HOMEWORK 2
(1) Section 2.1, Prob. 13. Find the solution of the initial value problem
y 0 − y = 2te2t ,
y(0) = 1.
Solution. Pick an integrating factor
µ(t) = e−t .
Multiplying µ to the both side of the equation, one has
(e−t y)0 = 2tet ,
and hence
−t
Z
2tet dt
Z
t
= 2(te − et dt)
e y =
= 2tet − 2et + C.
Therefore, the general solution is
y(t) = 2te2t − 2e2t + Cet .
Since y(0) = 1, one has
1 = −2 + C,
and then C = 3. So, the solution of the initial value problem is
y(t) = 2te2t − 2e2t + 3et .
(2) Section 2.1, Prob. 30. Find the value of y0 for which the solution of the
initial value problem
y 0 − y = 1 + 3 sin t,
y(0) = y0
remains finite as t → ∞.
Solution. Picking an integrating factor
µ(t) = e−t ,
and multiplying it to the both sides of the equation, one has
(e−t y)0 = e−t + 3e−t sin t.
1
2
HOMEWORK 2
Therefore
Z
(e−t + 3e−t sin t)dt
Z
t
t
= e (−e + 3 e−t sin tdt)
t
y(t) = e
3
= −1 − (sin t + cos t) + Cet .
2
By the initial condition y(0) = y0 , one has
3
y0 = −1 − + C,
2
and hence
5
C = + y0 .
2
So the solution is
5
3
y(t) = −1 − (sin t + cos t) + ( + y0 )et .
2
2
3
t
Since −1 − 2 (sin t + cos t) is bounded, and e → ∞ as t → ∞, the solution
remains finite if and only if
5
+ y0 = 0,
2
i.e.
5
y0 = − .
2
(3) Section 2.2, Prob. 9(a)(c). Find the solution of the given initial value
problem in explicit form. Determine the interval in which the solution is
defined.
Solution. Write the given equation as
1
dy = (1 − 2x)dx.
y2
Taking antiderivatives on both sides,
Z
Z
1
dy = (1 − 2x)dx,
y2
one has
1
− = x − x2 + C.
y
HOMEWORK 2
3
By the initial condition y(0) = − 16 , one has
6 = C,
and hence
1
− = x − x2 + 6.
y
So, the solution in explicit form is
1
1
=
.
y= 2
x −x−6
(x + 2)(x − 3)
The intervals in which the function is well defined are (−∞, −2), (−2, 3), and
(3, +∞). Since the initial value is given at x = 0, the interval in which the
solution is valid is
(−2, 3).
(4) Section 2.2, Prob. 24. Solve the initial value problem
y 0 = (2 − ex )/(3 + 2y),
y(0) = 0
and determine where the solution attains its maximum value.
Solution. Write the equation as
(3 + 2y)dy = (2 − ex )dx.
Taking antiderivatives on both sides, one has
3y + y 2 = 2x − ex + C.
Since y(0) = 0, one has
0 = −1 + C.
So the solution (in implicit form) is
y 2 + 3y + (ex − 2x − 1) = 0.
Hence
√
9 − 4(ex − 2x − 1)
−3 ± 13 − 4ex + 8x
y=
=
.
2
2
Since y(0) = 0 (the initial condition), one take the branch with “ + ”, and so
√
−3 + 13 − 4ex + 8x
.
y(x) =
2
The interval in which the solution is valid is exactly the interval containing
0 and in which the function
−3 ±
p
g(x) := 13 − 4ex + 8x
4
HOMEWORK 2
is positive. Also note that y(x) attains its maximum value if and only if
the function g(x) = 13 − 4ex + 8x attains its maximum value. So, in order
to find where y(x) attains its maximum value, one only need to find where
g(x) attains its maximum value in the interval containing 0 and where g(x)
is positive.
Note that
g 0 (x) = −4ex + 8.
Hence g(x) is increasing if x ≤ ln 2 and is decreasing if x ≥ ln 2. So, the
function y = g(x) attains the maximum value at x = ln 2, and x = ln 2 is in
the interval where g is positive (and so it is in the interval where the solution
y(x) is valid). So, the solution y(x) attains its maximum value at x = ln 2.