 ) + p(t)(y1 + y2 ) = (y10 + p(t)y1 ) + (y20 + p(t)y2 ) = 0 + g(t) = g(t)

Differential Equations Solutions Section 2.4

Related Documents

Description

Related Documents

Free up your schedule!

Take 5 seconds to unlock