Diff. Eqns.
Problem Set 2
Solutions
1. Find a function y(t) that solves the initial-value problem
ty 0 + (t + 1) y = 4t,
y(1) = 2,
t ≥ 1.
Multiply both sides by an integrating factor µ (t).
µ (t) ty 0 + µ (t) (t + 1) y = 4tµ (t)
We want to use (µy)0 = µ0 y + µy 0 on the left-hand side, so we divide the equation by t.
µ (t) y 0 + µ (t)
t+1
y = 4µ (t)
t
Again, because we want to use (µy)0 = µ0 y+µy 0 , we choose µ (t) to satisfy µ0 (t) = µ (t)
We solve this differential equation for µ (t):
Z 0
Z
µ (t)
t+1
dt =
dt
µ (t)
t
(1)
t+1
.
t
On the left we use substitution (u = µ (t)), and on the right we have two known integrals.
Z
Z
Z
1
1
du = (1) dt +
dt
u
t
ln |u| = t + ln |t| + c
µ (t) = ±et+ln(t)+c
= Ctet .
where C = ±ec is an arbitrary constant, and we choose µ (t) = tet . Then our differential
equation in (1) becomes
d
(µ (t) y) = 4µ (t)
dt
d
tet y = 4tet
dt
Z
tet y =
4tet dt
(use integration by parts)
Z
t
t
= 4 te − e dt
= 4 (t − 1) et + c
t−1
+ ct−1 e−t .
y (t) = 4
t
To find c, we use our initial condition.
2 = y (1) = ce−1
⇒ c = 2e.
So our solution is
y (t) = 4
t−1
+ 2t−1 e−t+1 .
t
2. Find a solution of the differential equation
y0 =
x2 y
.
x3 + 1
This differential equation is separable:
x2
y 0 (x)
= 3
y (x)
x +1
Z 0
Z
y (x)
x2
dx =
dx
3
y (x)
x +1
The left integral results in a natural log, while the right integral can be solved with the
substitution u = x3 + 1.
1 3
ln x + 1 + c
3
1/3
y (x) = C x3 + 1
ln |y (x)| =
where C is an arbitrary constant.
3. A tank with a capacity of 400 gal originally contains 100 gal of water with
80 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a
rate of 4 gal/min, and the mixture is allowed to flow out of the tank at a rate of
3 gal/min. Find the amount of salt in the tank at any time prior to the instant
when the solution begins to overflow. Find the concentration (in lb/gal) of salt
in the tank when it is on the point of overflowing.
Let g (t) be the gallons of water in the tank at time t. Taking the rate of flow in minus the
rate of flow out, we have
g 0 (t) = 4 − 3 = 1 gal/min
so g (t) = t + c. Also, we have the initial condition that g (0) = 100, so 100 = g (0) = c and
g (t) = t + 100 gal.
Let T be the time of overflow. Then
400 = g (T ) = T + 100
⇒
T = 300 min.
Now we can consider Q (t), the pounds of salt in the water at time t. From the information
given,
Q0 (t) = 1 lb/gal · 4 gal/min −
Q (t)
3Q (t)
· 3 gal/min = 4 −
g (t)
t + 100
with initial condition Q(0) = 80 lb. We multiply everything by an integration factor µ (t).
µ (t) Q0 (t) +
3
µ (t) Q (t) = 4µ (t) .
t + 100
µ (t) must solve µ0 (t) = 3/ (t + 100) µ (t), so
Z
3
µ (t) = exp
dt = exp (3 ln |t + 100| + c) = (t + 100)3
t + 100
(2)
(choose c = 0).
Then (2) becomes
d (t + 100)3 Q (t) = 4 (t + 100)3
dt
Z
(t + 100)3 Q (t) = 4
(t + 100)3 dt
(t + 100)3 Q (t) = (t + 100)4 + c
Q (t) = (t + 100) + c (t + 100)−3
Applying our initial condition,
80 = Q (0) = 100 + c (100)−3
⇒
c = −2 · 107 .
Our solution is then
Q (t) = (t + 100) − 2 · 107 (t + 100)−3 lb.
At the time of overflow, the amount of salt is
Q (T ) = Q (300) = 400 − 2 · 107 · 400−3 = 399.6875 lb
and the concentration is
Q (T )
400 − 2 · 107 · 400−3
=
≈ .9992 lb/gal.
g (T )
400