# 92.236 Engineering Differential Equations Practice Final Exam

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```					             92.236 Engineering Diﬀerential Equations   Practice Final Exam Solutions
Spring 2010

Problem 1. (10 pts.)
dy
Solve the following initial value problem: x2           = xy + 2, y(1) = 2.
dx
Express your solution y explicitly in terms of x.

This is a linear d.e. because y and dy/dx appear only to the ﬁrst power. 2 pts.
dy           1     2
First write the d.e. in standard form:          −           y = 2 . 1 pt.
dx           x    x
−1/x dx
The integrating factor is ρ(x) = e                  = e− ln(x) = x−1 . 2 pts.
Multiplying both sides of the standard form of the d.e. by the integrating factor, we have
dy     1               2                                       d
x−1     −      y = x−1       2
1 pt. ⇒ x−1 y − x−2 y = 2x−3 ⇒        x−1 y = 2x−3 ⇒
dx     x              x                                       dx
x−1 y =      2x−3 dx = −x−2 + c. 2 pts.

y(1) = 2 ⇒ 1−1 (2) = −1−2 + c ⇒ c = 3. 1 pt.

1
Therefore, x−1 y = −x−2 + 3 ⇒ y = 3x −                .     1 pt.
x

Problem 2. (10 pts.)
dy
Solve the following initial value problem: 2xy               + y 2 − 3x2 = 0, y(1) = 2.
dx

dy                   dy     3x2 − y 2
Method 1. 2xy     + y 2 − 3x2 = 0 ⇒     =            . dy/dx equals a rational function, and every
dx                   dx       2xy
term has the same degree (2).Therefore, this d.e. is homogeneous. 2 pts.
dy          dv
We introduce the new variable v = y/x. In the d.e. we replace                  by v + x    and we replace y
dx          dx
dy    3x2 − y 2         dv    3x2 − (xv)2
by xv:         =           ⇒ v+x       =              2 pts.
dx      2xy             dx      2x(xv)
x2    3 − v2    3 − v2      dv    3 − v2       3 − 3v 2     2v         1
=           2v
=         ⇒x     =         −v =           ⇒        2
dv = dx 2 pts.
2x           2v        dx      2v            2v      3 − 3v       x
1          2v            1         1        2
⇒                 dv =       dx ⇒ − ln 1 − v = ln(x) + c 2 pts.
3        1 − v2          x         3
⇒ ln     1 − v 2 = −3 ln(x) − 3c ⇒ 1 − v 2 = e−3 ln(x)−3c ⇒ 1 − v 2 = ±e−3c e−3 ln(x) = c1 x−3
c1
2
y
⇒ v 2 = 1 − c1 x−3 ⇒              = 1 − c1 x−3 . 1 pt.
x
2
2
The initial condition y(1) = 2 ⇒                 = 1 − c1 1−3 ⇒ c1 = −3 1 pt. .
1

3   x3 + 3           x3 + 3
Therefore, (y/x)2 = 1 + 3x−3 ⇒ y 2 = x2 1 + 3x−3 = x2 +                   =        ⇒ y=             .
x     x                x
dy      ∂M   ∂                  ∂N   ∂
Method 2. y 2 − 3x2 + 2xy       = 0.    =    y 2 − 3x2 = 2y.    =    [2xy] = 2y.
dx      ∂y   ∂y                 ∂x   ∂x
M        N
∂M   ∂N
Since    =    , the d.e. is exact. 2 pts.
∂y   ∂x

Therefore, the solution of the d.e. is f (x, y) = c, where the function f satisﬁes the conditions
∂f                       ∂f               ∂f
= M = y 2 − 3x2 and      = N = 2xy.        = y 2 − 3x2 ⇒ f =    y 2 − 3x2 dx = xy 2 − x3 + g(y)
∂x                       ∂y               ∂x
2 pts.

∂f
= 2xy ⇒ f =       (2xy) dy = xy 2 + h(x) 2 pts.
∂y
The two formulas for f must be equal: xy 2 − x3 + g(y) = xy 2 + h(x). Comparing the two for-
mulas, we see that g(y) = 0 and h(x) = −x3 . Therefore, f = xy 2 − x3 , so the solution of the d.e.
is xy 2 − x3 = c 3 pts. y(1) = 2 ⇒ (1)(22) − 13 = c ⇒ c = 3. 1 pt. Therefore, the solution of the

x3 + 3
initial value problem is xy 2 − x3 = 3 ⇒ y =
x

Problem 3. (15 points)
A tank initially contains 100 liters of water in which 1 kilogram of salt is dissolved. A salt solution
containing 0.2 kilograms of salt per liter is pumped into the tank at the rate of 10 liters per minute,
and the well-mixed solution is pumped out of the tank at the rate of 10 liters per minute.
Let t denote time (in minutes), and let x denote the amount of salt in the tank at time t (in
kilograms).
dx
a. Write down the diﬀerential equation        = something describing this mixing problem.
dt
dx
= rate in - rate out
dt
= (ﬂow rate in)(concentration in) - (ﬂow rate out)(concentration out), 3 pts. so
dx        liters          kg         liters      x kg
= 10              0.2       − 10                       .
dt       minute          liter      minute     100 liters
1 pt.          1 pt.        1 pt.        2 pts.
(The volume of liquid in the tank is constant at 100 liters because the ﬂow rates in and out are
dx     x
equal.) Therefore, the d.e. modeling this mixing problem is         =2−
dt     10
b. Draw the phase line for the d.e. from part a.
x
First ﬁnd the critical points: 2 −    = 0 ⇒ x = 20.
10
The critical point divides the phase line into 2 intervals: x > 20 and x < 20.
dP              30
=2−       = −1 < 0, so the direction arrow points up for x > 20.
dt x=30        10
dx              10
=2−       = 1 > 0, so the direction arrow points up for x < 20.
dt x=10         10
x

20

4 pts.

c. Use the initial condition x(0) = 1 and your phase line to ﬁnd the limiting value of x(t) as t
increases.

The phase line shows that x(t) → 20 as t increases.       3 pts.

Problem 4. (10 points) Find the general solution to each of the following diﬀerential equations:
a. y + 5y + 6y = 0

The characteristic equation is r 2 + 5r + 6 = 0 ⇒ (r + 3)(r + 2) = 0 ⇒ r = −3 or r = −3. 2 pts.

Therefore, y = c1 e−3x + c2 e−2x . 2 pts.

b. y (4) − 4y = 0

The characteristic equation is r 4 − 4r 2 = 0 ⇒ r 2 r 2 − 4 = 0 ⇒ r 2 (r + 2) (r − 2) = 0, which
gives r = 0 (double root), r = −2, and r = 2. 2 pts. Therefore, y = c1 e0x + c2 xe0x + c3 e−2x + c4 e2x,

or y = c1 + c2 x + c3 e−2x + c4 e2x . 4 pts.

Problem 5. (15 points)
Solve the following initial value problem:

y − 2y = −4x with y(0) = 0 and y (0) = 5.

Step 1. Find yc by solving the homogeneous d.e. y − 2y = 0.
Characteristic equation: r 2 − 2r = 0 ⇒ r(r − 2) = 0 ⇒ r = 0 or r = 2.
Therefore, yc = c1 e0x + c2 e2x = c1 + c2 e2x. 3 pts.
Step 2. Find yp using either the Method of Undetermined Coeﬃcients or the Method of Variation
of Parameters.
Method 1. Undetermined Coeﬃcients.
Since the nonhomogeneous term in the d.e. (−4x) is a polynomial of degree 1, we guess that yp
is a polynomial of degree 1: yp = Ax + B. The constant B in this guess duplicates the constant
term c1 in yc , so we need to multiply the guess by x: yp = x (Ax + B) = Ax2 + Bx. 3 pts.
y = Ax2 + Bx ⇒ y = 2Ax + B ⇒ y = 2A.
Therefore, the left side of the d.e. is
y − 2y = 2A − 2 (2Ax + B) = −4Ax + (2A − 2B). We want this to equal the nonhomogeneous
term −4x, so −4A = −4 and 2A − 2B = 0 ⇒ A = 1 and B = 1. Therefore, yp = x2 + x. 6 pts.
Method 2. Variation of Parameters.
From yc we obtain two independent solutions of the homogeneous d.e: y1 = 1 and y2 = e2x . The
y1 y2      1 e2x
Wronskian is given by W (x) =          =             = 1 2e2x − (0) e2x = 2e2x . 1 pt.
y1 y2      0 2e2x
−y2 f (x)         e2x (−4x)
u1 =              dx = −            dx = 2x dx = x2
W (x)               2e2x
3 pts.

y1 f (x)        1 (−4x)                           1              1              1
u2 =               dx =           dx = −2    xe−2x dx = −        ueu du = − (u − 1) eu = − (−2x − 1) e−2x =
W (x)             2e2x                            2              2              2
u=−2x        Integral Table #46
3 pts.
1 −2x 2x           1
Therefore, yp = u1 y1 + u2 y2 = x2 (1) +    x+     e   e = x2 + x +   2 pts.
2                  2
Step 3. y = yc + yp , so y = c1 + c2 e2x + x2 + x. 1 pt. (The constant 1/2 in the formula for yp
obtained by variation of parameters can be combined with the constant c1 in yc .)
Step 4. Use the intial conditions to ﬁnd c1 and c2 .
y = c1 + c2 e2x + x2 + x ⇒ y = 2c2 e2x + 2x + 1
y(0) = 0 ⇒ 0 = c1 + c2 e0 + 02 + 0 = c1 + c2 ⇒ c1 + c2 = 0
y (0) = 5 ⇒ 5 = 2c2 e0 + 2(0) + 1 = 2c2 + 1 ⇒ 2c2 = 4
Solving the system c1 + c2 = 0, 2c2 = 4, we ﬁnd that c1 = −2 and c2 = 2. 2 pts.

Therefore, y = −2 + 2e2x + x2 + x.

Problem 6. (15 points)
Consider a forced, undamped mass-spring system with mass m = 1 kg, spring constant k = 9 N/m,
and external force Fe (t) = 10 cos(2t) N. Find the position function x(t) if x(0) = 2 and x (0) = 3.

The d.e. modeling this system is mx + cx + kx = Fe (t), or x + 0x + 9x = 10 cos(2t). 2 pt.
Step 1. Find xc by solving the homogeneous d.e. x + 9x = 0. Characteristic equation: r 2 + 9 =
√
0 ⇒ r 2 = −9 ⇒ r = ± −9 = ±3i ⇒ xc = c1 cos(3t) + c2 sin(3t). 3 pts.
Step 2. Find xp using either the Method of Undetermined Coeﬃcients or the Method of Variation
of Parameters.
Method 1. Undetermined Coeﬃcients.
Since the nonhomogeneous term in the d.e. (10 cos(2t)) is a cosine, we guess that xp is a combination
of cosine and sine with the same angular frequency: xp = A cos(2t) + B sin(2t). 3 pts. No part
of this guess duplicates any term in xc , so there is no need to modify the guess. x = A cos(2t) +
B sin(2t) ⇒ x = −2A sin(2t) + 2B cos(2t) ⇒ x = −4A cos(2t) − 4B sin(2t).
Therefore, the left side of the d.e. is
x + 9x = −4A cos(2t) − 4B sin(2t) + 9 [A cos(2t) + B sin(2t)] A = 5A cos(2t) + 5B sin(2t). We want
this to equal the nonhomogeneous term 10 cos(2t), so 5A = 10 and 5B = 0 ⇒ A = 2 and B = 0.
Therefore, xp = 2 cos(2t). 4 pts.
Method 2. Variation of Parameters.
From xc we obtain two independent solutions of the homogeneous d.e: x1 = cos(3t) and x2 = sin(3t).
x1 x2
The Wronskian is given by W (t) =
x1 x2
cos(3t)      sin(3t)
=                         = cos(3t) (3 cos(3t)) − (−3 sin(3t)) sin(3t) = 3 cos2 (3t) + 3 sin2 (3t) = 3. 1 pt.
−3 sin(3t) 3 cos(3t)
−x2 Fe (t)         sin(3t) (10 cos(2t))        10
u1 =               dt = −                       dt = −       sin(3t) cos(2t) dt
W (t)                      3                  3
Int. Table # 31
10   cos ((3 − 2)t) cos ((3 + 2)t)  5        1
=−    −               −               = cos(t) + cos(5t) 2 pts.
3     2(3 − 2)       2(3 + 2)      3        3
x1 Fe (t)        cos(3t) (10 cos(2t))      10
u2 =                dt =                        dt =        cos(3t) cos(2t) dt
W (t)                     3                3
Int. Table # 30
10 sin ((3 − 2)t) sin ((3 + 2)t)        5         1
=                   +                  = sin(t) + sin(5t) 2 pts.
3     2(3 − 2)        2(3 + 2)         3         3
5          1                     5         1
Therefore, xp = u1 x1 + u2 x2 =     cos(t) + cos(5t) cos(3t) +       sin(t) + sin(5t) sin(3t)
3          3                     3         3
5                                   1
= [cos(3t) cos(t) + sin(3t) sin(t)] + [cos(5t) cos(3t) + sin(5t) sin(3t)]
3                                   3
5               1
= cos (3t − t) + cos (5t − 3t) = 2 cos(2t) 2 pts.
3               3
Step 3. x = xc + xp, so x = c1 cos(3t) + c2 sin(3t) + 2 cos(2t). 1 pt.
Step 4. Use the intial conditions to ﬁnd c1 and c2 .
x = c1 cos(3t) + c2 sin(3t) + 2 cos(2t) ⇒ x = −3c1 sin(3t) + 3c2 cos(3t) − 4 sin(2t)
x(0) = 2 ⇒ 2 = c1 cos(0) + c2 sin(0) + 2 cos(0) = c1 + 2 ⇒ c1 = 0
x (0) = 3 ⇒ 3 = −3c1 sin(0) + 3c2 cos(0) − 4 sin(0) = 3c2 ⇒ c2 = 1
2 pts.

Therefore, x = sin(3t) + 2 cos(2t).

Problem 7. (10 points)
a. (4 pts.) Find the Laplace transform of the function given by f (t) = t3 − e−t cos(2t).
n!                                     s−a
Using the table entries L {tn } =          and L eat cos(kt) =                        , we have
sn+1                                (s − a)2 + k2
3!             s − (−1)       6      s+1
L t3 − e−t cos(2t) = L t3 − L e−t cos(2t) =                     −             2      = 4−
s3+1         (s − (−1)) + 2 2   s  (s + 1)2 + 4
4 pts.

1
b. (6 pts.) Find the inverse Laplace transform of the function given by F (s) =                     .
(s − 2)3
n!
Using the table entry L tn eat =                  2 pts. we have
(s − a)n+1
1        1           2              1 2 2t
L−1               = L−1                   =     t e    2 pts.
(s − 2)3    2        (s − 2)3          2
2 pts.
Problem 8. (15 points)
Use the Laplace Transform to solve the following initial value problem:

x − 4x = 8t with x(0) = 1 and x (0) = 0.

Solutions to this IVP not using the Laplace transform method will not receive any
credit.

x − 4x = 8t ⇒ L x − 4x = L {8t} ⇒ L x − 4L{x} = 8L {t} 3 pts.
1
⇒ s2 L{x} − sx(0) − x (0) − 4L{x} = 8 2
s
2                             8
⇒ s L{x} − s · 1 − 0 − 4L{x} = 2 3 pts.
s
8        s3 + 8           s3 + 8
⇒ s2 − 4 L{x} = 2 + s =             ⇒ L{x} = 2 2      3 pts.
s          s2           s (s − 4)
s3 + 8
⇒ x = L−1                .
s2 (s2 − 4)
s3 + 8            s3 + 8       As + B    C   D
Use a partial fraction decomposition.            = 2                =        +    +
s2 (s2 − 4)   s (s + 2)(s − 2)     s2     s+2 s−2
⇒ s3 + 8 = (As + B) (s + 2)(s − 2) + Cs2 (s − 2) + Ds2 (s + 2)
⇒ s3 + 8 = (As + B) s2 − 4 + C s3 − 2s2 + D s3 + 2s2
⇒ s3 + 8 = As3 + Bs2 − 4As − 4B + Cs3 − 2Cs2 + Ds3 + 2Ds2
⇒ s3 + 8 = (A + C + D)s3 + (B − 2C + 2D)s2 − 4As − 4B
⇒ A + C + D = 1, B − 2C + 2D = 0, −4A = 0, and − 4B = 8 ⇒ A = 0, B = −2, C = 0, D = 1
−2     1               1            1
⇒ x = L−1     2
+        = −2L−1 2 + L−1
s    s−2              s          s−2

⇒ x = −2t + e2t . 6 pts.

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