Math 220, Differential Equations
Fall 2013 Exam 1 Solutions - 4-October-2013
Landon Kavlie
1. (25 pts) Find general solutions to the following differential equations:
(1) (10 points)
x
dy
sin(x)
+ 3y + x2 =
dx
x
(2) (15 points)
dy
2x + y 2 − cos(x + y)
.
=−
dx
2xy − cos(x + y) − ey
For the first one, this is a first-order linear problem. So we put it into standard
form first. That is,
dy
3
sin(x)
+ y=
− x.
dx x
x2
Then, we introduce the integrating factor
e
R
3
x dx
= e3 ln(x) = x3 .
We then multply the equation by this value.
x3
dy
+ 3x2 y = x sin(x) − x4 .
dx
d
The left-hand side reduces down to dx
[x3 y]. Thus, we integrate both sides giving
us that
Z
1
x3 y = x sin(x)dx − x5 + C.
5
Integration by parts then gives us that
1
x3 y = sin(x) − x cos(x) − x5 + C
5
Simplifying, we have that
y=
sin(x) cos(x) 1 2
C
−
− x + 3.
x3
x2
5
x
For the second one, this ends up being an exact equation. To see this, we crossmultiply and rearrange giving us that
(2xy − cos(x + y) − ey )dy + (2x + y 2 − cos(x + y))dx = 0.
Then, we see that
∂
(2xy − cos(x + y) − ey ) = 2y + sin(x + y)
∂x
∂
(2x + y 2 − cos(x + y)) = 2y + sin(x + y).
∂y
1
2
Thus, our equation is exact. So, we know that the full solution is given by F (x, y) =
C where F is some function whose partial derivatives are given by
∂F
= 2x − y 2 − cos(x + y)
∂x
∂F
= 2xy − cos(x + y) − ey .
∂y
We just have to pick one of these and integrate it. I choose the second one.
Z
F (x, y) = 2xy − cos(x + y) − ey dy = xy 2 − sin(x + y) − ey + g(x).
To find the function g(x), we then take the derivative with respect to x and compare.
That is,
y 2 − cos(x + y) + g 0 (x) = 2x + y 2 − cos(x + y).
This gives us that g 0 (x) = 2x and g(x) = x2 after integrating. Thus, we find that
F (x, y) = xy 2 − sin(x + y) − ey + 2x.
And our solution is given by F (x, y) = C That is,
xy 2 − sin(x + y) − ey + 2x = C.
2. (15 pts) Solve the initial value problem for the first-order differential equation:
(
y sin(x)
1 dy
x dx = y 2 +1
y(π) = 1.
This equation is separable. So, we just move all of the x terms to the right-hand
side and all of the y terms to the left-hand side. This gives us that
y2 + 1
dy = x sin(x)dx.
y
Now, we integrate. We integrate the right-hand side by parts giving us that
Z
x sin(x)dx = sin(x) − x cos(x) + C.
For the left-hand side, we split up the numerator giving us that
Z 2
Z
y +1
1
1
dy = y + dy = y 2 + ln(y) + C.
y
y
2
This gives us the equation
1 2
y + ln(y) = sin(x) − x cos(x) + C.
2
We now plug in y(π) = 1 to get that
1 2
1 + ln(1) = sin(π) − π cos(π) + C.
2
This simplifies using that ln(1) = sin(π) = 0 and cos(π) = −1. We get that
C = 12 − π. Thus, our solution is
1 2
1
y + ln(y) = sin(x) − x cos(x) + − π.
2
2
3
There is no way that I can see to solve for y.
3. (20 pts) Solve the initial-value problem for the second order differential equation

 4y 00 − 4y 0 + y = 0
y(0) = 2
 0
y (0) = 1.
This is a gimme. We just write down the associated equation
4r2 − 4r + 1 = 0.
This factors as (2r − 1)2 = 0 which gives us the double root r = 1/2. So, our
solution is given by
y = C1 e1/2t + C2 te1/2t .
We now use our initial conditions to find C1 and C2 . First, we use that y(0) = 2
to get that
2 = C1 e0 + C2 · 0 · e0 = C1 .
Next, we take a derivative of y getting that
y0 =
1
1
C1 e1/2t + C2 te1/2t + C2 e1/2t .
2
2
Using that y 0 (0) = 1 gives us that
1
C1 + C2 = 1 + C2 = 1.
2
Thus, we find that C2 = 0 and our solution is
y 0 (0) =
y = 2e1/2t .
4. (20 pts) A brine solution of salt flows at a constant rate of 4L/min into a
large tank that initially held 100L of pure water. The solution inside the tank is
kept well-stirred and flows out of the tank at a rate of 3L/min. The concentration
of salt entering the tank is 0.2kg/L.
(1) (10 points) Determine the mass of salt in the tank after t min.
(2) (10 points) Determine the concentration of salt in the tank when t = 2.
(Do not round your answer into a decimal number.)
The first problem is the standard one. Find the equation for the salt in the tank.
I hope you drew a lovely diagram of the problem. I will just skip to the good stuff.
First, we see that the flow rate into the tank is more than the flow rate out. So,
we set up a differential equation to figure out what the volume of liquid is in the
the tank at time t. For this, we do the usual compartmental analysis trick of
dV
= volume in − volume out = 4 − 3 = 1
dt
4
Where V (t) gives the volume of liquid in the tank at time t. Also, we have the
initial data V (0) = 100. So, we solve this and find that V (t) = 100 + t. Doing the
same thing for the salt, which I will label as A(t), we find that
dA
A
= salt in − salt out = 4 · 0.2 − 3
.
dt
100 + t
both of the “in” and the “out” are gotten by multiplying the flow rate times the
concentration. Adding the initial condition of A(0) = 0 (pure water) gives us the
initial-value problem
dA
3A
dt = 0.8 − 100+t
A(0) = 0.
This is a first-order linear problem that can be solved using integrating factors.
That is, we put it into standard form
dA
3A
+
= 0.8.
dt
100 + t
Then, we build the integrating factor
e
R
3
100+t dt
= e3 ln(100+t) = (100 + t)3 .
Multiplying the whole equation by this and simplifying gives us that
d
[(100 + t)3 A] = 0.8(100 + t)3 .
dt
Integrating gives us that
(100 + t)3 A = 0.2(100 + t)4 + C.
Simplifying gives
A = 0.2(100 + t) + C(100 + t)−3 .
Using the initial condition A(0) = 0 gives us that
0 = 0.2(100) + C(100)−3 .
This gives us that C = −2x107 . So, the solution to the first part is
A(t) = 0.2(100 + t) − 2x107 (100 + t)−3 .
Now for the concentration part. Remember that the concentration is given by
the amount of salt divided by the volume of the liquid. That is, C(2) = A(2)/V (2).
In full,
0.2(102) − 0.2x107 (102)−3
≈ −0.18152.
C(2) =
102
5. (20 pts) Consider the initial value problem
dy
dx = 1 + x sin(xy)
y(0) = 0.
(1) (5 points) Write down Euler’s method for yn+1 in terms of yn for a general
step-size h.
(2) (10 points) Use two steps of Euler’s method to approximate y(π).
(3) (5 points) Use one step of improved Euler’s method to approximate y(π).
5
Note: The recursive formula for Improved Euler’s method is
h
yn+1 = yn + [f (xn , yn ) + f (xn+1 , yn + hf (xn , yn ))].
2
The first one is easy. Just write down
yn+1 = yn + hf (xn , yn ).
If you want more, not that f (x, y) = 1 + x sin(xy). So, you can write
yn+1 = yn + h[1 + xn sin(xn yn )].
For the second one, we use two steps, so our step-size is h = (end−start)/(number of steps) =
(π − 0)/2 = π/2. Also, we use that y(0) = 0 to give us that y0 = 0 and x0 = 0.
Then, plugging stuff into the formula gives us that
y0 = 0
y1 = y0 + h[1 + x0 sin(x0 y0 )] = 0 + π/2[1 + 0 · sin(0 · 0)] = π/2
y2 = y1 + h[1 + x1 sin(x1 y1 )] = π/2 + π/2[1 + π/2 · sin(π/2 · π/2)] = π + π 2 /4 sin(π 2 /4).
For the third one, we again plug the numbers into the formula where h = π.
First, we notice that
y0 + hf (x0 , y0 ) = 0 + π[1 + 0 · sin(0 · 0) = π.
This will simplify things later. We then have that
y1 = 0 + π/2[(1 + 0 sin(0)) + (1 + π sin(π 2 ))] = π + π 2 sin(π 2 ).