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Math 285 - Spring 2012 - Review Problems for Exam 1
Problem 1. (1) State the Existence and Uniqueness Theorem for the first order differential
equations y0 = f (x, y).
(2) Show the initial value problem xy0 = 5y, y(0) = 0 has infinitely many solutions.
(3) Does this contradict the Existence and Uniqueness Theorem? Explain.
Answer:
(1) See Theorem 1 on page 24.
(2) It is a separable equation and the general solution is y = Cx5 .
(3) No, if we write the equation as y0 = f (x, y) = 3y
x , then f (x, y) is not continuous at (0, 0),
hence we can’t apply the above Theorem to conclude uniqueness of the solution.
Problem 2. (1) Is equation x2 y0 = 1 − 2x2 + y2 − 2x2 y2 separable? Explain.
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(2) Solve the equation y0 = x2 (y − 2) 3 .
(3) Are there any exceptional solutions not included in the general solution?
Answer:
(1) Yes, it can be written as y0 = ( x12 − 2)(y2 + 1).
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(2) It is separable, general solution is y = ( 13 x2 +C) 2 + 2.
(3) Yes, y = 2 in not included in the general solution.
Problem 3. (1) Solve (x2 + 1)y0 = x(y + 1) as a linear first order differential equation.
(2) Find the general solutions of the linear first order differential equation y0 + p(x)y = 0.
(3) State the Existence and Uniqueness Theorem for the first order differential equations.
(4) Show the IVP 3xy0 + y = 12x, y(0) = 1, has no solution. Does this contradict the EUT?
(5) Solve the (1 − 4xy2 )y0 = y3 .
Answer:
2 + 1) 21 .
(1) y = −1 +C(x
R
(2) y = Ce− p(x)dx .
(3) See Theorem 1 on page 51.
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(4) General solution is y = 3x +Cx− 3 , but none of them is defined at x = 0 except y = 3x which
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does not satisfy the initial condition y(0) = 1. Note that p(x) = 3x
is not continuous at x = 0,
so EUT does not apply here.
dy
(5) Think of x as a function of y, write the equation as (1 − 4xy2 ) dx
= y3 , then 1 − 4xy2 = y3 dx
dy
is a first order linear DE in x and the general solution is x = 21 y−2 +Cy−4 .
1
2
Problem 4. Solve the following differential equations.
(1) y0 = tan(2x + y + 3) − 2.
(2) 2xyy0 = x2 + 2y2 .
(3) x2 y0 + 2xy = 5y4 .
(4) xey y0 = 2(ey + x3 e3x ).
(5) (1 + yexy )dx + (2y + xexy )dy = 0.
(6) xy00 + y0 = 4x.
(7) y3 y00 = 1.
Answer:
(1) Consider the linear substitution u = 2x + y + 3. General solution y = sin−1 (Cex ) − 2x − 3
(2) It is homogeneous. Let u = y/x. General solutions y2 = x2 (ln x +C)
−1 +Cx6
(3) It is a Bernouli Equation. Let u = y−3 . General solutions y−3 = 15
7x
(4) Let u = ey . General solution is y = 2 ln x + ln(e2x +C)
(5) It is an Exact equation. x + y2 + xexy = C
(6) Missing y. Substitute u = y0 and u0 = y00 . General Solution y = x2 +C ln x +C0
(7) Missing x. Substitute u = y0 and y00 = u · du/dy. General solution Ay2 − (Ax + B)2 = 1
Problem 5. Consider a population model in which the death rate and birth rate per unit of
population are β(t) = 0.002 + 0.05P(t) and δ(t) = 0.03 respectively. Show
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, then population will extinct as t → ∞
(1) If P(0) = 10
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(2) If P(0) = 5 , then there will be a Doomsday.
(3) What happens if P(0) = 15 ?
P0
Answer:
= (β − δ)P = (−0.01 + 0.05P)P = −0.05( 51 − P)P, which is a separable equation.
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(1) If P(0) =
(2) If P(0) =
1
5
10 , then P(t) = 1+e0.01t
1
2
5
,
then
P(t)
=
1
5
1− e0.01t
which tends to zero as t → ∞.
which tends to ∞ as t → 100 ln 2 > 0.
2
(3) In this case, the population remains constant.
Problem 6. Suppose an object is shot upward from the ground with v0 = 10 m/s. Assume the
air resistance is proportional to v and ρ = 0.05.
(1) Compute the maximum height it can reach.
(2) How can we compute its velocity when the object hits the ground.
(3) Compute the Terminal Speed if the object could fall forever!
Answer:
(1) We have dv
dt = −ρv − g = −0.05v − 9.8, which is a separable equation and the solution
is v(t) = 206e−0.05t − 196. Now, by integration we find y(t) = −4120e−0.05t − 196t + 4120.
Maximum height is achieved at time t when v(t) = 0, which gives t = 0.995. Thus max height
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is y(0.995) = 4.93.
(2) First find the time t at which the object hits the ground. This is the time when y(t) = 0, using
calculator solve for t. Then plug it in v(t) to find the velocity at that time.
(3) The terminal speed is absolute value of limt→∞ v(t) = −196.
Problem 7. Consider the case of a movement in which air resistance is proportional to v2 .
Set up the differential equations governing the velocity of the object when
(1) the direction of the movement is upward.
(2) the direction of the movement is downward.
(3) Is there a terminal speed in this case in the downward movement?
Answer:
2
(1) dv
dt = −ρv − g.
2
(2) dv
dt = ρv − g.
q
√
(3) Yes, in this case the general solution for velocity is v(t) = − ρg tanh ρgt +C .
Since lim tanh(x) = 1 as t → ∞, then v(t) has a limit as
r
g
lim v(t) = −
t→∞
ρ