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EXAMPLE: PROBLEM 2.1.28 OF THE TEXT
110.302 DIFFERENTIAL EQUATIONS
PROFESSORRICHARDBROWN
Problem. Consider the Initial Value Problem
y′ + 2y = 1− 1t, y(0) = y .
3 2 0
Find the value of y0 for which the solution touches, but does not cross, the
t-axis.
Strategy. This ODE is linear, so the general solution can be found by using
an integrating factor. We solve for the general solution and then write the
general solution in terms of the initial value y . Then we use the properties
0
of the particular solution we are looking for the locate the proper value of y .
0
Solution. This ODE is linear and already in the standard form to read off
the coefficient function of y in the equation. We get p(t) = 2. Thus the
3
integration factor is R R
p(t) dt 2 dt 2t
e =e 3 =e3 .
Multiply the ODE through by this integrating factor and
2t ′ 2 1
e3 y + y = 1− t
3 2
2t ′ 2 2t 2t 1 2t
e3 y + e3 y = e3 − te3
3 2
d h 2t i 2t 1 2t
e3 y = e3 − te3 .
dt 2
Now integrate with respect to t to get
Z d h 2t i Z 2t 1 2t
e3 y dt = e3 − te3 dt
dt 2
2t 3 2t 1 3 2t 9 2t
e3 y = e3 − te3 − e3 +C.
2 2 2 4
Solving for y(t), we get
3 1 3 9 −2t 21 3 −2t
y(t) = − t − +Ce 3 = − t+Ce 3 .
2 2 2 4 8 4
1
2 110.302 DIFFERENTIAL EQUATIONS PROFESSOR RICHARD BROWN
This is a form of the general solution to the ODE, a 1-parameter family of
solutions parameterized by the constant of integration C. However, we can
also parameterize this family of curves directly by the initial data y(0) = y .
0
This is better, as it ties the curves directly to their y-intercept here. Here
21 3 −2(0) 21
y(0) = y0 = − (0)+Ce 3 = +C,
8 4 8
so that we can write our general solution directly using the initial data as
21 3 21 −2t
y(t) = − t+ y − e 3 .
8 4 0 8
At right are some representa-
tive curves for different values of
y . Notice that it does look like
0
there will be a curve that will
touch, but not cross, the t-axis.
We are looking for the value of
y of this particular curve.
0
Anytime one is trying to pick
out a particular curve in a pa-
rameterized family of them, one
must make use of distinguish-
ing data carefully. Here, we are
not given points on the curve di-
rectly, which is usually the case.
Instead, we are given certain
properties. Firstly, the curve
touches the t-axis at some point.
Wedonotknowwhere,sowecan Figure 1. A few representative
curves y(t).
write y(t ) = 0, where t is the
0 0
intersection of the curve with the t-axis, and the y-coordinate there would be
zero. Secondly, since the curve must be smooth, and touches but does not
cross the t-axis, the curve will have to be tangent to the curve at t . So the
0
second piece of data is y′(t ) = 0. So we have
0
21 3 21 −2t0
(1) y(t ) = 0 = − t + y − e 3 ,
0 8 4 0 0 8
′ 3 2 21 −2t0
(2) y (t ) = 0 = − − y − e 3 .
0 4 3 0 8
EXAMPLE: PROBLEM 2.1.28 OF THE TEXT 3
This is now two (non-linear) equations in two unknowns, which we can solve
to find the particular value of y .
0
One way to do this: Rearrange Equation 2 to get
9 21 −2t0
(3) − = y − e 3 .
8 0 8
Notice that in Equation 1 of the pair, a good piece of the RHS matches the
RHShere. Substitute in the LHS of this last equation into Equation 1 to get
21 3 21 −2t0
y(t ) = 0 = − t + y − e 3
0 8 4 0 0 8
21 3 9
0 = 8 − 4t0 + −8
=12−3t =0.
8 4 0
This leaves t = 2 as a solution.
0
To find the corresponding y , substitute this value of t back into either
0 0
Equation 1 or 2. In Equation 2, we get
3 2 21 −2(2)
0 = − − y − e 3
4 3 0 8
9 4 21 ∼
− e3 + =y =−1.64288.
8 8 0
Below, we have added the graph of the curve corresponding to this value of
y . As you can see, it fits the criteria and solves the problem.
0
Figure 2. The solution curve y(t) is in red.
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