You may answer the questions in the space provided or on your
own space. If you choose to use your own space, please provide a
CLEAR diagram with axes for problem 1b.
Answer ALL questions. SHOW your work if you wish to receive
credit.
Name:
Date:
Math 234 Test 1
Answer ALL questions. SHOW your work if you wish to receive credit.
dy
2 x y .
dx
(a) Calculate the slopes at each of the nine points shown on the x-y plane below.
Show the values of the slopes. (1 point)
1. Consider the differential equation
(t , 1)
(0, 1)
-
-
(1 : 1) e
( 1 0)
-
,
,
-
-
-
-
=
2 1
-
=
10 , 0) - 0
1
( 1 1)e
2(0) + ( 1) =
2(1) + ( 1) = 2 1
=
-
3
10 , 1)
-
=
2) 1) + 0 2
-
2) 1) + 1 = 3
-
-
,
-
-
-
-
e
(1 0) e
2( 1) + ( 1)
e
-
210)
1 1
+
=
(1 1) 2(1) + 1
-
-
=
,
-
2(1) + 0
=
1
2
(b) Create (draw) a slope field for the differential equation at each of the 9 points.
(3 points)
y
x
(c) Comment on the concavity of y = f(x) at the point (-1, 1). Show the work that
leads to your answer. (2 points)
d = 2( 1) + 1
=
hey
( 2x y)
+
:
of
2
,
=
3
x
+
(1 1)
-
=
-
2+ 3 =
up
v
,
concave
2. Find the value of k if the equation, y = xsin(x) + xcos(x), is a solution of the
d2y
differential equation
y 2cos(x) ksin(x) . (4 points)
dx 2
dy
dx
Sinx + Xrosx + CosX- Xsinx
=
:
d
d
Binx + cosx) +
(x(cosx-sinx)
&(xco(x) - (xsix)
=
cosx-sinx +
=
cosx-Sinx + cosX- XSinx-Sinx-XcosX
dX
=
(cosX-Isinx- XSinx-XrosX
20sX- Isinx + xpir
-
(x Posi
+
/xSi * + ((93X
=
+ KSinx
IcosX- Isinx = [losx
-
IsinX = KsinX
Xk
=
-
2
1cosX- Isux
dx
cos y 2 x
dy
2 y sin x
a) Show that this differential equation is exact. (2 points)
3. Consider the differential equation
= ,
2
b) Solve the differential equation. (4 points)
M
=
N
F(x,
2y SinX
-
(os(y) + 2x
=
y) fly-Sinx
=
dx---20((x) + Zxy
=
h
2x +
g'(y)
2x
cos(y)
g'(y)
gi(y) cosly)
2x +
+
=
=
g(y) Sin(y)
=
-
(os((x) +
+
g(y)
2xy Sin(y) =
+
4. Given that 𝑥
= 𝑥 𝑦 + 𝑦 , solve for y if y(1) = ½. (5 points)
a
d
ip) =
*
x de
d
=
+
=
=
u+
Ex
@y()
x(n + x ) x(ux) (ux]
Xu +
=
x+
2
1
mix
·
En a
12
:
In()-
-
:
-
=
:
(
5
2 =
-
C
u
C=
du
Side Side
-
=
1
In x
=
2(2)2
2
=
y2 :2)
-
=
-
-
(n (x) + 2
=
In (x) +
(x1 +
7
(x) 2)
:
+
(n (x) +
:
-In
dy y 1
for y = f(x). You may assume that the
dx x 1
solution function passes through the point (2, 5). (3 points)
5. Solve the differential equation
(d S
:
Inly-11
:
(n(x 1) +
-
(y 1) A(X 1
-
-
y -1
=
=A
+
5 = A(2 -1) + 1
A 4
=
y(x)
=
=
4(x 1) + 1
-
4x- 3
6. Consider the differential equation y1 y sin x dx 2 x 3 y cos x dy 0 .
a) Find an appropriate integrating factor for the differential equation.
(4 points)
MyA
m
=
(p (yysinx))
t
=
yu(2x 3ycosx)
N=
=
+
-
sinx
Ex +
Sinxdx-cy
My
M
-
=
e
cos(x)(y yz)mx)
N = e cos(X)(2x
=
My
=
e
c
-
+
3y(sx)
=
2
=
e
(1 2ySinx)
-
b) Solve the differential equation y1 y sin x dx 2 x 3 y cos x dy 0 .
(4 points)
-
e
Xy(1 ySm(x))dx
-
+ e
(2x + 3yasx)
dy 0
=
= M =y(lys
GF = N = e 2)(2x +
-
Ey
F-Se-y(1-ysinx)
U=
-
F =
y+
Ex =
3ycosx)
sinx
y Sedx-y29 e
de
cos(X) du= Sinx dx
ye
CSX
+
y2
-
cos(x)
+
(y
=
e
yz((s(x) gy2 y GX
=
+
-
Cs"(y yz) Cy
+
+
7. Solve the differential equation t
dQ
2Q sin t subject to the initial condition
dt
given as Q() = 0. (7 points)
+ :
Q
do P(t)@ get
+
:
M(t) eSPLEIdt In st
:
t 2 tQ EsirCt)
=
+
Se(ta) At
tQ
=
since) de
StSinCHbE
u= t
du = dt
tiQ
:
-
Ecost) +
t2Q
=
dv = sin(t) de
v=
frost de
:
-
-
cost
Ecost +Sint
-t cost + Sint
cost +
Q(t) :
+
Pa
Sit
8. For the differential equation
dy
y e x y 2 , solve for y = f(x) given the initial
dx
condition y(0) = 1. (7 points)
E P(x)y Q(x)y
=
-
V=
y
!"
-y t
=
x2d y
y
-
e
=
xy2
-Lie
-
du
-
V = e
.
Ex
+
X
V=
M(x) ex
=
eSld
=
etxx ev =
+
↑
(e)
=
+
1
eXV = x+ C
-
C x + c)
-
v= e
y
-
=
e
( x
-
y:
x
I
2= 1
y(x) =
dy x y 2
if the solution curve passes
dx x y 3
through (2, 0). Do not attempt to solve for y. Leave your answer in the form
given by f(x, y) = 0. (6 points)
9. Solve the differential equation
Y
= -
dat
=
1
X+
-
1 +
m
d
T
da-1
d su 3) du f
=
+
=
3
u
+ 3u =
)
(x
+
+
(20
-
d
x
+
3) x +y) + X =
-
+
@y(2)
-2) +=
C= 2
-
(
y(
2
x
+ 3)
-
y)
x+
+ x+
2 = 0
=
0
dy
2 x y with initial condition y(0) = 4.
dx
Use two equal steps of the Euler method to approximate y(1). (4 points)
10. Consider the differential equation
h
1 =.
=
Xo = 0 yo
f(x0 yo)
:
Y
=
4
,
y Yo
X, = 5
y
f(x,, )
y
yz y
=
hE(X0 ,yo)
+
=
=
2( 5) + 6 5
.
,
y (1)
.
.
6
y)
+ h.f(x
=
4 + ( 5 4) : 6
=
:
,
,
:
8 5
.
=
6 + ( 5 5)
.
.
=
85
.
11. Given the differential equation 𝑥
+ 4 = 0 and one solution y1 = A
(constant), use reduction of order to find a second solution, y2 . If 𝑦(1) =
3 𝑎𝑛𝑑 𝑦 (1) = 6, find the solution, y. (8 points)
yz v (x)y
=
d
=
AV(x) = y
=
,
Avl
=
Av(x)y(x) A) - (2) 4 .+x 4
=
d Ar"
+
=
y(x) 2 C
=
:
,
+
+
y'(x) =
XAr" + YAr" =
A(xv"
4r1)
+
xV" + 4)
=
-
=
y(1) 3 G Cy
0
=
(3 + (y 3
0
=
y'(l)
v
u= V , u = =
XU'
=
e
,
-
An
M(x)
=6
323
b =S
=
= 5
Cy = 3 C 2)
+ Yu 0
=
+
+
,
-
+5
y(x) =
(nx x4
x" d 4x3 0
=
+
(xn)
=
x n=
v=
0
= n
=
v
=
S Endx G(xdx G( E ) + = +
:
=
2
(3
=
=
2
r2
12. Solve the following differential equation by the method of undetermined
coefficients. (10 points)
8r
𝑑 𝑦
𝑑𝑦
+ 16 = 0
+
+8
+ 16𝑦 = 32𝑥 + 289sin (𝑥)
𝑑𝑥
𝑑𝑥
=0
+
( y)
r=
-
Yc(x)
4
:
C , e"*
32x yp (4)
,
+
(2xe-4
AxtB
=
,
289six , Yp2(x)
yp(x)
-
(sinx-pcsX
Yp'
=
Yp"
=
+ 8A +
=
:
Ax + B + (sinx
A + cosX-
(sinxe Dcos
(Snx
-
+ DX
DSimx
Drs(x)
+
16DcosX
Scrosx--8Dsinx + 16Ax+ 16B (6(sinx+
16Ax + Sa + 16B + (15( 84)(sinx) + (8c + 15D)(co(x)
32x + 289sinx
=
-
16H= 32
A=2
SA + 16 B =
B
8 C + 15D = 0
C=
-
15
J
=
-
D
15C-SD :289
1
0
Sl + 15D =
c = 15
0
yp(x)
y(x)
8
=
2x 1 + 155mx 82sX
-
=
Ge
=
**
+
-
(2x x + 2x 1 + 155mx 82sX
-
-
13. Given that at t = 0, y = 12 and = −7, solve the differential equation using the
method of undetermined coefficients: (10 points)
𝑑 𝑦
𝑑𝑦
+6
+ 25𝑦 = 195𝑐𝑜𝑠(2𝑡)
𝑑𝑡
𝑑𝑡
r2 + Gr + 25
-
0
y(0)
600
:
YcCH-e-it
=
12 ;
+ ( 0)
e (( -)
.
°
y
(c, cost + C2Si4t)
=
+7
.
1 +y
+ 7
12 C
=
yp(t) ABit + BSizt
=
4=5
2ASmit + IBcoSIE
Yp
YP" - 4AcosIt-YBSmIt
:
-
y(t) C cost (2(n4)t
+
,
:
7
up"+ byp' + 25yp 195c0s(2t)
=
·
*
+
cost sin
LACOSC-YBSinIt-17Asinit fliBcoSIE + 2 JACOS2t + 2JBSinIt
CosIt
=
y'(t)
21 A + /B
En
=
-+
e
-
) 34254t +(Sin Yt) (2 4()(4t) (i 4 snut)
14smit + Scot
Sin It : -12A + 21B
↳
B=
I
A=
,
y'(0)
-
2(A +
12 (EA) 195
:
.
@t --
7 + 4(1 =7
y(t)
7
B : (E)(7) : 4
20547 (2Sint) 7cos2 YsinIE
35
((
y(t)
=
,
-
15 + 4(z +8 = 7 + 42
+
=
= e
=
-
(2: 0
ITA 19
A=
+
-
+
+
+
=
e (JcsYt)
+
7 c1(27) + 4sin(e)
14. Solve the following Cauchy-Euler differential equation: 𝑥
(6 points)
y
xr
=
= ryrdx
↓
=
-Dy
X(r(r 1(x 2)
-
+
x
-
5x(rx
Y (r(r 1) 5r + 8)
-
r(r 1)xV 5rx + 8x
-
=
-
=
+
-
1) + 8x
0
0
-
d)
x (rEr-1) 5r +
=
0
-
+8
r(r 1) 5r
=
0
-
-
r2 r 55 + 8
=
t
-
(r 4)(r 2)
=
0
-
-
r = 2, 4
y(x)
=
4x + (2xY
=
0
− 5𝑥
+ 8𝑦 = 0.
15. Using the method of variation of parameters, solve the differential equation:
. (10 points)
𝑦" − 2𝑦′ + 𝑦 =
r2 2r + 1
-
=
un :
0
(r-12 0
=
u:
r=
Ye(X) (
=
yp
= n,
,
2
"
+ (exe
9
) - dxhi,
*
- * de -
xy1 + UzXY
:
=
U : y , + 4 yz
u,
d = +an(x)
=
'y; + vii
-
=
X
u2Xe
+
"
=
*
, e" + u2xe
u Xue
0
-
(-Xui)e* +
-
: Y(x)
=
=
ui(e+ xe*) =
uixe + uie * + hixe *
ni =
1 +X
)
+
-
+
+
0
niex +uile + XeY
u
+x
(fun (x))xeY
yp (-(n(1 x 2))e*
==
1
((n(1
=
=
Ge + Exe
+
-
an(x)
Le (1 +x2) + xe +
0
