Particular Solutions
If the differential equation is actually modeling something (like the
cost of milk as a function of time) it is likely that you will know a
specific value (like the fact that milk was $3.29/gallon on January
1, 2010.) This is what is called an initial condition.
So if we are trying to solve y 0 − 4y = 0 and we know that y = 3 if
x = 0 then we can use the general solution y = Ae4x together with
the initial condition y(0) = 3 to find a single solution (i.e. we can
find out what A is).
y = Ae4x and y = 3 at x = 0
⇒ 3 = Ae4·0 ⇒ 3 = A ⇒ y = 3e4x
More separation of variables:
In general you might be trying to solve an equation that looks like
dy
= f (x)g(y).
dx
For example y 0 = xy, where f (x) = x, and g(y) = y.
So we get all the x0 s on one side and all the y 0 s on the other side
and integrate:
Z
Z
1
dy = f (x)dx.
g(y)
Of course this relies on our ability to do these integrals, and in our
example we can and we have
Z
Z
1
x2
2
2
dy = xdx ⇒ ln |y| =
+ C ⇒ |y| = ex /2+C = Aex /2
y
2
Interpreting word problems:
Examples of differential equations come from all over the place,
from economics, biology, chemistry, business, etc. You must learn
to interpret certain words in problems. Most commonly if you see
the phrase “A is proportional to B” all that means is
A = kB
for some number k.
Also, all differential equations will involve rates of change because
that is exactly what the derivative is.
Investments
The rate of growth of an investment is proportional to the amount
in the investment at any time t. That is,
dA
= kA.
dt
What is A?
In this problem A represents the amount (in dollars say) of the
investment and A = A(t) i.e. A is a function of time, A may
increase or decrease with time (depending on the investment and
in this case the rate of growth k.)
What is the general solution?
Z
Z
1
dA = kdt ⇒ ln |A| = kt + c ⇒ A(t) = Cekt .
A
Investments cont.
We have A(t) = Cekt represents the rate of growth of an
investment. If we know that an initial investment of $1000 grew to
3320.12 in 10 years, what is the particular solution?
Now we have to find both C and k, what do we know?
Initially the investment was worth $1000 so A = 1000 at t = 0 or
A(0) = 1000. This tells me C :
A(0) = Cek·0 = Ce0 = C ⇒ C = 1000.
Now we know that at t = 10 we have A = 3320.12, this will tell us
what k is.
Before we try to figure out what k is, can you tell me if k should
be positive or negative?
Investments cont.
We have A(t) = 1000ekt and we know A(10) = 3320.12, we are
expecting k > 0, so let’s figure out what it is.
3320.12 = 1000ek·10 ⇒
ln
3320.12
= e10k
1000
ln 3320.12
3320.12
1000
= 10k ⇒ k =
= 0.12
1000
10
So
A(t) = 1000e0.12t
Newton’s Law of Cooling
Newton’s law of cooling states that the rate of change in the
temperature T of an object is proportional to the difference
between the temperature of the object (T ) and the surrounding
temperature T0 . This can be expressed by the differential equation:
dT
= k(T − T0 ).
dt
Note that here T0 is the ambient, or surrounding, temperature. k
is the rate of cooling and t is time. Since the object is going from
higher temperature to cooler temperature the rate of change will
be negative, so k < 0.
Newton’s Law of Cooling cont.
A room is kept at a constant temperature of 68◦ F. A pie is taken
out of a 350◦ oven and placed on the counter. If the pie has
reduced in temperature to 150◦ in 45 minutes, when will the pie
reach 80◦ ? First we must find the general solution:
Z
Z
1
dT = kdt.
(T − T0 )
ln(T − T0 ) = kt + c ⇒ T − T0 = ekt+c = Aekt ⇒ T (t) = T0 + Aekt
Newton’s Law of Cooling cont.
We have the general solution
T (t) = T0 + Aekt
We will measure time in hours, though we could use minutes if we
wanted. The ambient temperature here is 68◦ .
We know T (0) = 350 and T (.75) = 150 this will help us find k :
T (0) = 68 + Aek·0 = 68 + A = 350 ⇒ A = 350 − 68 = 282.
So now we have T (t) = 68 + 282ekt , and we know T (.75) = 150
so
150 − 68
82
150 = 68 + 282e.75k ⇒
= e.75k ⇒ ln
= .75k
282
282
⇒k=
82
ln 282
= −1.65 or k = −1.65
.75
Relation to slope and graphs:
We can also solve a problem like this:
1
At each point (x, y) on the graph, the slope is 2x/y.
2
The graph passes through the point (1, 1).
The equation to solve is
dy
= 2x/y,
dx
and the additional information we know is that y(1) = 1, in other
words when x = 1, then y = 1.
Relation to slope and graphs cont.:
dy
= 2x/y,
dx
First separate variables, then integrate:
Z
Z
ydy = 2xdx.
y2
= x2 + C
2
Writing it a little more simply we have y 2 − 2x2 = C, note that
we don’t have to write 2C we can just use C again for a different
constant. Now plug in the known values (x, y) = (1, 1) to find the
exact value for C :
12 − 2(1)2 = −1 = C ⇒ y 2 − 2x2 + 1 = 0 is the curve we are
trying to find.
⇒