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Page 1: Math 16B: Homework 4 Solutionsqadeer/teaching/Math16B/Homework 4... · Math 16B: Homework 4 Solutions Due: July 23 1. Verify that the given solution is a solution of the corresponding

Math 16B: Homework 4 Solutions

Due: July 23

1. Verify that the given solution is a solution of the corresponding differential equation:

(a) y = t + 1t

for ty′ = 2t− y:

Plugging in the solution into the left side of the differential equation gives

ty′ = t

(1− 1

t2

)= t− 1

t

Plugging it into the right side gives

2t− y = 2t−(t +

1

t

)= t− 1

t

Since both expressions are identical, y = t + 1t

is a solution of ty′ = 2t− y.

(b) y = sin(t) cos(t)− cos(t) for y′ + (tan(t))y = cos2(t):

Plugging in the solution into the left side of the differential equation gives

y′ + (tan(t))y = [cos(t)(cos(t)) + sin(t)(− sin(t)) + sin(t)] +

tan(t)(sin(t) cos(t)− cos(t))

= cos2(t)− sin2(t) + sin(t) + sin2(t)− sin(t)

= cos2(t)

which is the same as the right side of the differential equation. Thus, y =sin(t) cos(t)− cos(t) is a solution of y′ + (tan(t))y = cos2(t).

2. Sketch the slope field for the differential equation y′ = y(y − 1)(y + 1).

Note that:at y = 2, the slope is y′ = 6

at y = 1.5, the slope is y′ = 1.875

at y = 1, the slope is y′ = 0

at y = 0.5, the slope is y′ = −0.375

1

Page 2: Math 16B: Homework 4 Solutionsqadeer/teaching/Math16B/Homework 4... · Math 16B: Homework 4 Solutions Due: July 23 1. Verify that the given solution is a solution of the corresponding

at y = 0, the slope is y′ = 0

at y = −1, the slope is y′ = 0

at y = −1.5, the slope is y′ = −1.875

at y = −2, the slope is y′ = −6

The resulting slope field is shown in Figure 1.

Figure 1: Slope field for y′ = y(y − 1)(y + 1)

3. Solve the following differential equations using the appropriate technique. In the caseswith initial conditions, find the particular solutions.

(a) y′ = e2t+4y:

We use separation of variables. We have

dy

dt= e2t+4y

dy

dt= e2te4y∫

e−4y dy =

∫e2t dt

e−4y

−4=

e2t

2+ C

e−4y = −2e2t − 4C

−4y = ln(−2e2t − 4C)

y = −1

4ln(−2e2t − 4C)

2

Page 3: Math 16B: Homework 4 Solutionsqadeer/teaching/Math16B/Homework 4... · Math 16B: Homework 4 Solutions Due: July 23 1. Verify that the given solution is a solution of the corresponding

(b) xy′ + (5x + 1)y = e−5x:

We use the integrating factor method. The standard first-order linear form of theabove differential equation is

y′ +

(5x + 1

x

)y =

e−5x

x(1)

Then, a(x) = 5x+1x

= 5 + 1x

so

A(x) =

∫a(x) dx =

∫5 +

1

xdx = 5x + ln |x|.

Finally, the integrating factor is I(x) = eA(x) = e5x+ln |x| = e5xeln |x| = e5x|x|.Multiply (1) throughout by I(x) to get

e5x|x|y′ + (e5x|x|)(

5x + 1

x

)y = (e5x|x|)e

−5x

x

d

dx

(ye5x|x|

)= ±1

ye5x|x| =

∫±1dx

ye5x|x| = ±x + C

y =±x + C

e5x|x|

(c) ty′ = t2 sin(t) + y, t > 0:

We use the integrating factor method. The standard first-order linear form of theabove differential equation is

y′ −(

1

t

)y = t sin(t) (2)

Then, a(t) = −1t

so

A(t) =

∫a(t) dt =

−1

tdt = − ln(t).

Note that because t > 0, we do not need the absolute values inside the logarithmabove. Finally, the integrating factor is I(t) = eA(t) = e− ln(t) = 1

t. Multiply (2)

throughout by I(t) to get

1

ty′ − 1

t2y = sin(t)

d

dt

(1

ty

)= sin(t)

y

t=

∫sin(t) dt

y

t= − cos(t) + C

y = −t cos(t) + Ct

3

Page 4: Math 16B: Homework 4 Solutionsqadeer/teaching/Math16B/Homework 4... · Math 16B: Homework 4 Solutions Due: July 23 1. Verify that the given solution is a solution of the corresponding

(d) x2y′ = y − xy with initial condition y(−1) = −1:

Solution 1: We use separation of variables. We have

x2 dy

dx= y(1− x)∫

1

ydy =

∫1− x

x2dx∫

1

ydy =

∫1

x2− 1

xdx

ln |y| =−1

x− ln |x|+ C

|y| = e−1x−ln |x|+C

|y| = eCe−1x e− ln |x|

y = (±eC)(e−1x )

(−1

|x|

)y = −Ae−

1x

|x|

where A = ±eC is a real constant. Using the initial condition y(−1) = −1, wehave

−1 = −Ae−1

(−1)

| − 1|

1 =Ae1

1

A =1

e= e−1.

Thus, the specific solution of this initial value problem is

y = −e−1e−1x

|x|= −e−1−

1x

|x|

Solution 2: We use the integrating factor method. Writing the differential equa-tion in the standard first-order linear form,

y′ +

(x− 1

x2

)y = 0. (3)

Then, a(x) = x−1x2 so

A(x) =

∫x− 1

x2dx

=

∫1

x− 1

x2dx

= ln |x|+ 1

x

4

Page 5: Math 16B: Homework 4 Solutionsqadeer/teaching/Math16B/Homework 4... · Math 16B: Homework 4 Solutions Due: July 23 1. Verify that the given solution is a solution of the corresponding

Thus, the integrating factor is I(x) = eA(x) = eln |x|+1x = |x|e 1

x . Multiply (3) byI(x) to get

y′|x|e1x + |x|e

1x

(x− 1

x2

)y = 0

d

dx

(ye

1x |x|

)= 0

ye1x |x| =

∫0 dx

ye1x |x| = C

y =Ce−

1x

|x|where C is any real constant. This is the same general solution as in Solution 1above. Plugging in the initial condition therefore gives the same specific solution

y = −e−1−1x

|x|

(e) t ln(t)y′ + y = tet, t > e, with initial condition y(e2) = 0:

We use the integrating factor method. The standard first order linear form of thedifferential equation is

y′ +1

t ln(t)y =

et

ln(t)(4)

Then, a(t) = 1t ln(t)

. To integrate this and determine A(t), we use the substitution

u = ln(t)⇒ dudt

= 1t⇒ du = 1

tdt. Thus,∫

1

t ln(t)dt =

∫1

udu

= ln |u|= ln | ln(t)|

Note that because we are taking t > e, we have ln(t) > ln(e) = 1 > 0 soactually we can do away with the absolute value bars inside the logarithm to getA(t) = ln(ln(t)). Thus, the integrating factor is I(t) = eA(t) = eln(ln(t)) = ln(t).Multiply (4) throughout by I(t) to get

y′ ln(t) +1

ty = et

d

dt(y ln(t)) = et

y ln(t) =

∫et dt

y ln(t) = et + C

y =et + C

ln(t).

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Page 6: Math 16B: Homework 4 Solutionsqadeer/teaching/Math16B/Homework 4... · Math 16B: Homework 4 Solutions Due: July 23 1. Verify that the given solution is a solution of the corresponding

Finally, use the initial condition y(e2) = 0 to get

0 =ee

2+ C

ln(e2)

0 =ee

2+ C

2

C = −ee2

The specific solution therefore is

y =et − ee

2

ln(t)

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