University:
California State University, Northridge
Course:
MATH 351 | Differential Equations
Academic year:

2006
Views:

94

Pages:

5
Author:

Bunicco14j

Practice problems and their solutions Problem 1. Find values of m so that y = exp(mt) is a solution of (a) dy + 2y = 0, dt (b) d2 y − y = 0, dt2 d3 y d2 y dy − 3 2 + 2 = 0. 3 dt dt dt (c) Solution (a) m exp(mt)+2 exp(mt) = 0 ⇐⇒ (m+2) exp(mt) = 0 ⇐⇒ m+2 = 0 (since exp(mt) > 0) ⇐⇒ m = −2. (b) m2 exp(mt)−exp(mt) = 0 ⇐⇒ (m2 −1) exp(mt) = 0 ⇐⇒ m2 −1 = 0 (since exp(mt) > 0) ⇐⇒ m = ±1. (c) m3 exp(mt) − 3m2 exp(mt) + 2m exp(mt) = 0 ⇐⇒ (m3 − 3m2 + 2m) exp(mt) = 0 ⇐⇒ m3 − 3m2 + 2m = 0 (since exp(mt) > 0) ⇐⇒ m(m2 − 3m + 2) = 0 ⇐⇒ m = 0, 1, 2. Problem 2. Find the solution of the initial value problem y 0 = (1 − 2t)y 2 , y(0) = − 1 6 (1) and determine the interval in which the solution is defined. Solution Equation (1) is separable; integrating (when y 6= 0) we obtain Z Z dy 1 1 = (1 − 2t) dt, or − = t − t2 + c, or y = 2 , 2 y y t −t−c and 1 6 =⇒ 4 6 y(0) = − c = 6, 3 1 . −t−6 Remark 1: The graph of 1/(t2 − t − 6) consists of three branches, but only the middle one represents the solution to (1). (See Figure 1.) However, it is easy to see that the left branch can be a solution too, for example, when the initial value is given by y(−4) = 1/14. The same is true for the right branch, for example, when the initial value is y(4) = 1/10. y(t) = 2 t2 1 -6 -4 -2 2 -1 -2 -3 Figure 1. The graph of the function 1/(t2 − t − 6) Solution is defined in the interval containing 0 where y = 1/(t2 − t − 6) makes sense, i.e., −2 < t < 3. Note that y(−2) and y(3) are not defined. Problem 3. Solve the initial value problem 3t2 , 3y 2 − 4 and determine the interval in which the solution is valid. y0 = 1 y(1) = 0, (2) 2 Solution Equation (2) is separable; integrating (when 3y 2 − 4 6= 0) we obtain Z Z 2 (3y − 4) dy = 3t2 dt =⇒ y 3 − 4y = t3 + c =⇒ and y(1) = 0 =⇒ c = −1, and thus y 3 − 4y − t3 + 1 = 0. We look for a solution defined in the interval containing 1. The integral curve y 3 − 4y − t3 + 1 = 0 has vertical tangent lines at the points where y 0 is unbounded, when 3y 2 o − 4 = 0 (remember, y 0 = 3t2 /(3y 2 − 4)), or n √ i.e., √ √ √ 3 when y = ±2/ 3. This gives us t3 − 1 = ± 2/ 3 − 4 2/ 3 = ±16/(3 3). In other words, the sought √ √ interval is either the set of t for which |t3 − 1| < 16/(3 3) or the set of t for which |t3 − 1| > 16/(3 3). Only the first choice √ guarantees that t = 1 belongs to it. (Do you know why ?) Thus, the interval is given by 3 |t − 1| < 16/(3 3), or by 3 2 1 -3 −1.28 ≈ 16 1− √ 3 3 1/3 0. Solution r(r − 1)t2 tr−2 + 4rttr−1 + 2xtr = 0 ⇐⇒ [r(r − 1) + 4r + 2]tr = 0 ⇐⇒ r(r − 1) + 4r + 2 = 0 (since tr > 0 for t > 0) ⇐⇒ r = −1, −2. The solution are y1 (t) = c1 t−1 and y2 (t) = c2 t−1 , where c1 and c2 are arbitrary constants. Problem 5. Solve the initial value problem dy = y 2 cos(πx), dx 1 y(0) = − . 2 Solution The equation (3) can be written as dy = cos(πx)dx. y2 (3) 3 Integrating the left side with respect to y and the right side with respect to x yield: 1 π 1 . − = sin(πx) + c, or y(x) = − y π sin(πx) + πc 1 1 =⇒ c = 2. y(0) = − = − c 2 Thus the solution to the initial value problem (3) is π y(x) = − sin(πx) + 2π For the graph of the solution of (3) see Figure 3, below. -0.45 -0.475 -3 -2 -1 1 2 3 -0.525 -0.55 -0.575 Figure 3. The graph of the solution to equation (3). Problem 6. Solve the following differential equations by separation of variables. (a) y2 dy = , dx 1 + x 2 + y 2 + x2 y 2 (c) exp(x) dy = 2x, dx (b) dN + N = N t exp(t + 2), dt (d) y exp(x) dy = exp(−y) + exp(−2x − y). dx Solution 1 1 1 1 (a) From , or + 1 dy = dx we obtain − + y = tan−1 (x) + c, or y = 2 2 y 1+x y (1 − c) − tan−1 (x) 1 y= . c̄ − tan−1 (x) (b) From 1 dN N = (t exp(t + 2) − 1) we obtain log |N | = t exp(t + 2) − exp(t + 2) − t + c, or N = ± exp(c) exp [t exp(t + 2) − exp(t + 2) − t], or N = c̄ exp [t exp(t + 2) − exp(t + 2) − t]. (c) From dy = 2x exp(−x)dx we obtain y = −2x exp(−x) − 2 exp(−x) + c. (d) From y exp(y)dy = [exp(−x) + exp(−3x)] dx we obtain y exp(y) − exp(y) + exp(−x) + Problem 7. Verify that the indicated functions are solutions of the given differential equation. dy (a) 2ty + (t2 + 2y) = 0; t2 y + y 2 = c, where c ∈ R is arbitrary. dt 1 exp(−3x) = c. 3 4 HINT: Use implicit differentiation. Rt (b) y 0 + 2ty = 1; y = exp(−t2 ) 0 exp(s2 ) ds + c exp(−t2 ), where c ∈ R is arbitrary. Solution a) The differential equation has the form 2ty + (t2 + 2y)y 0 = 0. After implicitly differentiating t2 y + y 2 = c we obtain 2ty + (t2 + 2y)y 0 = 0. Rt (b) Differentiating y = exp(−t2 ) 0 exp(s2 ) ds + c exp(−t2 ) we obtain Z t Z t 2 2 2 0 2 2 2 exp(s2 ) ds − 2ct exp(−t2 ). exp(s ) ds − 2ct exp(−t ) = 1 − 2t exp(−t ) y = exp(−t ) exp(t ) − 2t exp(−t ) 0 0 Substituting into differential equation, we have Z t Z t 2 2 2 0 2 exp(s2 ) ds + 2ct exp(−t2 ) = 1. exp(s ) ds − 2ct exp(−t ) + 2t exp(−t ) y + 2ty = 1 − 2t exp(−t ) 0 0 Problem 8. In the theory of learning, the rate at which a subject is memorized is assumed to be proportional to the amount that is left to be memorized. In addition, the amount of material forgotten is proportional to the amount memorized at time t. Suppose that M denotes the total amount of a subject to be memorized and A(t) is the amount memorized at time t. (a) Determine a differential equations for the amount A(t). (b) Solve the differential equation from part (a) with the initial condition A(0) = A 0 > 0. Solution (a) The differential equation is dA = k1 (M − A) − k2 A, dt where k1 and k2 are positive constants. (b) This is the first order linear differential equation; its solution with the initial value A(0) = A0 is A(t) = A0 exp[−(k1 + k2 )t] + Remark 1. The limiting value of A(t), as t → ∞, memorized. M k1 {1 − exp[−(k1 + k2 )t]} . k1 + k 2 M k1 , is less than the amount M of the subject to be k1 + k 2 Problem 9. A thermometer is taken from an inside room to the outside, where the air temperature is 5 ◦ F. After 1 minute the thermometer reads 55◦ F, and after 5 minutes the reading is 30◦ F. What was the initial temperature of the room ? NOTE: Use Newton’s law of cooling. Solution The differential equation for temperature T of the thermometer is dT = k(T − 5), T (0) = T0 ⇐⇒ T (t) = (T0 − 5) exp(kt) + 5, (4) dt where both k and T0 (initial temperature of the room) are unknown. Our goal is to find T0 . We know that 55◦ = T (1) = (T0 − 5) exp(k) + 5, ◦ 30 = T (5) = (T0 − 5) exp(5k) + 5. (5) (6) 5 Equations (5)-(6) are two equations for two unknown, k and T0 . The solutions are ln 2 ln 2 k= − k=− =⇒ 4 4 T0 − 5 = 50 exp(−k) T0 = 5 + 50(2)1/4 Temperature of the room is T0 = 5 + 50(2)1/4 ≈ 64.46◦ . Problem 10. p 1 − y 2 , y(0) = −1, other than y(t) = −1. p (b) Show that y(t) = 1 is the only solution of the initial-value problem y 0 = t 1 − y 2 , y(0) = 1. (a) Find a solution of the initial-value problem y 0 = t Solution (a) It is easy to see that y(t) = −1 p is a solution of the initial-value problem. One can find another solution by 0 observing that the equation y = t 1 − y 2 is separable. The usual integration ! Z Z t2 1 p dy = t dt =⇒ sin−1 (y) = + c. 2 1 − y2 together with the initial value y(0) = −1 yield c = −π/2 and t2 π sin (y) = − , 2 2 −1 or equivalently The solution given in (7) exists for π t2 π ≤ , − 2 2 2 or equivalently for − √ 2π ≤ t ≤ 2 t y = − cos . 2 √ 2π. (7) (Do you know why ?) p (b) As in part (a) it is obvious that y(t) = 1 is a solution of the initial-value problem y 0 = t 1 − y 2 , y(0) = 1. Now, if φ(t) 6≡ 1 is another solution, then φ ↑ for t > 0 and φ ↓ for t < 0. Thus, since φ(0) = 1 and φ(t) 6≡ 1, there exists t1 such |φ(t1 )| > 1. We have reached a contradiction. p Remark 2. If one integrates y 0 = t 1 − y 2 , y(0) = 1 as in part (a), one obtains ! Z Z 1 t2 p dy = t dt =⇒ sin−1 (y) = + c. 2 1 − y2 The initial condition y(0) = 1 gives c = π/2 and we have 2 t or equivalently y = cos . 2 p However, it is easy to check that cos(t2 /2) does NOT satisfy the equation y 0 = t 1 − y 2 . t2 π sin (y) = + , 2 2 −1 Can you find a mistake in the above argument ?

Differential Equations,Solutions

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