Math 308 Exam I Practice Problems - Texas A&M Universityglahodny/Math308/Spring2018/Exam 1...

Math 308 Exam I Practice Problems

This review should not be used as your sole source for preparation for the exam. Youshould also re-work all examples given in lecture and all suggested homework problems.

1. Find the general solution of the differential equation

y′ + 3y = t+ e−2t

and use it to determine how solutions behave as t→∞.

2. Solve the initial value problem

dy

dt− y

t=

1

t+ 1, y(1) = 1.


y′ + y2 sinx = 0.


y′ = (cos2 x)(cos2 2y).

5. Find the solution of the initial value problem

dy

dx=y2

x, y(1) = 2

in explicit form and determine the interval in which the solution is defined.

6. A tank initially contains 50 lb of salt dissolved in 100 gal of water. Water containing0.25 lb of salt per gallon is poured into the tank at a rate of 3 gal/min, and the mixturedrains from the tank at the same rate. Find the amount of salt in the tank after a longtime. Find the time after which the salt level is within 1% of this limiting value.

7. Newton’s Law of Cooling states that the temperature of an object changes at a rateproportional to the difference between its temperature and that of its surroundings.Suppose that an object takes 40 minutes to cool from 30◦C to 24◦C in a room that iskept at 20◦C. How long will it take the object to cool down to 21◦C?

8. Suppose an object is launched straight up from the ground with initial velocity v0.Ignoring the effects of air resistance, determine the time at which the object willreturn to the ground. (Your answer should be in terms of v0 and the accelerationdue to gravity at the earth’s surface, g.)

9. Determine (without solving the problem) an interval in which the solution of the initialvalue problem

(t2 − 3t)y′ + 2ty = ln t, y(4) = 1

is certain to exist.

1

10. Consider the initial value problem

dy

dt=

3t2 + 4t+ 2

2(y − 1), y(0) = −1.

State where in the ty-plane a unique solution is certain to exist.


y′ + y3 = 0, y(0) = y0

and determine how the interval in which the solution exists depends on the value y0.

12. Consider the differential equation

y′ = y(1− y2).

Find all equilibrium solutions and determine their stability.


y′ = y2(1− y)2.


14. Consider the single-species population model

dN

dt= 2N(N − 150)

(1− N

400

),

with initial condition N(0) = N0 ≥ 0.

(a) Find all equilibrium solutions of this differential equation and determine theirstability.

(b) If N0 = 100, find limt→∞

N(t).

(c) If N0 = 200, find limt→∞

N(t).

(d) If N0 = 500, find limt→∞

N(t).


(y cosx+ 2xey) + (sin x+ x2ey − 1)y′ = 0.

16. Find an integrating factor for the differential equation

(x+ 2) sin y + (x cos y)y′ = 0.

Find the general solution of the differential equation.

2


y′′ + y′ − 2y = 0, y(0) = 1, y′(0) = 4

and describe how the solution behaves as t→∞.

18. Determine the longest interval in which the initial value problem

(t2 − 3t)y′′ + ty′ − (t+ 3)y = 0, y(1) = 2, y′(1) = 1

is certain to have a unique twice-differentiable solution.

19. Verify that the functions y1(t) = t and y2(t) = tet are solutions of the differentialequation

t2y′′ − t(t+ 2)y′ + (t+ 2)y = 0, t > 0.

Do they form a fundamental set of solutions?

20. Find the Wronskian of two solutions of Legendre’s equation

(1− x2)y′′ − 2xy′ + α(α + 1)y = 0

without solving the equation.


y′′ − 4y′ + 3y = 0, y(0) = 1, y′(0) = 1/3.

Describe the behavior of the solution as t→∞.


y′′ + 2y′ + y = 0, y(0) = 1, y′(0) = −3.



y′′ + 2y′ + 5y = 0, y(0) = 2, y′(0) = 1.


24. Verify that y1(t) = et is a solution of

(t− 1)y′′ − ty′ + y = 0, t > 1.

Use the method of reduction of order to find a fundamental set of solutions.

3

Solutions

Solutions may contain errors or typos. If you find an error or typo, please notify me [email protected].


y′ + 3y = t+ e−2t

and use it to determine how solutions behave as t→∞.

To solve, we first compute the integrating factor

µ(t) = exp

(∫3 dt

)= e3t.

Multiplying by µ(t), we obtain

e3ty′ + 3e3ty = te3t + et

d

dt(e3ty) = te3t + et.

Integrating both sides of this equation, we have

e3ty =

∫(te3t + et) dt

e3ty =1

3te3t − 1

9e3t + et + C.

Therefore, the general solution is

y(t) =1

3t− 1

9+ e−2t + Ce−3t.

As t→∞, solutions diverge to ∞.

4


dy

dt− y

t=

1

t+ 1, y(1) = 1.

To solve, we first compute the integrating factor

µ(t) = exp

(∫−1

tdt

)= e− ln t =

1

t.

Multiplying by µ(t), we obtain

1

ty′ − 1

t2y =

1

t(t+ 1)

d

dt

(yt

)=

1

t(t+ 1).

Integrating both sides of this equation, we have

y

t=

∫1

t(t+ 1)dt

y

t=

∫ (1

t− 1

t+ 1

)dt

y

t= ln(t)− ln(t+ 1) + C

y

t= ln

(t

t+ 1

)+ C


y(t) = t ln

(t

t+ 1

)+ Ct.

Applying the initial condition, we obtain

y(1) = ln1

2+ C = 1.

It follows that C = 1 + ln 2, and so

y(t) = t ln

(t

t+ 1

)+ (1 + ln 2)t.

5


y′ + y2 sinx = 0.

Using separation of variables, we obtain

dy

dx= −y2 sinx

−dyy2

= sin x dx

−∫

1

y2dy =

∫sinx dx

1

y= − cosx+ C.

Therefore, the general solution (in explicit form) is

y(x) =1

C − cosx.


y′ = (cos2 x)(cos2(2y)).


dy

dx= (cos2 x)(cos2(2y))

dy

cos2(2y)= (cos2 x) dx

(sec2(2y)) dy = (cos2 x) dx∫sec2(2y) dy =

∫cos2 x dx∫

sec2(2y) dy =1

2

∫(1 + cos(2x)) dx

1

2tan(2y) =

x

2+

1

4sin(2x) + C

1

2tan(2y)− x

2− 1

4sin(2x) = C

2 tan(2y)− 2x− sin(2x) = C.

The solution is defined implicitly.

6

5. Find the solution of the initial value problem

dy

dt=y2

t, y(1) = 2

in explicit form and determine the interval in which the solution is defined.


dy

y2=

dt

t

−1

y= ln t+ C

y(t) =1

C − ln t.


y(1) =1

C= 2.

It follows that C = 1/2, and so

y(t) =1

1/2− ln t=

2

1− 2 ln t.

The solution is defined if t > 0 and

0 < 1− 2 ln t

2 ln t < 1

ln t <1

2t < e1/2.

Thus, the solution is defined in the interval 0 < t <√e.

7

6. A tank initially contains 50 lb of salt dissolved in 100 gal of water. Water containing0.25 lb of salt per gallon is poured into the tank at a rate of 3 gal/min, and the mixturedrains from the tank at the same rate. Find the amount of salt in the tank after a longtime. Find the time after which the salt level is within 1% of this limiting value.

Let y(t) denote the amount (in lb) of salt in the tank at time t ≥ 0. Thus,

dy

dt= (0.25 lb/gal)(3 gal/min)−

( y

100lb/gal

)(3 gal/min)

dy

dt= 3

(0.25− y

100

)lb/min

dy

dt=

3

100(25− y).

Therefore, we have the initial value problem

dy

dt=

3

100(25− y), y(0) = 50.


dy

25− y=

3

100dt∫

1

25− ydy =

∫3

100dt

− ln |25− y| = 0.03t+ C

25− y = Ce−0.03t

y(t) = 25− Ce−0.03t.


y(0) = 25− C = 50.

Thus, C = −25 and the solution of the initial value problem is

y(t) = 25(1 + e−0.03t).

As t→∞, the the amount of salt in the tank approaches a limiting value of 25 lb. Tofind the time at which the salt level is within 1% of this limiting value, let

25(1 + e−0.03t) = (1.01)(25)

1 + e−0.03t = 1.01

e−0.03t = 0.01

−0.03t = ln(0.01)

t =ln(0.01)

−0.03t ≈ 153.5 min

t ≈ 2 hr 36 min.

Note: On the exam, you will not need to convert the unit of time.

8

7. Newton’s Law of Cooling states that the temperature of an object changes at a rateproportional to the difference between its temperature and that of its surroundings.Suppose that an object takes 40 minutes to cool from 30◦C to 24◦C in a room that iskept at 20◦C. How long will it take the object to cool down to 21◦C?

Let y(t) denote the temperature (in ◦C) at time t ≥ 0. By Newton’s Law of Cooling,

dy

dt= k(20− y), y(0) = 30.


dy

20− y= k dt∫

1

20− ydy =

∫k dt

− ln |20− y| = kt+ C

20− y = Ce−kt

y(t) = 20− Ce−kt.Applying the initial condition, we obtain

y(0) = 20− C = 30.

Thus, C = −10 and the solution of the initial value problem is

y(t) = 20 + 10e−kt.

We are given that y(40) = 24, so

20 + 10e−40k = 24

10e−40k = 4

e−40k = 0.4

−40k = ln(0.4)

k =ln(0.4)

−40.

Therefore,y(t) = 20 + 10eln(0.4)t/40.

To find the time at which the object reaches a temperature of 21◦C, let

20 + 10eln(0.4)t/40 = 21

10eln(0.4)t/40 = 1

eln(0.4)t/40 = 0.1ln(0.4)

40t = ln(0.1)

t =40 ln(0.1)

ln(0.4)t ≈ 100.5 min

t ≈ 1 h 41 min.

9

8. Suppose an object is launched straight up from the ground with initial velocity v0.Ignoring the effects of air resistance, determine the time at which the object will re-turn to the ground. (Your answer should be in terms of v0 and the acceleration due togravity at the earth’s surface, g.)

Let v(t) denote the velocity (in m/s) of the object at time t ≥ 0, and assume that apositive velocity corresponds to movement upward. By Newton’s Second Law,

F = ma = mdv

dt.

Ignoring air resistance, the only force acting on the object is the force due to gravity.That is,

F = −mg.

Therefore,

mdv

dt= −mg

dv

dt= −g.

Integrating with respect to t, the velocity of the object at time t is given by

v(t) = −gt+ v0.

Integrating with respect to t again, the height of the object above the ground (in m)at time t is given by

h(t) = −g2t2 + v0t.

To determine the time at which the object will return to the ground, let

−g2t2 + v0t = 0

t(v0 −

g

2t)

= 0

t =2v0g

s

Note: We take the positive value of t since t = 0 corresponds to the time at which theball was initially launced upward.

10

9. Determine (without solving the problem) an interval in which the solution of the initialvalue problem

(t2 − 3t)y′ + 2ty = ln t, y(4) = 1

is certain to exist.

In standard form, we have

y′ +2t

t2 − 3ty =

ln t

t2 − 3t.

So p(t) = 2t/(t2− 3t) and g(t) = ln t/(t2− 3t). Thus, for this equation, p is continuousfor t 6= 0, 3 and g is continuous for t > 0 and t 6= 3. The interval containing theinitial point t0 = 4 is (3,∞). By Theorem 2.4.1, the initial value problem has a uniquesolution on the interval (3,∞).

10. Consider the initial value problem

dy

dt=

3t2 + 4t+ 2

2(y − 1), y(0) = −1.

State where in the ty-plane a unique solution is certain to exist.

For this equation,

f(t, y) =3t2 + 4t+ 2

2(y − 1),

∂f

∂y(t, y) = −3t2 + 4t+ 2

2(y − 1)2.

Each of these functions is continuous for y 6= 1. That is, f and ∂f/∂y are continuouson the set

{(t, y)|y 6= 1}.

Since y(0) = −1, we consider the region in the ty-plane defined by

{(t, y)|y < 1}.

By Theorem 2.4.2, the initial value problem has a unique solution within the set{(t, y)|y < 1}.

11


y′ + y3 = 0, y(0) = y0

and determine how the interval in which the solution exists depends on the value y0.


dy

dt= −y3

−dyy3

= dt

−∫

1

y3dy =

∫1 dt

1

2y2= t+ C

1

y2= 2t+ C

y2 =1

2t+ C.


y20 =1

C.

It follows that C = 1/y20 and so

y2 =1

2t+ 1/y20

y2 =y20

2y20t+ 1

y =y0√

2y20t+ 1.

This solution is only valid if y0 6= 0. Therefore, if y0 6= 0, the solution exists if andonly if

2y20t+ 1 > 0

t > − 1

2y20.

If y0 = 0, then y′ = 0 and y(t) = 0 for all t ∈ R.

12


y′ = y(1− y2).


To find the equilibrium solutions, let

y′ = y(1− y2) = 0.

Thus, the equilibrium solutions are y = −1, 0, 1. To determine their stability, sketchthe graph of y′ versus y.

If y < −1, then solutions are increasing. If −1 < y < 0, then solutions are decreasing.If 0 < y < 1, then solutions are increasing. If y > 1, then solutions are decreas-ing. Therefore, y = −1 is asymptotically stable, y = 0 is unstable, and y = 1 isasymptotically stable.

13


y′ = y2(1− y)2.



y′ = y2(1− y)2 = 0.

Thus, the equilibrium solutions are y = 0, 1. To determine their stability, sketch thegraph of y′ versus y.

If y < 0, then solutions are increasing. If 0 < y < 1, then solutions are increasing. Ify > 1, then solutions are increasing. Therefore, y = 0 and y = 1 are both semistable.

14

14. Consider the single-species population model

dN

dt= 2N(N − 150)

(1− N

400

),

with initial condition N(0) = N0 ≥ 0.

(a) Find all equilibrium solutions of this differential equation and determine theirstability.


dN

dt= 2N(N − 150)

(1− N

400

)= 0.

Thus, the equilibrium solutions are N = 0, 150, 400. To determine their stability,sketch the graph of dN/dt versus N .

If 0 < N < 150, then solutions are decreasing. If 150 < N < 400, then solutionsare increasing. If N > 400, then solutions are decreasing. Therefore, N = 0 isasymptotically stable, y = 150 is unstable, and y = 400 is asymptotically stable.

(b) If N0 = 100, find limt→∞

N(t).

If N0 = 100, then N(t)→ 0.

(c) If N0 = 200, find limt→∞

N(t).

If N0 = 200, then N(t)→ 400.

(d) If N0 = 500, find limt→∞

N(t).

If N0 = 500, then N(t)→ 400.

15


(y cosx+ 2xey) + (sin x+ x2ey − 1)y′ = 0.

Let M(x, y) = y cosx+ 2xey and N(x, y) = sinx+ x2ey − 1. Thus,

My(x, y) = cos x+ 2xey = Nx(x, y).

Therefore, the equation is exact and there exists a function ψ(x, y) such that

ψx(x, y) = y cosx+ 2xey ψy(x, y) = sinx+ x2ey − 1.

Integrating the first equation with respect to x, we obtain

ψ(x, y) =

∫(y cosx+ 2xey) dx = y sinx+ x2ey + f(y).

Differentiating with respect to y, we obtain

ψy(x, y) = sinx+ x2ey + f ′(y).

Therefore,

sinx+ x2ey + f ′(y) = sinx+ x2ey − 1

f ′(y) = −1

f(y) = −y.

Then ψ(x, y) = y sinx+ x2ey − y and solutions are defined implicitly by

y sinx+ x2ey − y = C.

16

16. Find an integrating factor for the differential equation

(x+ 2) sin y + (x cos y)y′ = 0.

Find the general solution of the differential equation.

Upon calculating (My −Nx)/N , we obtain

My −Nx

N=

(x+ 2) cos y − cos y

x cos y=

(x+ 1) cos y

x cos y=x+ 1

x= 1 +

1

x.

Thus, there is an integrating factor given by

µ(x) = exp

[∫ (1 +

1

x

)dx

]= ex+lnx = xex.

Multiplying the equation by this integrating factor, we obtain

(x2 + 2x)ex sin y + (x2ex cos y)y′ = 0.

Since this equation is exact, there exists a function ψ(x, y) such that

ψx(x, y) = (x2 + 2x)ex sin y ψy(x, y) = (x2ex cos y).

Integrating the second equation with respect to y, we have

ψ(x, y) =

∫(x2ex cos y) dy = x2ex sin y + f(x).

Differentiating with respect to x, we obtain

ψx(x, y) = (2x+ x2)ex sin y + f ′(x).

Therefore,

(2x+ x2)ex sin y + f ′(x) = (x2 + 2x)ex sin y

f ′(x) = 0

f(x) = C = 0.

Then ψ(x, y) = x2ex sin y and solutions are defined implicitly by

x2ex sin y = C.

17


y′′ + y′ − 2y = 0, y(0) = 1, y′(0) = 4

and describe how the solution behaves as t→∞.

The characteristic equation is

r2 + r − 2 = 0

(r + 2)(r − 1) = 0

r = −2, 1.

Thus, the general solution (and its derivative) are

y(t) = c1e−2t + c2e

t,

y′(t) = −2c1e−2t + c2e

t.

Applying the initial conditions, we obtain

y(0) = c1 + c2 = 1,

y′(0) = −2c1 + c2 = 4.

It follows that c1 = −1 and c2 = 2. Therefore, the solution is

y(t) = −e−2t + 2et.

As t→∞, the solution diverges to ∞.

18. Determine the longest interval in which the initial value problem

(t2 − 3t)y′′ + ty′ − (t+ 3)y = 0, y(1) = 2, y′(1) = 1

is certain to have a unique twice-differentiable solution.


y′′ +

(1

t− 3

)y′ −

(t+ 3

t2 − 3t

)y = 0.

So p(t) = 1/(t− 3), q(t) = −(t+ 3)/(t2 − 3t), and g(t) = 0 are continuous for t 6= 0, 3.Therefore, the longest open interval containing t0 = 1 in which the coefficients arecontinuous is (0, 3). By Theorem 3.2.1, a unique solution exists in (0, 3).

18

19. Verify that the functions y1(t) = t and y2(t) = tet are solutions of the differentialequation

t2y′′ − t(t+ 2)y′ + (t+ 2)y = 0, t > 0.

Do they form a fundamental set of solutions?

If y1(t) = t, then y′1(t) = 1 and y′′1(t) = 0. So

t2y′′1 − t(t+ 2)y′1 + (t+ 2)y1 = t2(0)− t(t+ 2)(1) + (t+ 2)(t) = 0.

If y2(t) = tet, then y′2(t) = (1 + t)et and y′′2(t) = (2 + t)et. So

t2y′′2 − t(t+ 2)y′2 + (t+ 2)y2 = t2(2 + t)et − t(t+ 2)(1 + t)et + (t+ 2)tet = 0.

Thus, y1 and y2 are solutions of the differential equation. The Wronskian of y1 and y2is

W (y1, y2)(t) =

∣∣∣∣ t tet

1 (1 + t)et

∣∣∣∣ = (t+ t2)et − tet = t2et.

Since W 6= 0 for t > 0, y1 and y2 form a fundamental set of solutions.

20. Find the Wronskian of two solutions of Legendre’s equation

(1− x2)y′′ − 2xy′ + α(α + 1)y = 0

without solving the equation.


y′′ −(

2x

1− x2

)y′ +

(α(α + 1)

1− x2

)y = 0.

By Abel’s Theorem, the Wronskian of two solutions is

W (y1, y2)(x) = c exp

[∫2x

1− x2dx

]= ce− ln(1−x2) =

c

1− x2.

Note: To evaluate the integral above, we use a substitution of u = 1− x2.

19


y′′ − 4y′ + 3y = 0, y(0) = 1, y′(0) = 1/3.


If y = ert, then we obtain the characteristic equation

r2 − 4r + 3 = (r − 1)(r − 3) = 0.

The roots of the characteristic equation are r = 1 and r = 3. Therefore, the generalsolution is

y(t) = c1et + c2e

3t.

Differentiating, we obtainy′(t) = c1e

t + 3c2e3t.

Applying the initial conditions, we obtain the following equations

y(0) = c1 + c2 = 1

y′(0) = c1 + 3c2 = 1/3.

It follows that c1 = 4/3 and c2 = −1/3, so the solution of the initial value problem is

y(t) =4

3et − 1

3e3t.

As t→∞, the solution diverges to −∞.

0 0.5 1 1.5 2

t

-140

-120

-100

-80

-60

-40

-20

0

20

y

20


y′′ + 2y′ + y = 0, y(0) = 1, y′(0) = −3.



r2 + 2r + 1 = (r + 1)2 = 0.

The roots of the characteristic equation are r1 = r2 = −1. Therefore, the generalsolution is

y(t) = c1e−t + c2te

−t.

Differentiating, we obtain

y′(t) = −c1et + (1− t)c2e−t.


y(0) = c1 = 1

y′(0) = −c1 + c2 = −3.

It follows that c1 = 1 and c2 = −2, so the solution of the initial value problem is

y(t) = e−t − 2te−t = (1− 2t)e−t.

As t→∞, the solution converges to 0.

0 2 4 6 8 10

t

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

y

21


y′′ + 2y′ + 5y = 0, y(0) = 2, y′(0) = 1.



r2 + 2r + 5 = 0.

The roots of the characteristic equation are

r =1

2(−2±

√4− 20) = −1± 2i.


y(t) = c1e−t cos(2t) + c2e

−t sin(2t).

Differentiating, we obtain

y′(t) = −c1e−t cos(2t)− 2c1e−t sin(2t)− c2e−t sin(2t) + 2c2e

−t cos(2t).


y(0) = c1 = 2

y′(0) = −c1 + 2c2 = 1.

It follows that c1 = 2 and c2 = 3/2, so the solution of the initial value problem is

y(t) = 2e−t cos(2t) +3

2e−t sin(2t).

As t→∞, the solution exhibits a decaying oscillation approaching zero.

0 1 2 3 4 5

t

-0.5

0

0.5

1

1.5

2

2.5

y

22

24. Verify that y1(t) = et is a solution of

(t− 1)y′′ − ty′ + y = 0, t > 1.

Use the method of reduction of order to find a fundamental set of solutions.

If y1(t) = et, then y′1(t) = y′′1(t) = et. Then

(t− 1)y′′1 − ty′1 + y1 = (t− 1)et − tet + et = 0.

Thus, y1(t) = et is a solution of the differential equation.

Setting y(t) = v(t)et, then

y′(t) = v′et + vet,

y′′(t) = v′′et + 2v′et + vet.

Substituting y, y′, and y′′ in the differential equation and collecting terms, we obtain

(t− 1)y′′ − ty′ + y = (t− 1)(v′′et + 2v′et + vet)− t(v′et + vet) + vet

= (t− 1)etv′′ + (t− 2)etv′ + ((t− 1)et − tet + et)v

= (t− 1)etv′′ + (t− 2)etv′ = 0.

Note that the coefficient of v is zero, as it should be.

Setting w(t) = v′(t), then

(t− 1)etw′ + (t− 2)etw = 0.


dw

w=

2− tt− 1

dt∫1

wdw =

∫ (1

t− 1− 1

)dt

ln |w| = ln |t− 1| − t+ C

w(t) = C(t− 1)e−t.

Then

v(t) =

∫C(t− 1)e−t dt = −Cte−t +K.

It follows thaty(t) = v(t)et = −Ct+Ket,

where C and K are arbitrary constants. The second term is a multiple of y1(t) = et andcan be dropped, but the first term produces a new solution y2(t) = t. The Wronskianof y1 and y2 is

W (y1, y2)(t) =

∣∣∣∣ et tet 1

∣∣∣∣ = (1− t)et 6= 0 for t > 1.

Therefore, y1(t) = et and y2(t) = t form a fundamental set of solutions of the differentialequation for t > 1.

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