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Math 286 X1 — Midterm 1 Solutions
September 24, 2009
Question 1, Version A. (12 points.) In each of these problems, you should identify which kind of equation it is, then explain a method of solution. If the equation is of a type which we have not yet considered, explain this and say that we don’t yet have a method.
a]
dy dx
- x^4 y + x^19 = 0, b] y′^ = x^2 y, c] y′^ = x^2 + y,
d] y′^ = x^2 + y^2 , e] x^4 y′′^ + sin(x)y′^ + y = 0, f] y′′^ − 3 y′^ + 2y = 0.
Solution:
a. This is a first-order linear equation (here p(x) = x^4 , q(x) = −x^19 ) and so we use an integrating factor.
b. This equation is both separable and first-order linear, so we could separate or use integrating factor.
c. This equation is first-order linear, so we use integrating factor.
d. We have no method for this equation.
e. We have no method to solve this equation fully, however notice that if we did have two solutions to the equation, we would be able to proceed.
f. Second-order linear, constant coefficient: use exponential Ansatz, get two solutions from roots, take linear combination.
Question 1, Version B. (12 points.) In each of these problems, you should identify which kind of equation it is, then explain a method of solution. If the equation is of a type which we have not yet considered, explain this and say that we don’t yet have a method.
a]
dy dx
- y/x = 1, b] y′^ = xy^3 , c] y′^ = x^4 + y^2 ,
d] y′^ = x^2 , e] xy′′^ + exy′^ + x^5 y = 0, f] y′′^ − 2 y′^ + 5y = 0.
Solution:
a. First-order linear equation (here p(x) = 1/x, q(x) = 1) and so we use an integrating factor.
b. This equation is separable, so separate variables and integrate both sides.
c. We have no method for this equation.
d. The right-hand side is a function only of x, just integrate.
e. We have no method to solve this equation fully, however notice that if we did have two solutions to the equation, we would be able to proceed.
f. Second-order linear, constant coefficient: use exponential Ansatz, get two solutions from roots, take linear combination.
Question 2, Version A. (10 points.) Let P (t) be a solution to dP dt
= P (P − 1)(P − 2)eP^ (^2) sin(P 6 ) .
If we consider the solution with P (0) = 1/2, what is limt→∞ P (t)? Same question, but if P (0) = 3/2? Same question, but P (0) = 5/2?
Solution: Since this is a first-order autonomous equation P ′^ = f (P ), we need only know the sign and roots of f to answer these questions. In particular, a graph of f suffices. Also noting that the exponential term is always positive makes our job easier, since we need only graph P (P − 1)(P − 2) and the signs will be correct. Finally, to get this graph is easy, since it is clearly zero at P = 0, 1 , 2, and is a cubic with a positive leading order coefficient, so it must look like the graph in Figure 1. From this we see that P = 1 is stable and will
0.5 1.0 1.5 2.
**- 0.
0.
0.
Figure 1: Graph of P (P − 1)(P − 2).
attract any initial condition between 0 and 2; any IC which is < 0 goes to −∞, and any IC above 2 goes to ∞. Thus the answers to the three questions will be 1, 1 , ∞.
Question 2, Version B. (10 points.) Let P (t) be a solution to dP dt
= (P 2 − 1)(sin(P 2 ) + 17)^2.
If we consider the solution with P (0) = 0, what is limt→∞ P (t)? Same question, but if P (0) = 3? Same question, but P (0) = −17?
Solution: This problem is similar to version A. Notice that the last factor is always positive (the minimum value which sine can take is −1, thus sin(P 2 )+17 > 16 > 0. Therefore
- 1.0 - 0.5 0.5 1. - 0. - 0.
0.
0.
Figure 3: Graph of P (P 2 − 1).
Multiplying through gives
ex
3 y′(x) + 3x^2 ex
3 y(x) = 1, d dx
ex
3 y(x)
ex
3 y(x) = x + C, y(x) = xe−x
3
3 .
We need to satisfy the initial condition, plugging in gives
y(0) = C,
or C = 4, so the answer is y(x) = xe−x
3
3 .
Question 3, Version B. (20 points.) Solve
xy′^ + 3y = 5x^2 , y(1) = 2.
Solution: This is a first-order linear equation, so we will use an integrating factor. We first need to divide through by x to obtain the equation
y′^ +
x
y = 5x.
The integrating factor is
ρ(x) = e
R (^3) /x dx = e3 ln(x)^ = eln(x
(^3) ) = x^3.
Multiplying through gives
x^3 y′(x) + 3x^2 y(x) = 5x^4 , d dx
(x^3 y(x)) = 5x^4 ,
x^3 y(x) = x^5 + C, y(x) = x^2 + Cx−^3.
Plugging in the initial condition gives
y(1) = 1 + C,
so C = 1, and we have y(x) = x^2 + x−^3.
Question 3, Version C. (20 points.) Solve
y′^ − y = ex, y(0) = 2.
Solution: This is a first-order linear equation and we use an integrating factor. The integrating factor will be ρ = e
R (^) − 1 dx = e−x,
so we have
e−xy′(x) − e−xy(x) = 1, d dx
(e−xy(x)) = 1,
e−xy(x) = x + C, y(x) = xex^ + Cex.
Plugging in x = 0 gives y(0) = C,
or C = 2, so we get y(x) = xex^ + 2ex.
Question 4, Version A. (20 points.) Solve
dy dx
= sin(x)y^2 , y(0) = 2.
Solution: This equation is separable; separating gives dy y
= cos(t) dt,
ln |y| = sin(t) + C, y(t) = Cesin(t).
Plugging in the initial condition gives C = 5, so the solution is
y(t) = 5esin(t).
Question 5, Version A. (18 points.) The top of the bell tower of Altgeld Hall is 40m above the ground. How fast do you need to throw an object from ground level so that it goes this high? (Assume you’re throwing from exactly 0m height, and that the acceleration due to gravity is −10m/s^2 .)
Solution: The acceleration due to gravity is g = −10m/s^2. We know that acceleration is the derivative of velocity, so we have
dv dt
= g, or v(t) = gt + v(0),
and since velocity is the derivative of position, then
dx dt
= gt + v(0), or x(t) =
g 2
t^2 + v(0)t + x(0).
We assume that x(0) = 0, and we are looking for v(0). We first need to determine at which time the object reaches its maximum height, and this will be when the velocity is zero, so we set gt + v(0) = 0,
giving
t = −
v(0) g
Plugging this into the equation for x(t) gives
x(−v(0)/g) =
(v(0))^2 2 g
(v(0))^2 g
(v(0))^2 2 g
So we set this equal to 40m, and we get
(v(0))^2 2 g
= 40 m,
(v(0))^2 = 800 m^2 /s^2 , v(0) =
800 m/s.
Question 5, Version B. (18 points.) Let us say that we have two populations of bacteria whose growth is determined by the linear population model, but that they grow at different rates. Call these rates k 1 , k 2. We observe that population #1 starts off with 1000 bacteria and has 3000 bacteria one day later. Population #2 starts off with 500 bacteria and has 2500 bacteria one day later. Determine k 1 and k 2. Now, will the size of population #2 ever surpass that of population #1? If not, say why not, but if so, determine when it happens.
Solution: We will use the linear growth model, so population #1 grows like
dP 1 dt
= k 1 P 1 ,
and population #2 like dP 2 dt
= k 2 P 2.
First consider population #1. We can solve these equations to obtain
P 1 (t) = P 1 (0)ek^1 t, P 2 (t) = P 2 (0)ek^2 t
The problem tells us P 1 (0 days) = 1000, P (1 day) = 3000,
or
3000 = 1000ek^1 (1^ day),
3 = ek^1 (1^ day), k 1 (1 day) = ln 3, k 1 = ln 3 day−^1.
Plugging in the values to the other equation gives
k 2 = ln 5 day−^1.
Just looking at the numbers, we see that population #2 grows faster than population #1, but started smaller, so it should overtake at some point. We see if there is a point where the populations are equal:
1000 ek^1 t^ = 500ek^2 t, 2 = e(k^2 −k^1 )t, (k 2 − k 1 )t = ln 2,
t =
ln 2 k 2 − k 1
but then
v(10 s) = 300(1 − e−k(10^ s)) km/hr, = 300(1 − e−2 ln(3/2)) km/hr, = 300(1 − eln(4/9)) km/hr,
= 300(1 − 4 /9) km/hr = 300 ∗ 5 /9 km/hr =
km/hr.
Question 6, Version A. (20 points.) Solve
y′′^ − 4 y′^ + 3y = 0, y(0) = 2, y′(0) = − 2.
Solution: This is a second-order linear equation with constant coefficients. We make the exponential Ansatz y(x) = erx,
and we get r^2 − 4 r + 3 = 0.
This equation has two roots, r = 1, 3, so two solutions to this equation are
ex, e^3 x.
The general solution is y(x) = C 1 ex^ + C 2 e^3 x.
Plugging in initial conditions gives
y(0) = C 1 + C 2 = 2, y′(0) = C 1 + 3C 2 = − 2.
Subtracting the first equation from the second gives 2C 2 = −4 or C 2 = −2 and thus C 1 = 4. Thus the answer is y(x) = 4ex^ − 2 e^3 x.
Question 6, Version B. (20 points.) Solve
y′′^ − 3 y′^ + 2y = 0, y(0) = 1, y′(0) = − 1.
Solution: This is a second-order linear equation with constant coefficients. We make the exponential Ansatz y(x) = erx,
and we get r^2 − 3 r + 2 = 0.
This equation has two roots, r = 1, 2, so two solutions to this equation are
ex, e^2 x.
The general solution is y(x) = C 1 ex^ + C 2 e^2 x.
Plugging in initial conditions gives
y(0) = C 1 + C 2 = 1, y′(0) = C 1 + 2C 2 = − 1.
Subtracting the first equation from the second gives C 2 = −2 and thus C 1 = 3. The answer is y(x) = 3ex^ − 2 e^2 x.
Question 6, Version C. (20 points.) Solve
y′′^ − 6 y′^ + 8y = 0, y(0) = 2, y′(0) = 6.
Solution: This is a second-order linear equation with constant coefficients. We make the exponential Ansatz y(x) = erx,
and we get r^2 − 6 r + 8 = 0.
This equation has two roots, r = 2, 4, so two solutions to this equation are
e^2 x, e^4 x.
The general solution is y(x) = C 1 e^2 x^ + C 2 e^4 x.
Plugging in initial conditions gives
y(0) = C 1 + C 2 = 2, y′(0) = 2C 1 + 4C 2 = 6.
Subtracting twice the first equation from the second gives 2C 2 = 2 or C 2 = 1 and thus C 1 = 1. The answer is y(x) = e^2 x^ + e^4 x.
