AP Calculus BC Practice Exam 1 399
- Given the differential equation
dy
dx
=
2 xy
3
:
(A) Calculate slopes.
y=− 2 y=− 1 y= 0 y= 1 y= 2
x=− 3 4 2 0 − 2 − 4
x=− 2
8
3
4
3
0 −
4
3
−
8
3
x=− 1
4
3
2
3
0 −
2
3
−
4
3
x= 0 0 0 0 0 0
x= 1 −
4
3
−
2
3
0
2
3
4
3
x= 2 −
8
3
−
4
3
0
4
3
8
3
x= 3 − 4 − 2 0 2 4
Sketch the slope field.
(B)f( 0. 1 )=f( 0 )+ 0. 1
2 xy
3
∣∣
∣
∣x=0,y= 2
= 2 + 0. 1 ( 0 )= 2
f( 0. 2 )=f( 0. 1 )+ 0. 1
dy
dx
∣∣
∣∣
x= 0 .1,y= 2
= 2 + 0. 1
(
0. 4
3
)
= 2 +
0. 04
3
= 2. 013
f( 0. 3 )= f( 0. 2 )+ 0. 1
dy
dx
∣∣
∣∣
x= 0 .2,y= 2. 013
= 2. 013 + 0. 02684 = 2. 04017
≈ 2. 040
(C)
dy
dx
=
2 xy
3
1
y
dy=
2
3
xdx
ln|y|=
1
3
x^2 +c 1
y=c 2 ex^2 /^3
According to the initial condition,
2 =c 2 e^0 /^3 ⇒c 2 =2, so the particular
solution isy= 2 ex^2 /^3. Evaluate atx= 0. 3
andy( 0. 3 )= 2 e^0.^09 /^3 = 2 e^0.^03 ≈ 2 .061.
