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Math 308 Differential Equations Fall 2002
Solved Problems
Section 1.2/ 8, 10, 12, 28, 33, 35
1.2/8. I’m going to do this one in gory detail–you don’t have to do all these steps explicitly every time you solve a problem like this. First, we note that y(t) ≡ 2 is an equilibrium solution. Now suppose y(t) 6 = 2; in fact, let’s assume that y(t) < 2 Then
dy dt
= 2 − y =⇒
dy 2 − y
= dt =⇒ − ln | 2 − y| = t + C 1 =⇒ | 2 − y| = e−t−C^1 = C 2 e−t.
(Note that C 2 = e−C^1 > 0.) Now, since y(t) < 2, we have 2 − y > 0, thus | 2 − y| = 2 − y. Thus
| 2 − y| = C 2 e−t^ =⇒ 2 − y = C 2 e−t^ =⇒ y = 2 − C 2 e−t.
Suppose, on the other hand, that y(t) > 2; then 2 − y < 0, so | 2 − y| = −(2 − y). Then
| 2 − y| = C 2 e−t^ =⇒ −(2 − y) = C 2 e−t^ =⇒ y = 2 + C 2 e−t
The two cases (y(t) < 2 and y(t) > 2) have given us corresponding families of solutions,
y = 2 − C 2 e−t
and y = 2 + C 2 e−t,
where in each case, the constant C 2 is positive. We can combine the two cases into
y = 2 + C 3 e−t
where C 3 is an arbitrary constant (possibly positive or negative). In fact, by setting C 3 = 0, we obtain the equilibrium solution, so this formula gives us all the solutions.
1.2/10. First note that y(t) ≡ 0 is an equilibrium solution. Now suppose y 6 = 0:
dy dt
= (ty)^2 = t^2 y^2 =⇒
dy y^2
= t^2 dt =⇒
y
= t^3 /3 + C 1 =⇒ y = −
t^3 /3 + C 1
t^3 + C 2
So the solutions are
y(t) = −
t^3 + C 2
where C 2 is an arbitrary constant, or y(t) = 0.
1.2/12. y(t) ≡ 0 is an equilibrium solution. If y(t) 6 = 0,
dy dt
= t 3
y = ty^1 /^3 =⇒
dy y^1 /^3
= tdt =⇒
3 y^2 /^3 2
t^2 /2 + C 1 =⇒ y^2 /^3 = t^2 /3 + C 2 =⇒ y^2 = (t^2 /3 + C 2 )^3 =⇒ y = ±
(t^2 /3 + C 2 )^3. Thus the solutions are
y(t) = 0, or y(t) =
(t^2 /3 + C 2 )^3 , or y(t) = −
(t^2 /3 + C 2 )^3.
1.2/26. y(t) ≡ 0 is an equlibrium solution. If y(t) 6 = 0,
dy dt
= ty^2 + 2y^2 = (t + 2)y^2 =⇒
dy y^2
= (t + 2)dt =⇒ −
y
= t^2 /2 + 2t + C 1 =⇒ y = −
t^2 /2 + 2t + C 1
We require y(0) = 1, which gives us y(0) = − (^) C^11 = 1 =⇒ C 1 = −1. Thus
y(t) = −
t^2 /2 + 2t − 1
t^2 + 4t − 2
1.2/28. We see that x(t) ≡ 0 is an equilibrium solution. Now if x(t) 6 = 0,
dx dt
= −xt =⇒
dx x
= −tdt =⇒ ln |x| = −t^2 /2 + C 1 =⇒ |x| = e−t
(^2) /2+C 1 = C 2 e−t
(^2) / 2 =⇒ x = C 3 e−t
(^2) / 2 .
Thus the general solution is x(t) = C 3 e−t
(^2) / 2 ,
where C 3 is an arbitrary constant. We want the solution for which x(0) = 1/
π, so we have
x(0) = C 3 e^0 = C 3 = 1/
π.
So the solution to the initial value problem is
x(t) = e−t
(^2) / 2 √ π
1.2/33. First, recall from Calculus that
dy y^2 + 1
= arctan(y). Then
dy dt = (y^2 + 1)t =⇒
dy y^2 + 1 = tdt =⇒ arctan(y) = t^2 /2 + C 1 =⇒ y = tan
t^2 /2 + C 1
We want the solution where y(0) = 1, so we must have y(0) = tan (C 1 ) = 1 =⇒ C 1 = π/4. Thus the solution to the initial value problem is
y(t) = tan
t^2 /2 + π/ 4
1.2/35. Let S(t) be the amount of salt (measured in pounds) in the bucket at time t, where t is the time (in minutes). There are two processes here that are causing S(t) to change. First, we are dumping salt into the bucket at the rate 1/4 lbs/min; this will cause the amount of salt to increase at the rate 1/4 lbs/min. The amount of salt is also changing because of the water that is leaving through the spigot. We are told that the water leaves at the rate 1/2 gal/min. This water contains salt, so this causes the amount of salt to decrease, but at what rate? The concentration of the salt in the water is S(t)/5 lbs/gal (we are told that the bucket holds five gallons). The rate at which the salt is leaving is given by the formula (concentration)×(flow rate), which in this case gives us (S(t)/5) × (1/2) = S(t)/10. Thus, the bucket is gaining salt at the rate of 1/4 lbs/min, but is also losing it at the rate of S(t)/ 10 lbs/min. Therefore, the net rate of change of the salt is 1/ 4 − S/10, and the corresponding mathematical version of that statement is dS dt
S
We are told that initially the bucket is full of pure water, which means
S(0) = 0.
