Advanced Engineering Mathematics with MATLAB 6th Edition – Solutions Manual & Step-by-Step Problem Solutions | Dean G. Duffy

Worked Solutions
for
Classic Engineering Mathematics



Dean G. Duffy




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,Section 1.1

1. first-order, linear 2. first-order, nonlinear
3. first-order, nonlinear 4. third-order, linear
5. second-order, linear 6. first-order, nonlinear
7. third-order, nonlinear 8. second-order, linear
9. second-order, nonlinear 10. first-order, nonlinear
11. first-order, nonlinear 12. second-order, nonlinear
13. first-order, nonlinear 14. third-order, linear
15. second-order, nonlinear 16. third-order, nonlinear

Section 1.2

1. Because the differential equation can be rewritten e−y dy = x dx, integration immediately
gives −e−y = 12 x2 − C, or y = − ln(C − x2 /2).

2. Separating variables, we have that y 2 dy = cos(x) dx. Integrating this equation, we find
that y 3 /3 = sin(x) + C

3. The differential equation can be rewritten ey dy = 12x3 dx. Integration immediately
gives ey = 3x4 + C or y(x) = ln(3x4 + C).

4. Because the differential equation can be rewritten dy/y = 2x dx/(1 + x2 ), integration
immediately gives ln(y) − ln(C) = ln(1 + x2 ). Raising both sides of the equation to the
power e, we get the final result: y(x) = C(1 + x2 ).

5. Separating variables, we have that dx/(1 + x2 ) = dy/(1 + y 2 ). Integrating this equation,
we find that tan−1 (x) − tan−1 (y) = tan(C), or (x − y)/(1 + xy) = C.

6. Because the differential equation can be rewritten ln(x)dx/x = y dy, integration imme-
diately gives 12 ln2 (x) + C = 21 y 2 , or y 2 (x) − ln2 (x) = 2C.

7. Because the differential equation can be rewritten y 2 dy = (x + x3 ) dx, integration
immediately gives y 3 (x)/3 = x2 /2 + x4 /4 + C.

8. Because the differential equation can be rewritten y dy/(2 + y 2 ) = x dx/(1 + x2 ), inte-
gration immediately gives 21 ln(2 + y 2 ) = 12 ln(1 + x2 ) + 21 ln(C), or 2 + y 2 (x) = C(1 + x2 ).

9. Because the differential equation can be rewritten dy/y 1/3 = x1/3 dx, integration imme-

3/2
diately gives 23 y 2/3 = 34 x4/3 + 23 C, or y(x) = 21 x4/3 + C .

10. Because the differential equation can be rewritten e−y dy = ex dx, integration immedi-
ately gives −e−y = ex − C, or y(x) = − ln(C − ex ).

11. Because the differential equation can be rewritten dy/(y 2 + 1) = (x3 + 5)

dx, integration
immediately gives tan−1 (y) = 14 x4 + 5x + C, or y(x) = tan 41 x4 + 5x + C .

dx
12. Because the differential equation can be rewritten sin(y) cos(y) dy = x , integration
immediately gives 21 sin2 (y) = ln(x) + C/2, or sin2 (y) = 2 ln(x) + C.

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, 13. Because the differential equation can be rewritten cos(θ) dθ = k(a − x) dx, integration
immediately gives sin(θ) = kax − kx2 /2 + C.

14. Because the differential equation can be rewritten v dv = −µ dr/r2 , integration imme-
diately gives 12 v 2 = µ/r + C.

15. Because the differential equation can be rewritten −dy/(M − y) = −k dx, integration
immediately gives ln(M − y) = −kx + ln(C) or y = M − Ce−kx .

16. Because the differential equation can be rewritten y 2 dy/(b − ay 3 ) = dt, integration
y
immediately gives ln[b − ay 3 ] y = −3at, or (ay 3 − b)/(ay03 − b) = e−3at .
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17. Because the differential equation can be written du/u = dx/x2 , integration immediately
gives u = Ce−1/x or y(x) = x + Ce−1/x .

18. The governing differential equations is dT /dt = k[T (t)−Tm ], where T (t) is the tempera-
ture (in ◦ C), k is a proportionality constant, t is time in minutes, and Tm is the temperature
of the environment. Separation of variables gives the solution as T (t)−Tm = Kekt , where K
is the arbitrary constant of integration. Because T (0) = 100, K = 75. Since T (10) = 40◦ C,
k = −0.161/minute. Solving for the time when T (t) = 35◦ C, we find t = 12.5 minutes.
Therefore, it takes an additional 2.5 minutes to cool from 40◦ C to 35◦ C.

19. The governing differential equations is dT /dt = k[T (t) − Tm ], where T (t) is the tem-
perature (in ◦ F), k is a proportionality constant, t is time in minutes, and Tm is the tem-
perature of the environment. Separation of variables gives the solution T (t) − Tm = Kekt ,
where K is the arbitrary constant of integration. Because T (0) = 70◦ F, K = 60◦ F. Since
T (1) = 50◦ F, k = −0.182322/minute. Solving for the temperature at t = 2 minutes, we find
T (2) = 51.67◦ F. We next find the time at which T (t) = 30◦ F and find t = 6.025 minutes.

20. The governing differential equations is dT /dt = k[T (t)−Tm ], where T (t) is the tempera-
ture (in ◦ C), k is a proportionality constant, t is time in minutes, and Tm is the temperature
of the environment, 15.6◦ C. Separation of variables gives the solution T (t) − Tm = Kekt ,
where K is the arbitrary constant of integration. Because T (0) = 37.8◦ C, K = 22.2◦ C.
Since T (10) = 32.2◦ C, 10k = −0.2906896/minute. Now we want to find that time t
where T (t) = 26.7◦ C. Solving for t in our solution, we find that this temperature oc-
curs at t = 23.845 minutes. Therefore, it takes an additional 13.845 minutes to reach that
temperature after we reached a temperature of 32.2◦ C.

21. Because T (t) − Tm = Cekt , Tm = 72◦ F and T (0) = 82◦ F, C = 10◦ F. Furthermore, since
T (1) = 82◦ F, we find that k = − ln(0.8) = 0.223144. Finally, 98.6 − 72 = 26.6 = 10e−kt or
t = −4.38. Thus, death occurred at 7 : 37 am.

22. Because T (t) − Toutdoor = Cekt , T (0) − Toutdoor = 22 − Toutdoor = C, T (1) − Toutdoor =
26 − Toutdoor = Cek , and T (2) − Toutdoor = 28 − Toutdoor = Ce2k . Solving for Toutdoor , we
find that Toutdoor = 30◦ C.

23. Integrating the population growth equation dP/dt = kP , the equation describing the
population P (t) is P (t) = P0 ekt , where P0 is initial population of the sample (106 ). Since
the population grows to 1.5 million in 1 hour, k = ln[P (1 hr)/P0 ]/1 hr = ln(1.5)/hr =

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