Math 211 Fall 2007: Solutions: HW #2
Instructor: S. Cautis
1. section 2.4, #4
y ′ + 2ty = 5t
2
2
2
recognizing that (et y)′ = et (y ′ + 2ty) we multiply both sides by et to
get:
2
2
et (y ′ + 2ty) = et 5t which is:
2
2
(et y)′ = et 5t integrating both sides we get::
2
2
et y = 52 et + C so we have
y = 25 + eCt2
2. section 2.4, #8
(1 + x3 )y ′ = 3x2 y + x2 + x5
x2 +x5
3x2
y ′ − 1+x
3 y = 1+x3
R 3x2
3
− ln(1+x3 )
=
Recognize that 1+x
3 dx = ln(1+x ) we multiply both sides by e
1
1+x3 and simplify as before to get:
2
5
2
1
x +x
x
′
( 1+x
3 y) = (1+x3 )2 = 1+x3 integrating:
1
1
3 2
1+x3 y = 6 ln(1 + x ) + C giving
y = 13 (1 + x3 ) ln(1 + x3 ) + C(1 + x3 )
3. section 2.4, #14
Solve the initial value problem: y ′ = y + 2xe2x and y(0) = 3
(a) First, find the general solution of y ′ − y = 2xe2x
multiplying by the integrating factor (e−x ) and simplifying we get:
(e−x y)′ = 2xe2x−x integrating both sides and simplifying we get :
y = 2xe2x − 2e2x + Cex
(b) to solve the initial value problem C = 5
4. section 2.4, #18
Solve the initial value problem and discuss the interval of existence: xy ′ +
2y = sin(x) y( π2 ) = 0
(a) First, find the general solution of y ′ + x2 y = sin(x)
x
multiplying by the integrating factor (e2 ln(x) = x2 ) and simplifying we
get:
(x2 y)′ = x sin(x) integrating both sides and simplifying we get:
y = x12 (−x cos(x) + sin(x) + C)
(b) To solve the initial value problem C = −1
(c) The only problem point is x = 0. Taking into account the initial value,
the interval of existence is (0, ∞).
5. section 2.5, #4
initial state: 500 gal H2 0
concentration: .05 lb/gal (or 25 lb salt total)
500
Equation:A′ (t) = − rA(t)
500 solving this equation we get: A(t) = De
Given our initial condition we hav:
−rt
6r
= ln(5) so r = 50
A(t) = 25e 500 if we want A(60) = 5 then 50
6 ln(5)
−rt
6. section 2.5 #6
Note: this problem was written incorrectly in the text.
(a) the incorrect problem set up:
Initial: 100 gal H2 0, 5 lb. salt.
Flow in: 3 gal/min pure H2 O
Flow out: 2 gal/min
Equation: A′ (t) = −2A(t)
100+t
C
A(t) = (100+t)
2
given the initial condition C is 50,000.
(b) The correct problem:
Initial: 100 gal H2 0, 5 lb. salt.
Flow in: 2 gal/min pure H2 O
Flow out: 3 gal/min
What is the content of salt when there are 50 gal of solution left?
Equation:A′ (t) = −3A(t)
100−t
A(t) = C(100 − t)3
5
C = 100
3
A(50) = 85
7. section 2.5 #12
(a) Tank A: 100 gal H2 O, 20lb salt
Tank B: 200 gal H2 O, 40 lb salt
Flow into A: 5 gal pure H2 O
Flow out A: 5 gal
Flow into B: 5 gal from A
Flow out B: 2.5 gal
Equation describing the amount in Tank A:
A′ (t) = −5A(t)
100
t
A(t) = Ce −20
given our initial conditions C = 20
Equation describing the amount in Tank B:
2.5B(t)
B ′ (t) = 5A(t)
100 − 200+2.5t
t
Substituting for A(t): B ′ (t) =
((80 + t)B(t))′ = (80 + t)e
t
−20
20e −20
20
−
2.5
200+2.5t B(t)
t
= e −20 −
B(t)
80+t
t
t
t
1
[−20 ∗ 80e −20 + (−20)t(e −20 ) − 20 ∗ 20e −20 + C]
B(t) = 80+t
Using the initial condition, C = 5200 and so B(20) = 52 − 16
e −
8. section 2.6 #6
Find dF when F (x, y) = ln(xy) + x2 y 3
dF = ( x1 + 2xy 3 )dx + ( y1 + 3x2 y 2 )dy
9. section 2.6 #10
Find if the equation is exact solve it if it is.
Following the books notation: dP
dy = − sin(x)
and dQ
dx = − sin(x) so it is exact.
F (x, y) = x + y cos(x) + φ(y) where
φ′ (y) = cos(x) − cos(x) so
F (x, y) = x + y cos(x) + C
10. section 2.6 #14
Find if the equation is exact solve it if it is.
Following the books notation: dP
dy = 0
dQ
dx
= −1 so it is not exact.
4
e
−
4
e