MA 3972-MA-Book April 11, 2018 16:5
More Applications of Definite Integrals 333
(b) Sketch a possible graph for the particular solutiony=f(x) to the differential equation
with the initial conditionf(0)=3.
(c) Find, algebraically, the particular solutiony= f(x) to the differential equation with
the initial conditionf(0)=3.
Solution:
(a) Set up a table of values for
dy
dx
at the 15 given points.
x=− 2 x=− 1 x= 0 x= 1 x= 2
y= 12 1 0− 1 − 2
y= 24 2 0− 2 − 4
y= 36 3 0− 3 − 6
Then sketch the tangents at the various points as shown in Figure 14.5-6.
3
2
1
- 21 – 1 0 2
y
x
Figure 14.5-6
(b) Locate the point (0, 3) as indicated in the initial condition. Follow the flow of the field
and sketch the curve as shown Figure 14.5-7.
(c) Step 1: Rewrite
dy
dx
=−xyas
dy
y
=−xdx.
Step 2: Integrate both sides
∫
dy
y
=
∫
−xdxand obtain ln|y|=−
x^2
2
+C.
Step 3: Apply the exponential function to both sides and obtaineln|y|=e−
x 22 +C
.
Step 4: Simplify the equation and gety=
(
e
−x 22 )
(eC)=
eC
e
x 22.
Letk=eCand you havey=
k
e
x 22.