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Midterm Exam 1 Solution Key for Differential Equations I - Spring 2011 |, Exams of Differential Equations

Material Type: Exam; Class: Differential Equations 1 - Introduction; Subject: Mathematics; University: Skyline College; Term: Forever 1989;

Typology: Exams

2010/2011

Uploaded on 08/22/2011

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MATH 275 ORDINARY DIFFERENTIAL EQUATIONS
SPRING 2011
Midterm Examination #1 key
Name ______________________________ Date ___________________
Show all work. No work, no credit!!
()
2
1) Find the general solution, , for 1 . If there are
any singular solutions, provide them as well.
dz
zt z
dt +=
()()
()()
2
22
2
Solution: Find all solutions for the differential equation 1
11 11
1
Integrating both sides ...
For the right hand side we have
11
and
dz z
dt
dz dz dz dz
z z dt dt
dt dt z z
z
dz dt dt t c
zz
+=
+= = −⇒ = =
+−
==
+−
∫∫
()()
+
()()
()()
()()
using partial fractions for the left side we have
1
1 1 1
11 11
or
1 0 and 1
11
so ,
22
and then
11 1
ln 1
1121212
ABAz Bz
zz zz
ABz BA AB BA
ABB A
d z dz dz z
zz z z
=+⇒++=
+− +
++=+= =
⇒= = =
=− + =− + +
+− +
∫∫
()
()
2
2222
1ln 1
2
So
11
ln 1 ln 1
22
Then
ln 1 ln 1 2 2
11
ln 2 where
11
Solving for :
1 1 1 1
or
1
tk
tttt
z
zztc
zztkkc
zz
tk Ce Ce
zz
z
z z Ce z zCe Ce z zCe Ce
z
−++−=+
−++−=+ =
−−
=+ = =
++
−= + −= + = +
()
()
2t
()
2
22 2
1
11
1 is a singular solution not included in the general solution.
t
tt t
Ce
Ce Ce z t Ce
zt
+
=+ =
≡−
1
pf3
pf4
pf5
pf8
pf9
pfa

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Download Midterm Exam 1 Solution Key for Differential Equations I - Spring 2011 | and more Exams Differential Equations in PDF only on Docsity!

MATH 275 ORDINARY DIFFERENTIAL EQUATIONS

SPRING 2011

Midterm Examination #1 key

Name ______________________________ Date ___________________

Show all work. No work, no credit!!

2

  1. Find the general solution, , for 1. If there are

any singular solutions, provide them as well.

d z z t z dt

2

2 2 2

Solution: Find all solutions for the differential equation 1

Integrating both sides ...

For the right hand side we have 1 1

and

d z z dt

d z d z d z d z z z dt dt dt dt (^) z z z

d z dt dt t c z z

− +^ −

using partial fractions for the left side we have

1 1 1 1 1 1 1 1

or

1 0 and 1

so , 2 2

and then

1 1 1 ln 1 1 1 2 1 2 1 2

A B

A z B z z z z z

A B z B A A B B A

A B B A

d z dz dz z z z z z

2

2 2 2 2

ln 1 2

So

1 1 ln 1 ln 1 2 2

Then

ln 1 ln 1 2 2

ln 2 where 1 1

Solving for :

or

t k

t t t t

z

z z t c

z z t k k c

z z t k Ce C e z z

z

z z Ce z zCe Ce z zCe Ce

z

2 t

2 2 2 2

1 is a singular solution not included in the general solution.

t t t t

Ce Ce Ce z t Ce

z t

2

2 2 2 2

2

2 2 2 2

2

Alternate solution: Find all solutions for the differential equation 1

Let sin cos and sin cos 1 cos sin 1

cos

(^1) sin

d z z dt

d z d z d z d z z z dt dt dt dt z z

z dz d

d z d

z

2 2

cos sec 1 cos cos

and

sec ln sec tan (Constant of integration added later)

For the right hand side we have

And for the left hand side, since

sin cos 1 cos

d d d

d

dt t c

z z

2 2

2

1 2

2 2

2

1 sec 1

and tan we have 1

ln sec tan ln ln 1 1 1

or

ln ln ln 1 1 1

ln and so ln ln 2 2 2 1 2 1 1

Exponentiating

1

1

k

z z

z

z

z z

z z z

z z (^) z z

z (^) z z

z z z t c t k k c z z z

z e e z

2

2 2 2

2 2 2 2 2 2

Solving for

1 is a singular solution not included in the general solution.

t t k

t t t

t t t t t t

Ce C e

z

z z Ce z Ce Cze

Ce z Cze Ce z Ce Ce z Ce

z t

Slope field A Slope field B

Slope field C Slope field D

Here are the equations again for reference:

( )

2 2 2

2 2

dy dy dy dy y t y y dt dt dt dt

dy dy dy dy

2 y 1 y 6. 4 y 7. 4 y 8. t y dt dt dt dt

2

  1. Consider the IVP = , 1 3, 1 1.5, 0.

a) Use Euler's method with the given step size to approximate the solution

over the time interval specified. Fill in the table below with

dx t x x t t dt

t

x t

2 1

the results

of your calculations. Three decimal places is sufficient.

b) Verify that the exact solution for the initial value problem is given by

t x t t t e

− = − − − +

k tk^ xk

0 1.00 3.

1 1.10 3.

(^2) 1.20 3.

(^3) 1.30 4.

(^4) 1.40 4.

(^5) 1.50 5.

2 2 1 0 0 0

2 2 2 1 1 1

2 2 3 2 2 2

2 2 4 3 3 3

2 2 (^5 4 4 )

Solution:

a)

x x t x

x x t x

x x t x

x x t x

x x t x

2 1 1

b) Verify that the exact solution for the initial value problem is given by

2 2 8 Then 2 2 8

Substituting into the original differential equation giv

t t x t t t e x t t e

− −

1 2 2 1

es

2 2 8 2 2 8 which is an equality, thus

verifying the exact solution.

t t t e t t t e

− − − − + = + − − − +

3

  1. Solution

a) Sketch the phase line for the differential equation , where and

are positive constants.

b) Identify and classify (sink, source, node) all equilibrium points for this

dx ax bx a b dt

3 2

differential

equation. Label your graph appropriately.

For the equilibria set 0 0 0,

We have four intervals to examine:

, , , 0 , 0, , and

dx a ax bx x a bx x x dt b

a a a

b b b

⎜ −∞ −^ ⎟ ⎜ − ⎟ ⎜ ⎟

2 2 2 2

2 2 2 2

2 2

  1. If , then 0 and , , , 0,

so that 0

  1. If 0, then 0 and , , , 0,

so that 0

  1. If 0 , then 0 and ,

a

b

a a x x x bx a bx a a bx b b

dx

dt

a a x x x bx a bx a a bx b b

dx

dt

a a x x x bx b b

2 2

2 2 2 2

so that 0

  1. If , then 0 and , , , 0,

so that 0

a bx a a bx

dx

dt

a a x x x bx a bx a a bx b b

dx

dt

  1. Use the method of undetermined coefficients (guessing) to find the general

solution for the equation 5 sin 3.

dy y t dt

( )

5

Solution: This equation is linear and nonhomogeneous. To find the general

solution, note that is the general solution of the

associated homogeneous equation

t yh t ke

( )

To get a particular solution of the nonhomogeneous equation, we

guess

cos 3 sin 3.

Substituting this guess into the n

y p t = α t +β t

( ) ( )

( )

onhomogeneous equation gives

5 3 sin 3 3 cos 3 5 cos 3 5 sin 3

5 3 cos 3 5 3 sin 3

In order for to be a solution, we

p p

p

dy y t t t t dt

t t

y t

must solve the simultaneous

equations

5 3 0

5 3 1

From these equations we get 3 and^5 34 34

So the gener

( ) ( ) ( )

5

al solution is

cos 3 sin 3 34 34

t y t yh t y (^) p t ke t t

− = + = − +

7) Given the following predator-prey model of two species

of microorganisms (where , and are positive parameters).

a) Which variable, or

x

x

a b c

dx

ax by

dt

dy

cy

dt

x

, represents the predator population?

Which variable represents the prey population?

b) What happens to the predator population if the prey is

extinct?

y

Solution:

a) The effect of on is through the term. Because this term

is negative ( and are positive), a larger population decreases ,

so is the prey. Similarly

dx

y by x

dt

dx

x y y

dt

x

, the term gives the effect of on

and a larger gives a larger. So is the predator.

b) If 0, 0, so the predator population is constant. The predator

must ha

dy

cy x x

dt

dy

x y

dt

dy

x

dt

ve alternate sources of food or be able to hibernate without

need for prey.

( )

8) a. Use an integrating factor to find the general solution for the differential equation

b. Find the particular solution using the initial condition , where

dy x

x y e

dx

y a b

a and b are constants and a ≠0.

( ) ( )

Solution: a. Use an integrating factor to find the general solution for the differential equation

So the integrating factor is exp exp ln

Multi

x

dy x dy x dy e

x y e x y e y

dx dx dx x x

dx

x x x

x

β

( )

[ ] (^ )

( )

plying both sides of the differential equation by , we have

b. Find the particular solution using the initial condition , where

and are constan

x x x x

x

d e C

xy e xy e dx e C y x

dx x

y a b

a b

β

( )

( )

ts and 0.

The condition implies that

and we have the particular solution

a a

x a

a

e C

y a b b C ab e

a

e ab e

y x

x