Georgia Tech Lorraine
Summer 2026
Heinrich Matzinger
Solution for homework 1 for math2552
Name:
1)
GT-number:
The speed of a parachute at time t is denote by v(t). It follows the differential equation
dv(t)
= 9.81 − 0.1v(t)
dt
Assume that at time t = 0 the speed is 0.
a) What is v(t) equal to? Answer: The general solution is
v(t) = Ce−0.1t + 98.1
where C is a constant which will be determined by the inital condition. So, to determine C we
put v(0) = 0 and hence
Ce0 + 98.1 = 0
which leads to C = −98.1 Hence the speed at time t is
v(t) = −98.1e−0.1t + 98.1.
b) What is the speed after a long time going to be equal to? Answer: Note that as t goes to
infinity, we have that e−0.1t goes to 0 and hence the speed v(t) on the long run becomes close
(converges to) 98.1.
2) The speed of a rocket being lunched from an airplane downwards is denote by v(t). Assume
the differential equation for v(t) is
dv(t)
= 9.81 − 0.1v(t) + t2
dt
At time t = 0 the speed is 0. What is the speed at time t = 3? Answer: Take v(t) = f (t) · e−0.1t
Plug into the differential equation:
df (t)e−0.1t
= 9.81 − 0.1f (t)e−0.1t + t2
dt
and hence applying the rule for the derivative of a product we get
f ′ (t)e−0.1t + f (t)(−0.1e−0.1t ) = 9.81 − 0.1f (t)e−0.1t + t2
Now the expression −0.1f (t)e−0.1t is to be found on both sides of the last equation above and
hence cancels out. Thus, we find
f ′ (t)e−0.1t = 9.81 + t2
so that
f ′ (t) = e0.1t (9.81 + t2 )
and hence by integrating
Z
f (t) =
e0.1t (9.81 + t2 )dt + C
(1)
where C is a constant to accomodate the initial condition. Now by intergration by part one
could solve the above integral, but I am not going to do it.... so finally the solution is
v(t) = f (t)e−0.1t
where f (t) is obtained by integrating the integral on the right side of equation 1.
3)
Find the general solution to the differential equation:
dy
= x3 · y.
dx
What is the value of y(2) if y(1) = 5? Answer: we have
1
dy = x3 dx
y
and hence integrating on both sides of the above equation:
Z
Z
1
dy = x3 dx
y
so that
x4
+ C.
4
Now we take epxonential on both sides of the above equation and find the general solution
ln(y) =
4
y(x) = kex /4
where :exp(C) = k is a constant which will be determined by the initial conditions. In the
present case we know that y(1) = 5. So, 5 = ke0.25 and hence k = 5/e0.25 . thus for x = 2 we find
y(2) = 5e4−0.25 = 5e3.75
4)
Solve the differential equation:
dy
= x3 · y + cos(x).
dx
What is the value of y(2) if y(1) = 5? Answer: Take the solution to problem 3 and multiply it
4
with f (x). Call the solution of problen 3: g(x) so that g(x) = kex /4 . Then take y(x) = f (x)g(x).
Plug it into the current equation you get
f ′ (x)g(x) + f (x)g ′ (x) = x3 f (x)g(x) + cos(x)
and hence
f ′ (x)g(x) = cos(x)
(Here we used the fact that g ′ (x) = x3 g(x) and also f (x)g ′ (x) = f (x)x3 g(x).) thus
f ′ (x) = cos(x)/g(x)
and hence
Z
f (x) =
4
cos(x)e−x /4 dx
so the general solution is
Z
y(x) = f (x)g(x) =
4
4
cos(x)e−x /4 dx · ex /4
5) Let X(t) be the number of fish in a given lake. Every minute out of thousand there are
X(t)(4 − X(t)) which are born and 4 which die. X(t) is measured in millions and the time unit
is the minute.
a) Write a differential equation for X(t).Answer: the change in the number of fish per time
unit is the derivatie. It is [X(t)(4 − X(t)) − 4]X(t)0.001 and hence
dX
= [X(t)(4 − X(t)) − 4]X(t)0.001
dt
b) Assume that you start with 1 million fish. What is going to happen on the long run? Answer:
There is a solution where the number of fish remain constant for X = 0 and another such
solution for when the number of fish is equal to X = 2. For these constant values we find that
the expression on the right of the differential equation is 0. Between these values the derivative
is negative. So, the number of fish will decrease. Hence, the number of fish will converge to 0
and hence the fish will die out.
6)
Consider the differential equation
dX
= (X 2 − 5X + 6) cos(X · t).
dt
Assume that X(0) = 2.5. What happens on the long run with X(t)? Answer: This differential
equation is maybe difficult to solve. But you can see that the function f (x) = x2 − 5x + 6 is
equal to 0 for x = 2 and x = 3. Hence, two solutions are the horizontal lines given by X(t) = 2
and X(t) = 3. But we start with X(0) = 2.5 so we will never get out of the interval [2, 3]. It is
easy to see that the long term solution X(t) will fluctuate between 2 and 3 indefinitely. (Draw
the direction field to see it!)
7)
Consider the differential equation
dX
t + cos(2t)
=
.
dt
X +3
Assume that X(0) = 0. What is the solution to the above equation? Answer: This is a separable
differential equation. We find:
(X + 3)dX = (t + cos(2t))dt
and hence integrating on both sides
Z
Z
(X + 3)dX = (t + cos(2t))dt
so that
0.5X 2 + 3X = 0.5t2 + 0.5sin(2t) + C
where C is a constant determined by the initial condition. Now since X(0) = 0, we find C = 0.
Hence
0.5X 2 + 3X − 0.5t2 − 0.5sin(2t) = 0
so that
X 2 + 6X − t2 − sin(2t) = 0
The above equation is a quadratic equation in X once t is given. Using the formula for solving
a quadrativ equation we find
p
−6 + 36 + 4(t2 + 0.sin(2t)
X(t) =
.
2