Calculus IV - HW 2

MA 214

Due 6/29

Section 2.5

3. (Problems 16 from B&D) Another equation that has been use to model populationgrowth is the Gompertz equation

dy

dt

= ry ln(K/y)

where r and k are positive constants.

(a) Sketch the graph of f(y) versus y, find the critical points, and determine whethereach is asymptotically stable or unstable.

(b) For 0 y K, determine where the graph of y versus t is concave up and whereit is concave down.

(c) For each y in 0 < y K, show that dy

dt

as given by the Gompertz equation is

never less than dy

dt

as given the the logistic equation dy

dt

= r

�1� y

K

�y.

Solution:

(b) Remember that d

2y

dt

2 = d

dt

dy

dt

= d

dt

f(y) = f

0(y)dydt

= f

0(y)f(y). Note that

concavity can only change where d

2y

dt

2 = 0. However this would imply eitherf

0(y) = 0 or f(y) = 0. Since the solutions to f(y) = 0 are just our criticalpoints, we need only find where f 0(y) = 0. Our course we must first determinef

0(y). Applying the product rule, we get

f

0(y) =d

dy

f(y) =d

dy

✓ry ln(K/y)

◆

= r ln(K/y) + ry

✓y

K

· �K

y

2

◆= r ln(K/y)� r

1

Thus we need to solve the equation

0 = r ln(K/y)� r =) r = r ln(K/y) =) 1 = ln(K/y)

=) e =K

y

=) y =K

e

Take a value, say K

2e , in the interval 0 < y <

K

e

. Then

d

2y

dt

2

⇣K

2e

⌘= r

⇣K

2e

⌘ln

0

@ K⇣K

2e

⌘

1

A · r�ln

0

@ K⇣K

2e

⌘

1

A� 1�

= r

2⇣K

2e

⌘ln(2)

�ln(2)� 1

�> 0

Thus the graph of y versus t is concave up on 0 < y <

K

e

. Similar computa-tions show that the graph is concave down on K

e

< y < K.

(c) This amounts to showing that

ry ln(K/y) � ry

�1� y

K

�

which is equivalent to showing

ln(K/y) ��1� y

K

�

Here we use a trick which involves considering the di↵erence of the two func-tions

D(y) = ln(K/y)��1� y

K

�= ln(K/y) +

y

K

� 1

and taking the derivative getting

D

0(y) = �1

y

+1

K

Setting it equal to 0 we get

0 = �1

y

+1

K

=) y = K

Thus y = K is either a maximum or minimum for the di↵erence functionD(y) However, we note that D0(y) < 0 for all y < K. Thus we we see y = K

must be a minimum of the di↵erence equation. However,

D(K) = ln(K/K) +K

K

� 1 = 0 + 1� 1 = 0

Page 2

Thus we see

D(y) = ln(K/y) +y

K

� 1 � 0 for 0 < y K

.=) ln(K/y) � 1� y

K

=) ry ln(K/y) � ry

�1� y

K

�

4. (Problem 17a from B&D) Solve the Gompertz equation

dy

dt

= ry ln(K/y)

subject to the initial condition y(0) = y0.

Hint: You may want to let u = ln(y/K).

Solution:

This equation is separable so we write it as

1

y lnK/y

= rdt

Thus we need to integrate both sides. For the left side, we let u = lnK/y which,applying the chain rule, gives du = 1

K

Y

· �K

y

2 dy = � 1y

dy. Thus

Z1

y ln(K/y)dy =

Z �1

u

du = � ln |u| = � ln | ln(K/y)|

so ln | ln(K/y)| = �rt + C implying | ln(K/y)| = C1e�rt. Note that ln(K/y) only

switches from negative to positive at the equilibrium solutions K. Since solutionsdo not cross the equilibrium solutions, we may drop the absolute values, gettingln(K/y) = C1e

�rt. Furthermore, applying our initial conditions implies C1 =± ln(K/y0), with we choose the positive if y0 < k and negative if y0 > k. Assuming

y0 < K, this implies K

y

= e

ln(K/y0)e�rt

=⇣

K

y0

⌘e

�rt

meaning

y = K

⇣K

y0

⌘�e

�rt

Similarly, if y0 < k, we have

y = K

⇣K

y0

⌘tge

�rt

Page 3

5. Optional Hard Problem (Problems 18 from B&D) A pond forms as water collectsin a conical depression of radius a and depth h. Suppose that water flows in at aconstant rate k and is lost through evaporation at a rate proportional to the surfacearea.

(a) Show that the volume V (t) of water in the pond at time t satisfies the di↵erentialequation

dV

dt

= k � ↵⇡(3a

⇡h

)23V

23

where ↵ is the coe�cient of evaporation.

(b) Find the equilibrium depth of water in the pond. Is the equilibrium asymptoticallystable?

(c) Find a condition that must be satisfied if the pond is not to overflow.

Solution:

(a) First, let’s clarify notation. The radius a and height h are fixed constantsgiving the dimensions of the pond, and thus serving to provide us with afixed ratio between height and radius. We will denote the radius at any giventime as r and the height as g. We note that dV

dt

is just the rate water entersminus the rate it leaves. It is given to us that water enters at a rate k. Thuswe need to find the rate at which it evaporates. To do this, we need to findthe surface area. Remember that the volume of a cone is V (r) = 1

3⇡r2g and

the surface area of the top is just S(r) = ⇡r

2. Note that there is a linearrelationship between g and r. In particular, g = r · h

a

(consider the similartriangles). Thus the volume equation becomes

V =1

3⇡r

2 · r · ha

=⇡h

3ar

3

. If we now solve the volume equation for r, we get r =�3a⇡h

� 13V

13 . Plugging

this into the surface area equation gives

S = ⇡r

2 = ⇡

� 3a⇡h

� 23V

23

Thus multiply by our coe�cient of evaporation, we get that

dV

dt

= k � ↵⇡

� 3a⇡h

� 23V

23

Page 4

(b) Note that the equilibrium solution occurs when dy

dt

= 0 or equivalently

k = ↵⇡

� 3a⇡h

� 23V

23

. We now want to express V in terms of g. Using similar triangles again,we can find that r = ag

h

. Plugging this into the volume equation gives V =13⇡

a

2g

2

h

2 · g = 13⇡

a

2g

3

h

2 . We can now write the equation as

k = ↵⇡

� 3a⇡h

� 23

⇣13⇡

a

2g

3

h

2

⌘ 23= ↵⇡

� 3a⇡h

� 23

⇣13⇡

a

2

h

2

⌘ 23g

2

Solving for g we get

g =

k

↵⇡

�3a⇡h

� 23

⇣13⇡

a

2

h

2

⌘ 23

! 12

=

pk

↵⇡

�3a⇡h

� 13

⇣13⇡

a

2

h

2

⌘ 13

=

pk

p↵⇡

�a

3

h

3

� 13

=

rk

↵⇡

h

a

Thus we need only determine if this solution is asymptotically stable or un-stable. However, it is fairly clear that dV

dt

is negative if g is greater thanthe equilibrium solution and positive if it is smaller. Thus the solution isasymptotically stable.

(c) In order for the pond not to overflow, we need the g value of the equilibriumsolution to be less than h. Thus we require

rk

↵⇡

h

a

h =)r

k

↵⇡

a

=) k ↵⇡a

2

Page 5

6. Suppose we are given an autonomous di↵erential equation dy

dt

= f(y) with the followinggraph of f(y) vs y.

1.jpg

(a) Find all the critical points, draw the phase lines of the graph, and determinewhether each critical point cooresponds to a stable, unstable, or semistable equi-librium solution.

Solution:

40 Unstable

80 Stable

0 Stable

(b) Between each pair of critical points, estimate the y value at which solutions whoseinitial values are in that interval switch concavity.

Solution: Given that dy

dt

= f(y). we have d

2y

(dt)2 = d

dt

f(y) = f

0(y)dydt

=

f

0(y)f(y). Now we know that when f(y) = 0 we have an equilibrium solu-tion, so no solutions change concavity at these points. It follows that the onlypossible y values for which concavity is going to change are when f

0(y) = 0.From the graph we can see that this happens at approximately y = 17 andy = 63. Furthermore, we can see that these will, in fact, be inflection points.

Page 6

(c) Use parts (a) and (b) to sketch solutions to the di↵erential equation in the ty

plane given di↵erent initial conditions y(0) = y0. Include several solutions withinitial conditions y(0) = y0 between each pair or equilibrium. solutions.

Solution:

(d) Recall that a threshold for a population is value above which the population growsbut below which the population becomes extinct. Use part (c) to determine thethreshold and carrying capacity for the population described by this di↵erentialequation.

Solution: From the above plot we can see that the threshold is 40 and thecarrying capacity is 80.

(e) Based o↵ your investigations and the the discussion on page 86 and 87 of B&D,give a function for f(y).

Solution: Something like f(y) = �y( y

40 � 1)( y

80 � 1)

Page 7

Section 2.6

7. (Problems 1,5,6,8 from B&D) For the following problems, determine whether or notthe equations are exact. You do not need to solve the equations.

(a) (2x+ 3) + (2y � 2)y0 = 0

(b) dy

dx

= �ax+by

bx+cy

(c) dy

dx

= �ax�by

bx�cy

(d)�e

x sin(y) + 3y�dx�

�3x� e

x sin(y)�dy = 0

Solution: We have proven that showing an equation is exact is equivalent to show-ing that d

dy

M(x, y) = d

dx

N(x, y)

(a) d

dy

(2x+ 3) = 0 = d

dx

(2y � 2) so it is exact

(b) We must first put this in the standard form:

dy

dx

= �ax+ by

bx+ cy

=) (bx+cy) y0 = �(ax+by) =) (ax+by)+(bx+cy)y0 = 0

Then d

dy

(ax+ by) = b = d

dx

(bx+ cy) so it is exact.

(c) Putting this in the standard form gives

dy

dx

= �ax� by

bx� cy

=) (bx�cy)y0 = �(ax�by) =) (ax�by)+(bx�cy)y0 = 0

Then d

dy

(ax� by) = �b and d

dx

(bx� cy) = b so it is only exact if b = 0.

(d) d

dy

�e

x sin(y) + 3y�= e

x cos(y) + 3 6= 3 + e

x sin(y) = d

dx

�� 3x + e

x sin(y)�so

it is not exact.

Page 8

8. Show that the following equations are exact and use this to find an implicit solutionfor y.

(a) (2xy + 5y3) + (x2 + 15xy2 + 15)y0 = 0

(b) (� sin(x)y2 + 30xy2 + e

x) + (2y cos(x) + 30x2y)y0 = 0

Solution: We leave it to the reader to check that the equations are in fact exact.(To see the method for this, reference problem 1.)

(a) We want a function =RM(x, y)dx+h(y) =

R2xy+5y3dx+h(y). Integrat-

ing gives = x

2y + 5xy3 + h(y). However we also know that @

@y

= N(x, y)

which means x2+15xy2+h

0(y) = x

2+15xy2+15. Thus h0(y) = 15 meaningh(y) = 15y + C1. Putting this all together gives us

= x

2y + 5xy3 + 15y + C1 = C2

=) x

2y + 5xy3 + 15y = C3

(b) Let’s solve this one in a slightly di↵erent manner. This time we will integrateN(x, y) with respect to y. We let =

RN(x, y)dy + g(x) =

R2y cos(x) +

30x2y)dy + g(x) = y

2 cos(x) + 15x2y

2 + g(x). However, we want it to betrue that @

@x

= M(x, y) meaning � sin(x)y2 + 30xy2 + g

0(x) = � sin(x)y2 +30xy2 + e

x. It follows that g0(x) = e

x implying g(x) = e

x + C1. Thus

= y

2 cos(x) + 15x2y

2 + e

x + C1 = C2

y

2 cos(x) + 15x2y

2 + e

x = C3

9. Put the equationy

0 = e

2x + y � 1

into the standard form for exact equations and show that the equation is not exact.Determine which of the following integration factors make the equation exact. Youdon’t need to solve the equation.

(a) µ(t) = e

y

(b) µ(t) = e

�x

(c) µ(t) = y

Page 9

Solution: Subtracting everything over we get (1 � y � e

2x) + y

0 = 0 which isstandard form. It is easy to check that the equation is not exact. From here, it issimply a matter of multiplying through by each and checking to see which resultin an exact equation. We see that the correct answer is (b).

10. (Problems 25 and 27 from B&D) For each of the following, find an integrating factorand solve the given equation.

(a)�3x2

y + 2xy + y

3)dx+ (x2 + y

2)dy = 0

(b) dx+�x

y

� sin(y)�dy = 0

Solution: We have learned methods for attempting to find integrating factors

only in terms of x, namely µ

x

= e

RM

y

�N

x

N

dx, or only in terms of y, namely µ

y

=

e

RN

x

�M

y

M

dy. Since the numerator in the integral only changes by a negative sign,we are essential picking whether to divide by M or N . Thus it makes sense to firsttry the simpler one.

(a) Here we haveMy

= 3x2+2x+3y2 andN

x

= 2x. Then µ

x

= e

R 3x2+2x+3y2�2xx

2+y

2 dx

=e

R3dx = e

3x. Multiplying through we get

e

3x�3x2

y + 2xy + y

3)dx+ e

3x(x2 + y

2)dy = 0

Thus we want to set

=

Ze

3x(x2 + y

2)dy + g(x)

= e

3x(x2y +

1

3y

3) + g(x)

We then set @

@x

= M(x, y) implying

@

@x

= 3e3xx2y + e

3xy

3 + e

3x2xy + g

0(x) = e

3x�3x2

y + 2xy + y

3�+ g

0(x)

= e

3x�3x2

y + 2xy + y

3) = M(x, y)

Thus g0(x) = 0 meaning g(x) = C1. Thus

= e

3x(x2y +

1

3y

3) + C1 = C2

=) e

3x(x2y +

1

3y

3) = C3

Page 10

(b) We have M

y

= 0 and N

x

= 1y

. Then µ

y

= e

R 1y = e

ln |y| = ±y However, weonly need one integrating factor so we choose µ

y

= y. Then the equation

becomes y dx+⇣x� y sin(y)

⌘dy = 0. Thus we let

=

Zydx+ h(y) = yx+ h(y)

Then@

@y

= x+ h

0(y) = x� y sin(y) = N(x, y)

so h

0(y) = �y sin(y). Integration by parts gives

h(y) = y cos(y)�Z

cos(y)dy = y cos(y)� sin(y) + C1

=) = yx+ y cos(y)� sin(y) + C1 = C2

=) yx+ y cos(y)� sin(y) = C3

11. Consider the di↵erential equation

(3x2y + 10ex + 4y2) + (x3 + cos(y) + 8xy)y0 = 0

with initial condition y(0) = y0.

(a) Show that the equation is exact.

Solution: Noting that M(x, y) = 3x2y + 10ex + 4y2 and N(x, y) = x

3 +cos(y) + 8xy, we have that M

y

= N

x

= 3x2 + 8y.

(b) Solve the equation in the following ways:

i. Integrate M(x, y) with respect to x and find the function h(y) such that =

RM(x, y)dx+ h(y).

Solution: Letting

=

ZM(x, y) dx+ h(y) = x

3y + 10ex + 4y2x+ h(y)

and taking the partial of with respect to y we have

y

= x

3 + 8yx+ h

0(y).

Page 11

Setting

y

= N(x, y)

we can conclude that h0(y) = cos(y) and h(y) = sin(y)+ c. This leaves uswith a final solution of

x

3y + 10ex + 4y2x+ sin(y) = c.

ii. Integrate N(x, y) with respect to y and find the function g(x) such that =

RN(x, y)dy + g(x).

Solution: Now let

=

ZN(x, y) dy + g(x) = yx

3 + sin(y) + 4xy2 + g(x).

Then

x

= 3yx2 + 4y2 + g

0(x).

Setting

x

= M(x, y)

we can conclude that g0(x) = 10ex and g = 10ex + c. This leaves us withthe same final solution of

yx

3 + sin(y) + 4xy2 + 10ex = c.

12. Consider the di↵erential equation y dx+(2xy�e

�2y) dy = 0. Show that it is not exact,find an integrating factor, and use this to find an implicit solution to the equation.

Solution: Noting that M(x, y) = y and N(x, y) = 2xy � e

�2y, we have

M

y

= 1 6= 2y = N

x

.

It follows that the equation is not exact. Noticing that

N

x

�M

y

M

=2y � 1

y

we see that there is an integrating factor, µ, in terms of y alone. It follows that

µ = exp

✓Z2y � 1

y

dy

◆= e

2y�ln |y| =e

2y

y

.

Page 12

We dropped the absolute values here because (as you will see below) the integratingfactor still works without them. Multiplying through by µ, we have the exactequation

e

2y +

✓2xe2y � 1

y

◆y

0 = 0.

Using the ‘new’ M and N and letting

=

ZM(x, y) dx+ h(y) = xe

2y + h(y)

and setting

y

= 2xe2y + h

0(y) = N(x, y)

we can conclude that h

0(y) = � 1y

and h(y) = � ln |y| + c. This leaves us with afinal solution of

xe

2y � ln |y| = c

Section 2.7

13. (Problem 1 from B&D) Consider the initial value problem

y

0 = 3 + t� y y(0) = 1

(a) Find approximate values of the solution of the given initial value problem att = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h = 0.1.

(b) Repeat part (a) with h = 0.05. Compare the results with those found in (a).

(c) Find the solution y = �(t) of the given problem and evaluate �(t) at t = 0.1, 0.2,0.3, and 0.4. Compare these values with the results of (a) and (b).

Solution:

(a) We know the formula yn

= hf(tn�1, yn�1)+y

n�1 where here f(t, y) = 3+t�y.Thus y

n

= 0.3 + 0.1tn�1 + +0.9y

n�1. This allows us to build the followingtable inductively.

i t

i

y

i

0 0 11 0.1 1.22 0.2 1.393 0.3 1.5714 0.4 1.7439

Page 13

(b) We now use h = 0.05 and proceed similarly to get the table

i t

i

y

i

0 0 11 0.05 1.12 0.10 1.19753 0.15 1.29264 0.20 1.38555 0.25 1.47626 0.30 1.56497 0.35 1.65178 0.40 1.7366

While the solutions do not di↵er drastically, those produced using h = 0.05are all slightly less than those produced using h = 0.1.

(c) To solve the equation y

0 = 3 + t � y, we note that it is a first order linearequation. Putting it in the standard form, we have y

0 + y = t + 3 Then wehave µ(t) = e

Rdt = e

t so the equation becomes y0et+ye

t = (t+3)et. Howeverthe left side is just d

dt

ye

t. Therefore integrating both sides with respect to t

gives

ye

t =

Z(t+ 3)etdt

. Integration by parts givesZ

(t+ 3)etdt = (t+ 3)et �Z

e

t

dt = (t+ 2)et + C

Therefore we gety = t+ 2 + Ce

�t

Applying our initial condition that y(0) = 1, we see that C = �1. Thereforewe have our solution to the initial value problem

�(t) = t+ 2� e

�t

Making a final table we have

i t

i

�(ti

)0 0 11 0.05 0.0.9987292 0.10 0.9948293 0.15 0.9881664 0.20 0.9785975 0.25 0.9659756 0.30 0.9501417 0.35 0.9309328 0.40 0.908175

Page 14

Obviously, these results di↵er quite a bit from our approximations we gotusing h = 0.1 and h = 0.05. It appears we would need to use much smallervalues of h to approximate this curve well.

14. (Problem 12 from B&D) Consider the initial value problem

y

0 = y(3� ty) y(0) = 0.5

Use Euler’s method to find approximate values of the solution at t = 0.5, 1.0, 1.5, 2.0

(a) with h = 0.5

(b) with h = 0.25

Solution:

(a) Again, we make a table

i t

i

y

i

0 0 0.51 0.5 1.252 1 2.73443 1.5 3.09754 2.0 0.5478

(b) With h = .25 we have the following table

i t

i

y

i

0 0 0.51 0.25 0.87502 0.5 1.48343 0.75 2.32094 1 3.05165 1.25 3.01226 1.5 2.43597 1.75 2.03778 2 1.7494

Page 15

15. Optional Hard Problem Solve the di↵erential equation

�3xy + y

2�+�x

2 + xy

�y

0 = 0

using the integrating factor µ(x, y) = [xy(2x + y)]�1. Verify that the solution is thesame as that obtained using a di↵erent integrating factor in the lecture notes (example4 from B&D).

Solution:

(a) Multiplying by the integrating factor gives

3xy + y

2

xy(2x+ y)+

x

2 + xy

xy(2x+ y)y

0 = 0

We need to find

=

Z3xy + y

2

xy(2x+ y)dx+ h(y)

To do this we need to use partial fractions as follows:

3xy + y

2

xy(2x+ y)=

A

xy

+B

2x+ y

=2Ax+ Ay +Bxy

xy(2x+ y)

Thus we must have A = y which necessitates that B = 1 We now have

Z3xy + y

2

xy(2x+ y)dx =

Zy

xy

+1

2x+ y

dx =

Z1

x

+1

2x+ y

dx

so = ln |x|+ 1

2ln |2x+ y|+ h(y)

Then we consider

@

@y

=1

2· 1

2x+ y

+ h

0(y) =x

2 + xy

xy(2x+ y)

Thus

h

0(y) =x

2 + xy

xy(2x+ y)� 1

2· 1

2x+ y

=x

2 + xy

xy(2x+ y)�

12xy

xy(2x+ y)

=x

2 + 12xy

xy(2x+ y)=

x+ 12y

y(2x+ y)=

1

2y

Page 16

so h(y) = 12 ln |y|. Therefore = ln |x|+ 1

2 ln |2x+ y|+ 12 ln |y| = C. We can

simply this somewhat by multiply through by 2 then using the log propertiesto get

2 ln |x|+ ln |2x+ y|+ ln |y| = C1

=) ln |x2|+ ln |2x+ y|+ ln |y| = C1

=) ln |x2(2x+ y)y| = C1

The putting both sides as the exponent of e we get

|x2(2x+ y)y| = e

C1

=) 2x3y +

1

2x

2y

2 = C2

=) x

3y +

1

2x

2y

2 = C3

which was the solution we found using the other integrating factor.

(Note that we dropped the absolute values sense we previously had to decidea domain for x and y as the integrating factor had both in the denominator.Thus we can account for the sign which cannot change with the constant.)

16. (Problem 2 from B&D) Consider the initial value problem

y

0 = 2y � 1 y(0) = 1

(a) Find approximate values of the solution of the given initial value problem att = 0.1, 0.2, 0.3, and 0.4 using the Euler method with h = 0.1.

Solution: Let f(t, y) = 2y � 1 and note that according to Euler’s method,y

i+1 = y

i

+ hf(ti

, y

i

) = y

i

+ h(2yi

� 1) = (1 + 2h)yi

� h. Making a computerdo the computations, we have the following.i t

i

y

i

0 0 11 .1 1.12 .2 1.223 .3 1.3644 .4 1.5368

(b) Repeat part (a) with h = 0.05. Compare the results with those found in (a).

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Solution:i t

i

y

i

0 0 11 0.05 1.052 0.10 1.10503 0.15 1.1655004 0.20 1.232050005 0.25 1.3052550006 0.30 1.3857805007 0.35 1.4743585508 0.40 1.571794405

(c) Find the solution y = �(t) of the given problem and evaluate �(t) at t = 0.1, 0.2,0.3, and 0.4. Compare these values with the results of (a) and (b).

Solution: The di↵erential equation

y

0 = 2y � 1

is both separable and linear. We will solve it as a linear equation. Rearrangingyields

y

0 � 2y = �1.

Then the integrating factor µ = e

�2t and we have

µy =

Z�µ dt =

Z�e

�2tdt =

1

2e

�2t + c.

The general solution is then

y =1

2+ ce

2t.

Plugging the initial conditions in gives

1 = y(0) =1

2+ ce

2·0 =1

2+ c.

So c = 12 and the final solution is

y =1

2(1 + e

2t).

Plugging numbers in we havet y(t).1 1.110701379.2 1.245912349.3 1.411059400.4 1.612770464

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Now compare the results...

17. Consider the autonomous equation dy

dt

= �ry where r > 0. If we apply Euler’s methodwith step size h, we notice that the equation y

n

= h · f(tn�1, yn�1) + y

n�1 becomesy

n

= h · f(yn�1) + y

n�1. Suppose we are given the initial condition y(to

) = y0.

(a) Show that yi

= y

i�1(1� rh).

Solution: From Euler’s method we know that

y

i

= y

i�1 + hf(ti�1, yi�1).

where f(t, y) = �ry. Plugging this in we have

y

i

= y

i�1 + h(�ry

i�1) = y

i�1(1� hr).

(b) Use (a) to show y

n

= y0(1� rh)n.

Solution: Letting i = 1 in our equation from part (a) we have

y1 = y0(1� rh)

and the formula is true in the case that n = 1. Now if we assume that it istrue for n� 1 and show that this implies that the formula holds in the n case,we are done. Why is this? We would then have that the formula holds forn = 0, which implies that it holds for n = 1, which implies that it holds forn = 2, etc. . .So assume y

n�1 = y0(1� rh)n�1. From part (a) we have

y

n

= y

n�1(1� rh)

. Plugging in our above equation we have

y

n

= y0(1� rh)n�1(1� rh) = y0(1� rh)n.

(c) Optional Hard Problem We say a family of solution converges if as t ! 1,the solutions approach the same value regardless of initial condition.

i. For what values of h do the approximations given by Euler’s method converge?

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Solution: From our above solution, we have that the approximation con-verges if and only if

limn!1

y0(1� rh)n 6= ±1.

This happens if and only if |(1�rh)| < 1. Assuming h is positive we havethat this happens if and only if 0 < rh < 2 if and only if 0 < h <

2r

.

ii. Solve the di↵erential equation explicitly. You should note that all its solutionsconverge.

Solution: The equation y

0 = �ry is linear (and separable). So we have

y

0 + ry = 0.

Using the integrating factor µ = e

rt our solution is

y = ce

�rt

.

iii. Why is this not true for the approximations given by Euler’s method if h istoo large? (It may be helpful to consider the slope fields of the equation.)

Solution: If h is too large, successive iterations of Euler’s method pro-duce points that are farther and farther away from the equilibrium solu-tion (alternating positive and negative y values). For large values of y theslope of the tangent lines to the solution curves become larger and largerin magnitude and therefore successive iterations get ‘thrown’ increasingfarther. This is why the method fails.

18. See problem 7

19. See problem 8

20. See problem 9

21. See problem 10

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Section 3.1

22. Find the general solution to the di↵erential equation y

00 + y

0 � 30y = 0.

Solution: The characteristic equation for the above di↵erential equation is

r

2 + r � 30 = (r + 6)(r � 5).

This corresponds to the two solutions e

5t and e

�6t. Our general solution thenbecomes

c1e5t + c2e

�6t.

23. Find the solution to the following initial value problems:

(a) y

00 + 4y0 + 3y = 0; y(0) = 2, y

0(0) = 0

Solution: The characteristic equation is

r

2 + 4r + 3 = (r + 3)(r + 1)

and the general solution is

y = c1e�3t + c2e

�t

Plugging in the initial conditions we have

c1 + c2 = 2

�3c1 � c2 = 0.

Solving the system we have c1 = �1 and c2 = 3 and our final solution is

y = �e

�3t + 3e�t

(b) y

00 � 6y0 + 8y = 0; y(0) = 3, y

0(0) = �2

Solution: Here we have a characteristic equation on

r

2 � 6r + 8 = (r � 2)(r � 4).

Then the general solution is

y = c1e2t + c2e

4t

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and our initial conditions leave us with

c1 + c2 = 3

2c1 + 4c2 = �2.

Solving this gives that c1 = 7, c2 = �4 so our final solution is

y = 7e2t � 4e4t.

24. Find a second order linear equation with constant coe�cients whose general solutionis:

(a) C1e2t + C2e

�5t

Solution: This corresponds to the characteristic equation

(r � 2)(r + 5) = r

2 + 3r � 10,

which corresponds to the di↵erential equation

y

00 + 3y0 � 10y = 0.

(b) C1e3t + C2e

t

Solution: This corresponds to a characteristic equation

(r � 3)(r � 1) = r

2 � 4r + 3,

which corresponds to the di↵erential equation

y

00 � 4y0 + 3y = 0.

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