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Solutions to Differential Equations in Math 308: Section 1.2, Study notes of Differential Equations

The detailed solutions to selected problems in math 308: differential equations, focusing on section 1.2. It covers various methods to find equilibrium solutions and families of solutions for given differential equations.

Typology: Study notes

Pre 2010

Uploaded on 08/18/2009

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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
2y=dt = ln |2y|=t+C1= |2y|=etC1=C2et.
(Note that C2=eC1>0.) Now, since y(t)<2, we have 2 y > 0, thus |2y|= 2 y. Thus
|2y|=C2et=2y=C2et=y= 2 C2et.
Suppose, on the other hand, that y(t)>2; then 2 y < 0, so |2y|=(2 y). Then
|2y|=C2et= (2 y) = C2et=y= 2 + C2et
The two cases (y(t)<2 and y(t)>2) have given us corresponding families of solutions,
y= 2 C2et
and
y= 2 + C2et,
where in each case, the constant C2is positive. We can combine the two cases into
y= 2 + C3et
where C3is an arbitrary constant (possibly positive or negative). In fact, by setting C3= 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 y6= 0:
dy
dt = (ty)2=t2y2=dy
y2=t2dt =1
y=t3/3 + C1=y=1
t3/3 + C1
=3
t3+C2
.
So the solutions are
y(t) = 3
t3+C2
,
where C2is an arbitrary constant, or
y(t) = 0.
1.2/12. y(t)0 is an equilibrium solution. If y(t)6= 0, dy
dt =t3
y=ty1/3=dy
y1/3=tdt =3y2/3
2=
t2/2 + C1=y2/3=t2/3 + C2=y2= (t2/3 + C2)3=y=±p(t2/3 + C2)3.Thus the solutions are
y(t) = 0,or y(t) = p(t2/3 + C2)3,or y(t) = p(t2/3 + C2)3.
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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.