4.1. We look for an integrating factor µ(t), i.e., such a function that y 0 µ− 23 yµ =
0
(yµ)0 . The latter is equivalent to µ0 = − 32 µ, i.e., to µµ = − 32 . Integrating, we get
3
log µ = − 23 t + C, so that we can take µ(t) = e− 2 t . Then, after multiplication by
µ, our original equation becomes
3
(ye− 2 t )0 = 2e−t/2 .
Integrating the right-hand side, we get
3
ye− 2 t = −4e−t/2 + C,
3
y = −4et + Ce 2 t .
The initial condition gives
y0 = −4 + C,
C = y0 + 4,
so that solution of the initial value problem is
3
y(t) = −4et + (y0 + 4)e 2 t .
3
The function e 2 t grows faster than et , hence the sign of y(t) will be eventually the
same as the sign of y0 + 4 (if it is not zero). It follows that y grows positively when
y0 > −4, and negatively when y0 < −4. If y0 = 4, then y(t) = −4et , and y grows
negatively.
4.2. It is separable, and can be written
x2
dx,
1 + x3
or
Z
Z
x2
y dy =
dx,
1 + x3
1
y 2 /2 = ln |1 + x3 | + C.
3
The last equation is the general solution.
4.3. It is again a separable equation, which can be written
y dy =
(3 + 4y) dy = (e−x − ex ) dx,
which after integration becomes
3y + 2y 2 = −e−x − ex + C.
Substituting the initial condition, we get
3 + 2 = −2 + C,
hence C = 7. Let us solve the equation with respect to y:
3
−e−x − ex
7
y2 + y =
+ ,
2
2
2
3
9
−e−x − ex
7
9
y2 + y +
=
+ + ,
2
16
2
2 16
2
−e−x − ex
65
3
y+
=
+
4
2
16
r
3
−e−x − ex
65
y=− ±
+ .
4
2
16
1
2
Again substituting the initial condition, we get
r
3
65
3 7
1 = − ± −1 +
=− ± ,
4
16
4 4
which is correct only if we choose plus sign.
Consequently, the solution of the initial value problem is
r
3
−e−x − ex
65
y=− +
+ .
4
2
16
It is defined by the condition
65
−e−x − ex
+
> 0.
2
16
4.4. The equation can be rewritten as
(4x + 2y) dx + (bx + 3y) dy = 0,
which is exact if and only if ∂(4x+2y)
= ∂(bx+3y)
, i.e., when 2 = b. If b = 2, then
∂y
∂x
∂P
we want to find P (x, y) such that ∂x = 4x + 2y and ∂P
∂y = 2x + 3y. From the first
2
equation we have P (x, y) = 2x + 2xy + C(y). Substituting this into the second
equation, we get 2x + C 0 (y) = 2x + 3y, so that C 0 (y) = 3y, and we can take
C(y) = 23 y 2 . Hence, the general solution is
3
2x2 + 2xy + y 2 = C.
2
4.5. If Q(t) is the amount of salt (in lb) in the tank in the first 10 minutes, then
Q(0) = 0, and
1
Q
Q0 = 2 · − 2 ·
,
2
100
i.e.,
1
Q0 = 1 − Q.
50
or
0
Q
= 1,
1
1 − 50
Q
−50 ln |1 − Q/50| = t + C1 ,
ln |1 − Q/50| = −t/50 + C2 ,
1 − Q/50 = C3 e−t/50 ,
Q = 50 + Ce−t/50 .
From the initial condition we get:
0 = 50 + C,
C = −50,
hence in the first ten minutes the amount of salt is changing according to the law
Q(t) = 50 − 50e−t/50 = 50(1 − e−t/50 ).
At the end of 10 minutes, the amount of salt will be 50(1 − e−0.2 ) lb.
After that the differential equation describing the amount of salt is
Q
Q0 = − ,
50
hence
Q(t) = Ce−t/50 ,
3
and from the initial condition Q(10) = 50(1 − e−0.2 ) we get
50(1 − e−0.2 ) = Ce−0.2 ,
C = 50(e0.2 − 1),
hence
Q(t) = 50(e0.2 − 1)e−t/50 ,
and after 10 more minutes, we get
Q(20) = 50(e0.2 − 1)e−0.4
pounds of salt.
4.6. The characteristic polynomial is λ2 + λ − 2. Its roots are 1 and −2, hence
the general solution is
y(t) = C1 et + C2 e−2t .
4.7. The characteristic polynomial is λ2 + 2λ + 2 = (λ + 1)2 + 1. Its roots are
λ = −1 ± i, hence the general solution is
y(t) = C1 e−t cos t + C2 e−t sin t.
Its derivative is
y 0 (t) = −C1 e−t cos t − C1 e−t sin t − C2 e−t sin t + C2 e−t cos t.
Substituting the initial conditions, we get the system
(
√
√
2
C1 e−π/4 22√+ C2 e−π/4 22 =
−C1 e−π/4 2
= −2
√
We get from the second equation that C1 = eπ/4 2. Substituting it into the first
equation, we get
√
−π/4 2
1 + C2 e
= 2,
√ 2
2
C2 e−π/4
= 1,
2√
C2 = eπ/4 2.
Consequently, solution of the initial value problem is
√
y(t) = 2e−t+π/4 (cos t + sin t).
4.8. Characteristic polynomial is 4λ2 + 4λ + 1 = (2λ + 1)2 , so that its one
repeated root is λ = −1/2, and the general solution is
y(t) = (C1 + C2 t)e−t/2 .
4.9. Was solved in class.