Math 204
Homework 2.1
1) Which of the following DE has the direction field shown in the figure?
a)
b)
c)
d)
ππ¦
ππ₯
ππ¦
ππ₯
ππ¦
ππ₯
ππ¦
ππ₯
= π₯2 β π¦2
=π₯
= β2π¦
=
π₯
π¦
2) Which of the following DE has the direction field shown in the figure?
a)
b)
c)
d)
ππ¦
ππ₯
ππ¦
ππ₯
ππ¦
ππ₯
ππ¦
ππ₯
= π₯2 β π¦2
=π₯
= β2π¦
=
π₯
π¦
3) Use the following direction field to for the differential equation
ππ¦
ππ₯
= π(π₯, π¦) to
identify where the function is positive, negative or zero. Is f (x,y) a function of x
alone, y alone, or a function of both variables together? Find a function f (x,y)
whose vector field looks like this.
4) Draw the vector field for
ππ¦
ππ₯
= π¦ β 1 and sketch an appropriate solution curves
passing through the points
a) (0,0)
b) (1,2).
5) Find the critical points and draw the phase portrait of the given autonomous
differential equations. Classify each critical point as asymptotically stable,
unstable, or semi-stable. Sketch the equilibrium solutions and typical solution
curves in the different regions in the x-y plane.
a)
b)
c)
ππ¦
ππ₯
ππ¦
ππ₯
ππ¦
ππ₯
= π¦3 β π¦2
= π¦ 3 β 4π¦
= π¦ 2 β 3π¦ β 10
Homework 2.2
1) In problems 1-5 determine whether the given differential equation is separable.
1)
2)
ππ¦
ππ₯
ππ¦
ππ₯
3) π₯
4)
=
π¦ 2 +π¦
π₯ 2 +π₯
1
=
ππ¦
ππ₯
π₯(π₯βπ¦)
= π¦π π₯/π¦ β π₯
π¦ ππ¦
π₯ 2 ππ₯
+ cos(π₯ + π¦) = 0
5) (π₯ + 4)ππ¦ = (π₯ 2 π¦ β 8 + 4π₯ 2 β 2π¦)ππ₯
2) In problems 6-10 solve the given differential equation by separation of variables.
6)
ππ§
ππ€
= π€π 3π€+2π§
7) π₯ sin2 π¦
ππ¦
ππ₯
= (π₯ + 1)2
2
8) (π₯ + π₯π¦ 2 )ππ₯ + π π₯ π¦ππ¦ = 0
9) (π₯ + 1)2 ln π¦
10) π₯
ππ¦
ππ₯
=
ππ¦
ππ₯
=
π₯
π¦2
π₯ 2 βπ₯β2
π₯π¦+π₯+π¦+1
3) In problems 11-13solve the given initial value problem.
11)
12)
ππ¦
ππ₯
ππ¦
ππ₯
= 3π₯ 2 π βπ¦ ,
=
π¦
π₯(π₯+1)
π¦(0) = 1
,
13) π¦ β² = 2π₯ cos2 π¦,
π¦(1) = 3
π¦(0) = π/4
4) Proceed as in example 5 to find an explicit solution of the given initial value problem.
ππ¦
ππ₯
= π¦ 2 sin π₯ 2 ,
1
π¦(β2) = 3
Homework 2.3
I. In problems 1-6 determine whether the given equation is separable, linear, neither, or
both.
ππ¦
1) π₯ 2 ππ₯ + cos π₯ = y
2)
ππ₯
ππ‘
+ π₯π‘ = π π₯
ππ₯
3) π₯ ππ‘ + π‘ 2 π₯ = sin π‘
4) 3π‘ = π π‘
ππ¦
ππ‘
+ π¦ ln π‘
ππ¦
5) (π‘ 2 + 1) ππ‘ = π¦π‘ β π‘
ππ
6) 3π = ππ β π 3
II. In problems 7-17 find the general solution of the given differential equation. Give the
largest interval I over which the solution is defined. Determine whether there are any
transient terms in the general solution.
1 ππ¦
2π¦
7) π₯ ππ₯ β π₯ 2 = π₯ cos π₯
8) π¦ β² + π¦ = β1 + cos 2π₯
9)
10)
ππ
ππ
ππ¦
ππ₯
+ π tan π = sec π
π¦
= π₯ + 2π₯ + 1
11) (π‘ + π¦ + 1)ππ‘ β ππ¦ = 0
12)
ππ¦
ππ₯
= π₯ 2 π β3π₯ β 4π¦
13) 4π₯ 3 π¦ + π₯ 4 π¦ β² = sin3 π₯
14) π¦ππ₯ β 2(π₯ + π¦ 4 )ππ¦ = 0
ππ¦
15) (π₯ 2 + 1) ππ₯ = π₯ 2 + 2π₯ β 1 β 4π₯π¦
16) cos 2 π₯ sin π₯
ππ¦
ππ₯
+ (cos3 π₯)π¦ = 1
III. In problems 17-19 solve the given initial value problem.
17) π₯π¦ β² + π¦ = π π₯ ,
18) sin π₯
ππ¦
ππ₯
+ π¦ cos π₯ = π₯ sin π₯,
ππ¦
19) (π₯ + 1) ππ₯ + π¦ = ln π₯,
π¦(1) = 2
π
π¦ (2 ) = 2
π¦(1) = 10
5) In problem 20 proceed as in example 6 to solve the initial value problem
ππ¦
20) ππ₯ + 2π₯π¦ = π(π₯), π¦(0) = 2,
π₯ 0β€π₯<1
where π(π₯) = {
0
π₯β₯1
Homework 2.4
I.
In problems 1-7 classify the equation as separable, linear, exact, or none of these.
Some equations may have more than one classification.
1) (π₯ 2 π¦ + π₯ 4 cos π₯)ππ₯ β π₯ 3 ππ¦ = 0
2) π₯π¦ππ₯ + ππ¦ = 0
3)
ππ¦
ππ₯
=
π₯βπ¦
π₯
4) [2π₯ + cos(π₯π¦)]ππ₯ + [sin(π₯π¦) + 2π¦]ππ¦ = 0
5) π¦ 2 ππ₯ + (2π₯π¦ + cos π¦)ππ¦ = 0
6) π₯π¦π¦ β² = 2π₯π π¦
7) (π¦π π₯π¦ + 2π₯)ππ₯ + (π₯π π₯π¦ β 2π¦)ππ¦ = 0
II. In problems 8-14 determine whether the differential equation is exact. If it is exact
solve it.
8) (2π₯π¦ β sec 2 π₯)ππ₯ + (π₯ 2 + 2π¦)ππ¦ = 0
9) (π₯ 2 β π¦ 2 )ππ₯ + (π₯ 2 β 2π₯π¦)ππ¦ = 0
10) (1 + π π₯ π¦ + π₯π π₯ π¦)ππ₯ + (π₯π π₯ + 2)ππ¦ = 0
π‘
11) (π¦) ππ¦ + (1 + ln π¦)ππ‘ = 0
12) (tan π₯ β sin π₯ sin π¦)ππ₯ + cos π₯ cos π¦ ππ¦ = 0
1
π₯
13) (π¦) ππ₯ β (3π¦ β π¦ 2 ) ππ¦ = 0
1
2
14) [ β1βπ₯ 2 + π¦ cos(π₯π¦)]ππ₯ + [π₯ cos(π₯π¦) β π¦ β3 ]ππ¦ = 0
III. In problem 15 and 16 find the value of k so that the equation is exact
15) (ππ₯ 2 π¦ + π π¦ )ππ₯ + (π₯ 3 + π₯π π¦ β 2π¦)ππ¦ = 0
16) (6π₯π¦ 3 + cos π¦)ππ₯ + (2ππ₯ 2 π¦ 2 β π₯ sin π¦)ππ¦ = 0
IV. In problems 17-19 solve the given differential equation by finding an appropriate
integrating factor as in example 4.
17) (2π₯ 2 + π¦)ππ₯ + (π₯ 2 π¦ β π₯)ππ¦ = 0
18) (π¦ 2 + 2π₯π¦)ππ₯ β π₯ 2 ππ¦ = 0
2
19) cos π₯ ππ₯ + (1 + π¦ ) sin π₯ ππ¦ = 0
Homework 2.5
I. In problems 1-9 solve the following differential equations using an appropriate
substitution.
1)
ππ¦
ππ₯
= π¦ β π₯ β 1 + (π₯ β π¦ + 2)β1
2) (π₯π¦ + π¦ 2 + π₯ 2 )ππ₯ β π₯ 2 ππ¦ = 0
3)
ππ¦
5
β 5π¦ = β 2 π₯π¦ 3
ππ₯
4) βπ¦ππ₯ + (π₯ + βπ₯π¦)ππ¦ = 0
5)
6)
ππ¦
ππ₯
ππ
ππ
=
=
1 βπ₯βπ¦
π₯+π¦
π 2 +2ππ
π2
7) cos(π₯ + π¦)ππ¦ = sin(π₯ + π¦)ππ₯
8) 3(1 + π‘ 2 )ππ¦ = 2π‘π¦(π¦ 3 β 1)ππ‘
9) π₯ππ¦ β π¦(ln π¦ β ln π₯ + 1)ππ₯ = 0
II. In problems 10-12solve the following initial value problem.
10) (π₯ + π¦π π¦/π₯ )ππ₯ β π₯π π¦/π₯ ππ¦ = 0,
ππ¦
11) ππ₯ = cos(π₯ + π¦),
ππ¦
3π₯+2π¦
12) ππ₯ = 3π₯+2π¦+2 ,
π¦(1) = 0
π¦(0) = π/4
π¦(β1) = β1
III. Find a one-parameter family of solutions for the differential equation
ππ¦
π¦
= π₯ 3 (π¦ β π₯)2 +
ππ₯
π₯
Where π¦1 = π₯ is a known solution of the equation.
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