QUIZ 14 - SOLUTIONS
1. Solve
x
Answer:
dy
+ 2y = 3x3
dx
y(1) = 2
dy
+ 2y = 3x3
dx
dy 2
+ y = 3x2
dx x
R 2
Then the integrating factor is I(x) = e x dx = e2 ln(x) = x2 .
dy 2
+ y = 3x2
dx x
dy
+ 2xy = 3x4
x2
dx
d ! 2 "
x · y = 3x4
dx
#
#
d ! 2 "
x · y dx = 3x4 dx
dx
3
x2 · y = x5 + C
5
3
y = x3 + Cx−2
5
x
Then we solve for C.
3
y = x3 + Cx−2
5
3
2= +C
5
7
C=
5
Therefore
3
7
y = x3 + x−2
5
5
2
QUIZ 14 - SOLUTIONS
2. Solve
x
Answer:
$
dy
+y
dx
%
= xe−x − y
y(2) = 0
$
%
dy
x
+ y = xe−x − y
dx
dy
x
+ xy = xe−x − y
dx
dy
x
+ xy + y = xe−x
dx
dy
x
+ (x + 1)y = xe−x
dx $
%
dy
1
+ 1+
y = e−x
dx
x
R
1
Then the integrating factor is I(x) = e 1+ x dx = ex+ln(x) = ex eln(x) = xex .
$
%
1
x dy
x
xe
+ xe 1 +
y = xex e−x
dx
x
d
[xex · y] = x
dx
#
#
d
x
[xe · y] dx = x dx
dx
x2
xex · y =
+C
2
x
C
y= x+ x
2e
xe
Then we solve for C.
2
C
+
2e2 2e2
C = −2
0=
Therefore
y=
x
−2
+ x
x
2e
xe
Name:
QUIZ 15
1. Explain what section 11.3 is about.
2. Draw a vector field (it is sufficient to use 9 points) for the differential equation
dy
= x − y.
dx
Use the Euler approximation with h = ∆x = 1 to draw a solution curve with the
initial value y(−1) = 0.