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MATH 275 ORDINARY DIFFERENTIAL EQUATIONS
SPRING 2011
Midterm Examination #1 key
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2
- 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
- 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
- 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
- If , then 0 and , , , 0,
so that 0
- If 0, then 0 and , , , 0,
so that 0
- 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
- 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
- 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
