2.1. The equation is separable: we can write it y −1 dy = − sin x dx, which
after integration becomes −y 2 /2 = cos x + C. This is the general solution.
2.2. It is also separable, and we can write: (y + ey ) dy = (x − e−x ) dx,
which after integration becomes y 2 /2 + ey = x2 /2 + e−x + C.
2.3. It is also separable, and can be written (1 + 2y) dy = 2x dx, or, after
integration, y + y 2 = x2 + C. The initial condition gives 0 = 4 + C, so that the
solution is
y 2 + y = x2 − 4.
2.4. Let Q(t) be the amount of salt in the tank at moment t. Then we have
Q0 = 2γ − 2Q/120 with the initial value Q(0) = 0. It is a separable equation,
dQ
= dt, which after integration is −60 ln |2γ−Q/60| =
that can be written 2γ−Q/60
t + C1 , or 2γ − Q/60 = C2 e−t/60 , or
Q = Ce−t/60 + 120γ.
Taking into account the initial condition, we get that C = −120, so that the
solution is Q(t) = 120γ − 120e−t/60 .
2.5. Writing down sum of the forces, we get: (here positive velocity is going
down, so that initial condition is v(0) = −v0 , one can write it the other way
around, then the gravity force is −mg, not mg):
mv 0 = mg − kv,
or
v0 = g −
k
v,
m
which is separable, and we get
k m − ln g − v = t + C1 ,
k
m
or
g−
or
k
k
v = C2 e− m t ,
m
mg
.
k
The initial condition gives us −v0 = C + mg
k , hence C = −v0 −
formula for velocity is
mg mg − k t
v=
e m .
− v0 +
k
k
k
v = Ce− m t +
Height will be maximal when v = 0, i.e., when
mg − k t
mg 0=
− v0 +
e m ,
k
k
mg mg − k t
= v0 +
e m ,
k
k
1
mg
k ,
and the
k
e− m t =
mg
,
kv0 + mg
k
mg
t = ln
,
m
kv0 + mg
m
kv0
m kv0 + mg
ln
=
ln 1 +
.
tmax =
k
mg
k
mg
Rt
In order to get the maximal height, we have to compute 0 max v dt, which is
−
Z
0
tmax
t
mg − k t max
mg − mg t mgt m − v0 +
+
e k
v0 +
e m =
dt =
k
k
k
k
k
t=0
mgtmax
mg − k tmax
m
v0 +
(e m
− 1) =
+
k
k
k
!
−1
kv0
mg m2 g
m
kv0
ln 1 +
v0 +
+
1+
−1 =
k2
mg
k
k
mg
m2 g
mv0
kv0
−
.
ln 1 +
2
k
mg
k
mg
It follows that the final answer is
xmax =
mv0
m2 g
kv0
− 2 ln 1 +
.
k
k
mg
We had to change the sign, since our coordinate system was oriented down.
2.6. If we write the equation in the standard form, we get
y0 +
2t
3t2
y
=
.
4 − t2
4 − t2
The functions are continuous everywhere except for the points where 4 − t2 = 0,
i.e., when t = ±2. The equation is linear, so the solution exists throughout any
interval on which the functions are continuous. It follows that the initial value
problem has a solution on (−2, 2).
2