## Math 53 - Spring First Midterm Exam

Math 53 - Spring 2009 - First Midterm Exam Name: Student ID: Signature: Instructions: Print your name and student ID number, write your signature to...
Author: Derick McDowell
Math 53 - Spring 2009 - First Midterm Exam

Name: Student ID: Signature:

Instructions: Print your name and student ID number, write your signature to indicate that you accept the honor code. During the test, you may not use computers, phones, or any other electronic device. Read each question carefully, and show all your work. Justify all your answers. You have 120 minutes to answer all the questions. At most two bonus points may be awarded for quality of the presentation.

Question

Score

Maximum

1

10

2

15

3

15

4

20

5

20

6

20

Total

100

Problem 1. (10 pts.) Solve for t < 0 the differential equation t3 y ′ + 4t2 y = e−t ,

y(−1) = 0.

Solution: The ODE is linear and explicit: 4 e−t y′ + y = 3 , t t

y(−1) = 0

and its coefficients are continuous on (−∞, 0). We can use the method of integrating factors: µy ′ +

d 4µ y = (µy) = µy ′ + µ′ y t dt

iff

4µ t 4 which is satisfied by µ(t) = t . We hence study the equivalent ODE µ′ =

d 4 (t y) = t4 y ′ + 4t3 y = te−t dt yielding 4

t y(t) = C +

Z

t −s

se ds = C + 0

[−se−s ]t0

+

Z

t

0

e−s ds = C − te−t − e−t + 1

where C is an integration constant. (Note that despite the fact that 0 does not belong to the interval of study, we may use it as a bound for the integral, in order to find a primitive of se−s , which is well-defined at 0.) Finally  y(t) = t−4 C + 1 − (t + 1)e−t and using y(−1) = 0 yields C = −1, so

y(t) = −

t + 1 −t e . t4

1

Problem 2. Consider the ODE (y + 1)

dy = cos x, dx

y(0) = 2.

a) (5 pts.) Check that existence and uniqueness of solutions are guaranteed. Solution: dy cosx = f (x, y) = . The function f dt y+1 cos x and ∂f /∂y = − (y+1) 2 are continuous, except when y = −1. Given the initial condition y(0) = 2 we have existence and uniqueness, at least on some interval (−ǫ, ǫ).

The ODE is explicit and nonlinear when written

b) (10 pts.) Find the solution and its interval of definition. Solution: We recognize a separable equation and integrate   2 y + y = sin x + C where C is an integration constant. 2 This implicit√equation can be made explicit by solving the degree 2 equation, √ yielding y(x) = −1 ± 2C + 1 + 2 sin x. Using the initial condition, we obtain −1 ± 2C + 1 = 2 which can only √ be fufilled when the square root is nonnegative. Hence C = 4 and y(x) = −1 + 9 + 2 sin x. The function is thus defined for all x ∈ R.

2

Problem 3. (15 pts.) Find the solution for x > 0 of (ln x − 3)

dy y + + 3ex = 0, dx x

y(1) = e,

and determine its interval of definition. Solution: We first check that it is an exact equation of the form N(x, y)y ′ +M(x, y) = 0 for N = ln x−3 and M = y/x + 3ex . Indeed ∂N 1 ∂M = = . ∂x x ∂y We now look for primitives of M in the x variable: they are of the form ψ(x, y) = y ln x + 3ex + h(y) where h is any differentiable function of y. Using now N = ∂ψ/∂y = ln x + h′ (y) we see that h′ (y) = −3 so that h(y) = −3y + C and ψ(x, y) = y(ln x − 3) + 3ex + C. Since y(1) = e, we have C = 0 and 3ex y(x) = 3 − ln x defined on the interval (0, e3 ) (we can check that 1 indeed lies in that interval).

3

Problem 4. Find a general solution of the ODE x2

dy = x2 + 3xy + y 2 dx

using the following sequence of steps: a) (5 pts.) Let w be the function defined by w(x) = y(x)/x. Which ODE does w satisfy? Solution: dy dw dy = w + x . Replacing y and in the ODE by the above dx dx dx expression yields x2 (w + xw ′ ) = x2 + 3x2 w + x2 w 2 , which we can divide by x2 Since y = xw, we have

xw ′ = 1 + 2w + w 2 = (w + 1)2 which is a separable equation. b) (10 pts.) Solve this ODE in the w function. Solution: 1 w′ = 2 (w + 1) x so that −1/(w + 1) = ln |x| + C and w(x) = −1 − 1/(C + ln |x|). c) (5 pts.) Give the solution in terms of the function y. Solution: y(x) = xw(x) = −x −

4

x C + ln |x|

Problem 5. Determine the intervals of definition for the following equations. a) (10 pts.) t2 dy/dt =

1 + cos t and y(1) = 0. Same question if instead y(2π) = 0.

Solution: The ODE is actually linear, and the functions of t are continuous except at t = 0 (since 1 + cos t ≥ 0) so a solution with y(1) = 0 exists on the interval (0, +∞). The same result holds for y(2π) = 0, because both 1 and 2π lie in (0, +∞). One might also remark that this ODE is simply the equation of a derivative, so that Z t√ 1 + cos s y(t) = C + ds s2 t0 where t0 may chosen anywhere in (0, +∞). Obviously a constant C can be chosen for any initial condition. b) (10 pts.) dy/dt =

1 , and y(0) = 2. (y − 1)(y + 2)

Solution: The ODE is separable and can be integrated by writing (y − 1)(y + 2)y ′ = (y 2 + y − 2)y ′ = 1 y3 y2 + −2y = t+ C. The initial condition implies that C = 2/3. The solution 3 2 will be defined in some interval containing 0 (as can also be derived from the existence theorem).

yielding

2

1

1

-1

2

-1

-2

-3

The derivative cannot be infinite: y 6= 1 and y 6= −2, so y takes values in (−∞, −2), (−2, 1) and (1, +∞). Only the third interval contains the initial value 2, so we need only determine for which value of t we have y = 1: 1/3 + 1/2 − 2 = t + 2/3, so that we want to exclude t = −11/6. Therefore the interval of definition is (−11/6, +∞). Note that y = −2 is not a problem (see picture), indeed for t ∈ (−11/3, 8/3) there is another solution, which is decreasing and goes from 1 to −2. However that solution does not take the value 2 at t = 0.

5

Problem 6. Consider a meteor falling vertically in a uniform gravitational field, and losing its mass in the process. Denote the meteor’s speed by v(t), and let the gravitational force be F (t) = m(t)g. Both v(t) and F (t) are taken to be positive, even though they correspond to vectors pointing down. Assume that for some constant α > 0, the meteor’s mass obeys dm (t) = −αv(t)m(t), dt

m(0) = M.

Newton’s second law for bodies with varying mass is not simply F = ma, instead it is F (t) =

d [m(t)v(t)]. dt

Take the meteor to be at rest at time t = 0, and disregard that it could hit the surface of the Earth. a) (10 pts.) Solve for v(t). Do you recover v(t) = gt in the limit α → 0? Solution: We have two equations dm = −αvm (1) dt

and F = mg =

d dm dv [mv] = v+m dt dt dt

(2).

Replacing dm/dt in (2) by its expression we obtain: mg = m

dv − αv 2 m dt

and dividing by m (which is valid as long as the meteor exists), v satisfies the following dv explicit nonlinear separable ODE = g + αv 2 , which we write as dt dv/dt = 1 + αv 2 /g

dv/dt  p 2 = g. 1 + v α/g

r

p g arctan(v(t) α/g), so that α r g √ tan(C + t αg). v(t) = α

We recognize the derivative of

√ The initial value yields C = 0. For a given time t, when α → 0, t αg → 0, so that r r g g √ √ tan(t αg) ∼ t αg = gt. v(t) = α α

6

b) (10 pts.) Solve for m(t). When does the meteor turn to dust (m = 0)? Solution: Substituting the expression of v from above in equation (1) we have dm √ √ = −m gα tan(t αg) dt √ which is again a separable equation yielding ln m(t) = ln(cos(t gα)) + C ′ , ie m(t) = √ ′ eC cos(t gα) for some constant C ′ . Using the initial value m(0) = M, we see that √ √ ′ eC = M so m(t) = M cos(t gα). The meteor turns to dust at time T = π/2 gα. c) (Bonus for 2pts, only if you have time.) Criticize the model for m(t), based on your expressions for v(t) and m(t).

7