question 1 - university of notre dameapilking/math10560/work/old exams... · 2020. 3. 16. ·...

Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 Question 11 Question 12 Question 13 Question 14

Question 1

1. Evaluate the improper integralZ ∞4

(x − 2)(x − 3)dx .

I Using a partial fraction expansionR1

(x−2)(x−3)dx =

x−3− 1

idx = ln | x−3

x−2|+ C .

I ThereforeR∞

(x−2)(x−3)dx = limt→∞ ln | t−3

t−2| − ln | 1

limt→∞ | t−3t−2|i

+ ln 2 = ln(1) + ln 2 = 0 + ln 2.

Annette Pilkington Improper Integrals

Question 1

(x − 2)(x − 3)dx .

(x−2)(x−3)dx =

x−3− 1

idx = ln | x−3

x−2|+ C .

I ThereforeR∞

(x−2)(x−3)dx = limt→∞ ln | t−3

t−2| − ln | 1

+ ln 2 = ln(1) + ln 2 = 0 + ln 2.

Question 1

(x − 2)(x − 3)dx .

(x−2)(x−3)dx =

x−3− 1

idx = ln | x−3

x−2|+ C .

I ThereforeR∞

(x−2)(x−3)dx = limt→∞ ln | t−3

t−2| − ln | 1

+ ln 2 = ln(1) + ln 2 = 0 + ln 2.

Question 2

2.What can be said about the integrals

x2dx ;

Z ∞1

cos2 x

I Integral (i) diverges by the Comparison Theorem since the integrand isgreater than 1

I Integral (ii) converges by the Comparison Theorem since the integrand isless than 1

I Therefore (i) diverges and (ii) converges.

Question 2

x2dx ;

Z ∞1

cos2 x

Question 2

x2dx ;

Z ∞1

cos2 x

Question 2

x2dx ;

Z ∞1

cos2 x

Question 3

3. Which of the following is an expression for the arclength of the curvey = cos x between x = −π

2and x = π

−π2

p1 + sin2 x dx . (b)

−π2

p1− cos2 x dx .

p1 + 2 sin2 x dx . (d)

−π2

p1− sin2 x dx .

(e) π2

I The arclength formula gives the answer asR b

p1 + (f ′(x))2dx .

I =R π

2−π

p1 + (−sinx)2dx =

R π2−π

p1 + sin2 x dx .

Question 3

2and x = π

−π2

p1− cos2 x dx .

p1 + 2 sin2 x dx . (d)

−π2

p1− sin2 x dx .

(e) π2

p1 + (f ′(x))2dx .

I =R π

2−π

p1 + (−sinx)2dx =

R π2−π

p1 + sin2 x dx .

Question 3

2and x = π

−π2

p1− cos2 x dx .

p1 + 2 sin2 x dx . (d)

−π2

p1− sin2 x dx .

(e) π2

p1 + (f ′(x))2dx .

I =R π

2−π

p1 + (−sinx)2dx =

R π2−π

p1 + sin2 x dx .

Question 4

4. Use Euler’s method with step size 0.1 to estimate y(1.2) where y(x) is thesolution to the initial value problem

y ′ = xy + 1 y(1) = 0.

I x0 = 1, y0 = 0

I x1 = x0 + h = 1.1, y1 = y0 + h(x0y0 + 1) = 0 + (0.1)(1 · 0 + 1) = 0.1

I x2 = x1 + h = 1.2, y2 = y1 + h(x1y1 + 1) = 0.1 + (0.1)((1.1) · (0.1) + 1)

I = 0.1 + 0.1(0.11 + 1) = 0.1 + 0.1(1.11) = 0.1 + 0.111 = 0.211

I y(1.2) ≈ 0.211.

Question 4

y ′ = xy + 1 y(1) = 0.

I x0 = 1, y0 = 0

I x1 = x0 + h = 1.1, y1 = y0 + h(x0y0 + 1) = 0 + (0.1)(1 · 0 + 1) = 0.1

I x2 = x1 + h = 1.2, y2 = y1 + h(x1y1 + 1) = 0.1 + (0.1)((1.1) · (0.1) + 1)

I = 0.1 + 0.1(0.11 + 1) = 0.1 + 0.1(1.11) = 0.1 + 0.111 = 0.211

I y(1.2) ≈ 0.211.

Question 4

y ′ = xy + 1 y(1) = 0.

I x0 = 1, y0 = 0

I x1 = x0 + h = 1.1, y1 = y0 + h(x0y0 + 1) = 0 + (0.1)(1 · 0 + 1) = 0.1

I x2 = x1 + h = 1.2, y2 = y1 + h(x1y1 + 1) = 0.1 + (0.1)((1.1) · (0.1) + 1)

I = 0.1 + 0.1(0.11 + 1) = 0.1 + 0.1(1.11) = 0.1 + 0.111 = 0.211

I y(1.2) ≈ 0.211.

Question 4

y ′ = xy + 1 y(1) = 0.

I x0 = 1, y0 = 0

I x1 = x0 + h = 1.1, y1 = y0 + h(x0y0 + 1) = 0 + (0.1)(1 · 0 + 1) = 0.1

I x2 = x1 + h = 1.2, y2 = y1 + h(x1y1 + 1) = 0.1 + (0.1)((1.1) · (0.1) + 1)

I = 0.1 + 0.1(0.11 + 1) = 0.1 + 0.1(1.11) = 0.1 + 0.111 = 0.211

I y(1.2) ≈ 0.211.

Question 4

y ′ = xy + 1 y(1) = 0.

I x0 = 1, y0 = 0

I x1 = x0 + h = 1.1, y1 = y0 + h(x0y0 + 1) = 0 + (0.1)(1 · 0 + 1) = 0.1

I x2 = x1 + h = 1.2, y2 = y1 + h(x1y1 + 1) = 0.1 + (0.1)((1.1) · (0.1) + 1)

I = 0.1 + 0.1(0.11 + 1) = 0.1 + 0.1(1.11) = 0.1 + 0.111 = 0.211

I y(1.2) ≈ 0.211.

Question 4

y ′ = xy + 1 y(1) = 0.

I x0 = 1, y0 = 0

I x1 = x0 + h = 1.1, y1 = y0 + h(x0y0 + 1) = 0 + (0.1)(1 · 0 + 1) = 0.1

I x2 = x1 + h = 1.2, y2 = y1 + h(x1y1 + 1) = 0.1 + (0.1)((1.1) · (0.1) + 1)

I = 0.1 + 0.1(0.11 + 1) = 0.1 + 0.1(1.11) = 0.1 + 0.111 = 0.211

I y(1.2) ≈ 0.211.

Question 5

5. Which of the following gives the direction field for the differential equation

y ′ = y 2 − x2

Note the letter corresponding to each graph is at the lower left of the graph.

I We draw the direction vectors at a few points around the origin, say(0, 0), (0, 1), (1, 0), (1, 1), (−1, 0), (0,−1), (−1,−1) to get a feel for whatthe vector fiels might look like. (we will do this on paper)

I We note and sketch a few general patterns, if x = ±y , then y 2 − x2 = 0and the direction vector is horizontal.

I If |x | > |y | the slope is negative and if |y | > |x | the slope is positive.

I These and other observations allow us to eliminate some possibili-ties and form a picture to see that the direction field is the one shown below.

Question 5

y ′ = y 2 − x2

Question 5

y ′ = y 2 − x2

Question 5

y ′ = y 2 − x2

Question 5

y ′ = y 2 − x2

Question 6

6. Find the solution of the differential equation

p1− y 2

1 + x2

with initial condition y(0) = 0.

I Separating variables here givesR

dy√1−y2

dx1+x2 .

I Integrating both sides gives arcsin y = arctan x + C .

I substituting y(0) = 0 we find C = 0.

I Therefore y = sin(arctan x) = x√1+x2

(from triangle).

x1 + x2

Question 6

p1− y 2

1 + x2

dy√1−y2

dx1+x2 .

(from triangle).

x1 + x2

Question 6

p1− y 2

1 + x2

dy√1−y2

dx1+x2 .

(from triangle).

x1 + x2

Question 6

p1− y 2

1 + x2

dy√1−y2

dx1+x2 .

(from triangle).

x1 + x2

Question 6

p1− y 2

1 + x2

dy√1−y2

dx1+x2 .

(from triangle).

x1 + x2

Question 7

7. Find a general solution, valid for −π2

< x < π2

, of the differential equation

dx− (tan x)y = 1.

I The integrating factor here is I = eR− tan xdx = e ln(cos x) = cos x

I Multiplying the equation by the integrating factor, we get(cos x) dy

dx− (cos x)(tan x)y = cos x or

d(y cos x)

dx= cos x .

I so a general solution is given by y cos x =R

cos xdx = sin x + C

I Therefore y = sin x+Ccos x

Question 7

< x < π2

dx− (tan x)y = 1.

d(y cos x)

dx= cos x .

cos xdx = sin x + C

Question 7

< x < π2

dx− (tan x)y = 1.

d(y cos x)

dx= cos x .

cos xdx = sin x + C

Question 7

< x < π2

dx− (tan x)y = 1.

d(y cos x)

dx= cos x .

cos xdx = sin x + C

Question 7

< x < π2

dx− (tan x)y = 1.

d(y cos x)

dx= cos x .

cos xdx = sin x + C

Question 8

8. Find the limit: limn→∞

(−1)nn2

n2 + n + 1

I limn→∞n2

n2+n+1= limn→∞

11+1/n+1/n2 = 1 6= 0

I Therefore

limn→∞

(−1)nn2

n2 + n + 1does not exist.

Question 8

(−1)nn2

n2 + n + 1

I limn→∞n2

n2+n+1= limn→∞

11+1/n+1/n2 = 1 6= 0

I Therefore

limn→∞

(−1)nn2

Question 8

(−1)nn2

n2 + n + 1

I limn→∞n2

n2+n+1= limn→∞

11+1/n+1/n2 = 1 6= 0

I Therefore

limn→∞

(−1)nn2

Question 9

9. Calculate

limn→∞

(ln n)2

I Use L’Hospital’s rule (twice):

I limx→∞(ln x)2

x= limx→∞

((ln x)2)′

I = limn→∞2 1

I = limx→∞(2 ln x)′

I = limx→∞2x1

Question 9

9. Calculate

limn→∞

(ln n)2

I limx→∞(ln x)2

x= limx→∞

((ln x)2)′

I = limn→∞2 1

I = limx→∞2x1

Question 9

9. Calculate

limn→∞

(ln n)2

I limx→∞(ln x)2

x= limx→∞

((ln x)2)′

I = limn→∞2 1

I = limx→∞2x1

Question 9

9. Calculate

limn→∞

(ln n)2

I limx→∞(ln x)2

x= limx→∞

((ln x)2)′

I = limn→∞2 1

I = limx→∞2x1

Question 9

9. Calculate

limn→∞

(ln n)2

I limx→∞(ln x)2

x= limx→∞

((ln x)2)′

I = limn→∞2 1

I = limx→∞2x1

Question 9

9. Calculate

limn→∞

(ln n)2

I limx→∞(ln x)2

x= limx→∞

((ln x)2)′

I = limn→∞2 1

I = limx→∞2x1

Question 10

10. Find∞Xn=1

3 · 5n−1.

I This is a geometric series of forP∞

n=1 arn−1 = a + ar + . . . .

n=122n

3·5n−1 = 43

+ 1615

+ . . .

I a = 43

and r = 1615

I ThereforeP∞

n=122n

3·5n−1 = a1−r

= 4/31−(4/5)

= 4/31/5

= 203.

I AlternativelyP∞

n=122n

3·5n−1 =P∞

n=14·4n−1

3·5n−1 =43

P∞n=1

´n−1= 4

1− 45

”= 20

Question 10

10. Find∞Xn=1

3 · 5n−1.

n=1 arn−1 = a + ar + . . . .

n=122n

3·5n−1 = 43

+ 1615

+ . . .

I a = 43

and r = 1615

I ThereforeP∞

n=122n

3·5n−1 = a1−r

= 4/31−(4/5)

= 4/31/5

= 203.

I AlternativelyP∞

n=122n

3·5n−1 =P∞

n=14·4n−1

3·5n−1 =43

P∞n=1

´n−1= 4

1− 45

”= 20

Question 10

10. Find∞Xn=1

3 · 5n−1.

n=1 arn−1 = a + ar + . . . .

n=122n

3·5n−1 = 43

+ 1615

+ . . .

I a = 43

and r = 1615

I ThereforeP∞

n=122n

3·5n−1 = a1−r

= 4/31−(4/5)

= 4/31/5

= 203.

I AlternativelyP∞

n=122n

3·5n−1 =P∞

n=14·4n−1

3·5n−1 =43

P∞n=1

´n−1= 4

1− 45

”= 20

Question 10

10. Find∞Xn=1

3 · 5n−1.

n=1 arn−1 = a + ar + . . . .

n=122n

3·5n−1 = 43

+ 1615

+ . . .

I a = 43

and r = 1615

I ThereforeP∞

n=122n

3·5n−1 = a1−r

= 4/31−(4/5)

= 4/31/5

= 203.

I AlternativelyP∞

n=122n

3·5n−1 =P∞

n=14·4n−1

3·5n−1 =43

P∞n=1

´n−1= 4

1− 45

”= 20

Question 10

10. Find∞Xn=1

3 · 5n−1.

n=1 arn−1 = a + ar + . . . .

n=122n

3·5n−1 = 43

+ 1615

+ . . .

I a = 43

and r = 1615

I ThereforeP∞

n=122n

3·5n−1 = a1−r

= 4/31−(4/5)

= 4/31/5

= 203.

I AlternativelyP∞

n=122n

3·5n−1 =P∞

n=14·4n−1

3·5n−1 =43

P∞n=1

´n−1= 4

1− 45

”= 20

Question 10

10. Find∞Xn=1

3 · 5n−1.

n=1 arn−1 = a + ar + . . . .

n=122n

3·5n−1 = 43

+ 1615

+ . . .

I a = 43

and r = 1615

I ThereforeP∞

n=122n

3·5n−1 = a1−r

= 4/31−(4/5)

= 4/31/5

= 203.

I AlternativelyP∞

n=122n

3·5n−1 =P∞

n=14·4n−1

3·5n−1 =43

P∞n=1

´n−1= 4

1− 45

”= 20

Question 11

11. Calculate the arc length of the curve if y = x2

4− ln(

√x), where 2 ≤ x ≤ 4.

I Recall

p1 + (y ′)2dx .

I Note

y ′ =x

2− 1√

“x − 1

I Thus

1 + (y ′)2 = 1 +1

“x − 1

= 1 +1

“x2 − 2x

”= 1 +

“x2 − 2 +

”= 1 +

4x2 − 1

“x2 + 2x

“x +

I Therefore

p1/4(x + 1/x)2dx =

”dx =

2+ ln x

2ln 2.

Question 11

4− ln(

√x), where 2 ≤ x ≤ 4.

I Recall

p1 + (y ′)2dx .

I Note

y ′ =x

2− 1√

“x − 1

I Thus

1 + (y ′)2 = 1 +1

“x − 1

= 1 +1

“x2 − 2x

”= 1 +

“x2 − 2 +

”= 1 +

4x2 − 1

“x2 + 2x

“x +

I Therefore

p1/4(x + 1/x)2dx =

”dx =

2+ ln x

2ln 2.

Question 11

4− ln(

√x), where 2 ≤ x ≤ 4.

I Recall

p1 + (y ′)2dx .

I Note

y ′ =x

2− 1√

“x − 1

I Thus

1 + (y ′)2 = 1 +1

“x − 1

= 1 +1

“x2 − 2x

”= 1 +

“x2 − 2 +

”= 1 +

4x2 − 1

“x2 + 2x

“x +

I Therefore

p1/4(x + 1/x)2dx =

”dx =

2+ ln x

2ln 2.

Question 11

4− ln(

√x), where 2 ≤ x ≤ 4.

I Recall

p1 + (y ′)2dx .

I Note

y ′ =x

2− 1√

“x − 1

I Thus

1 + (y ′)2 = 1 +1

“x − 1

= 1 +1

“x2 − 2x

”= 1 +

“x2 − 2 +

”= 1 +

4x2 − 1

“x2 + 2x

“x +

I Therefore

p1/4(x + 1/x)2dx =

”dx =

2+ ln x

2ln 2.

Question 11

4− ln(

√x), where 2 ≤ x ≤ 4.

I Recall

p1 + (y ′)2dx .

I Note

y ′ =x

2− 1√

“x − 1

I Thus

1 + (y ′)2 = 1 +1

“x − 1

= 1 +1

“x2 − 2x

”= 1 +

“x2 − 2 +

”= 1 +

4x2 − 1

“x2 + 2x

“x +

I Therefore

p1/4(x + 1/x)2dx =

”dx =

2+ ln x

2ln 2.

Question 12

12. (a) Circle the letter below alongside the trapezoidal approximation to

ln 3 =

xdx using n = 8

xdx ≈ 1

”+2“2

”+2“4

”+2“1

”+2“4

”+2“2

”+2“ 4

”+“1

xdx ≈ 1

”+2“2

”+4“4

”+2“1

”+4“4

”+2“2

”+4“ 4

”+“1

xdx ≈ 1

h1+“4

”+“2

”+“4

”+“1

”+“4

”+“2

”+“ 4

”+“1

I ∆x = b−a8

I T8 = ∆x2

[f (1) + 2f ( 54) + 2f ( 3

2) + 2f ( 7

4) + 2f (2) + 2f ( 9

4) + 2f ( 5

2f ( 114

) + f (3)] = A (f (x) = 1x

Question 12

ln 3 =

xdx using n = 8

xdx ≈ 1

”+2“2

”+2“4

”+2“1

”+2“4

”+2“2

”+2“ 4

”+“1

xdx ≈ 1

”+2“2

”+4“4

”+2“1

”+4“4

”+2“2

”+4“ 4

”+“1

xdx ≈ 1

h1+“4

”+“2

”+“4

”+“1

”+“4

”+“2

”+“ 4

”+“1

I ∆x = b−a8

I T8 = ∆x2

[f (1) + 2f ( 54) + 2f ( 3

2) + 2f ( 7

4) + 2f (2) + 2f ( 9

4) + 2f ( 5

2f ( 114

) + f (3)] = A (f (x) = 1x

Question 12

ln 3 =

xdx using n = 8

xdx ≈ 1

”+2“2

”+2“4

”+2“1

”+2“4

”+2“2

”+2“ 4

”+“1

xdx ≈ 1

”+2“2

”+4“4

”+2“1

”+4“4

”+2“2

”+4“ 4

”+“1

xdx ≈ 1

h1+“4

”+“2

”+“4

”+“1

”+“4

”+“2

”+“ 4

”+“1

I ∆x = b−a8

I T8 = ∆x2

[f (1) + 2f ( 54) + 2f ( 3

2) + 2f ( 7

4) + 2f (2) + 2f ( 9

4) + 2f ( 5

2f ( 114

) + f (3)] = A (f (x) = 1x

Question 1212.(b) Recall that the error ET in the trapezoidal rule for approximatingR b

af (x)dx satisfies ˛̨̨ Z b

af (x)dx − Tn

˛̨̨= |ET | ≤

K(b − a)3

whenever |f ′′(x)| ≤ K for all a ≤ x ≤ b. Use the above error bound todetermine a value of n for which the trapezoidal approximation to

ln 3 =

xdx has an error |ET | ≤ 1

310−4.

I f (x) = 1x, f ′(x) = −1

x2 , f ′′(x) = 2x3

I Since |f ′′(x)| = 2x3 is decreasing on the interval 1 ≤ x ≤ 2, we have

|f ′′(x)| ≤ f ′′(1) = 2 for 1 ≤ x ≤ 2. Hence, we can use K = 2 in the errorbound above.

I For the trapezoidal approximation Tn, we have

|ET | ≤K(b − a)3

2(3− 1)3

I If we find a value of n for which 1310−4 ≥ 4

3n2 , then we will have

|ET | ≤ 1310−4.

I 1310−4 ≥ 4

3n2 → n2 ≥ 4 · 104 → n ≥ 2 · 102 = 200.

af (x)dx − Tn

˛̨̨= |ET | ≤

K(b − a)3

ln 3 =

310−4.

I f (x) = 1x, f ′(x) = −1

x2 , f ′′(x) = 2x3

|ET | ≤K(b − a)3

2(3− 1)3

|ET | ≤ 1310−4.

I 1310−4 ≥ 4

3n2 → n2 ≥ 4 · 104 → n ≥ 2 · 102 = 200.

af (x)dx − Tn

˛̨̨= |ET | ≤

K(b − a)3

ln 3 =

310−4.

I f (x) = 1x, f ′(x) = −1

x2 , f ′′(x) = 2x3

|ET | ≤K(b − a)3

2(3− 1)3

|ET | ≤ 1310−4.

I 1310−4 ≥ 4

3n2 → n2 ≥ 4 · 104 → n ≥ 2 · 102 = 200.

af (x)dx − Tn

˛̨̨= |ET | ≤

K(b − a)3

ln 3 =

310−4.

I f (x) = 1x, f ′(x) = −1

x2 , f ′′(x) = 2x3

|ET | ≤K(b − a)3

2(3− 1)3

|ET | ≤ 1310−4.

I 1310−4 ≥ 4

3n2 → n2 ≥ 4 · 104 → n ≥ 2 · 102 = 200.

af (x)dx − Tn

˛̨̨= |ET | ≤

K(b − a)3

ln 3 =

310−4.

I f (x) = 1x, f ′(x) = −1

x2 , f ′′(x) = 2x3

|ET | ≤K(b − a)3

2(3− 1)3

|ET | ≤ 1310−4.

I 1310−4 ≥ 4

3n2 → n2 ≥ 4 · 104 → n ≥ 2 · 102 = 200.

af (x)dx − Tn

˛̨̨= |ET | ≤

K(b − a)3

ln 3 =

310−4.

I f (x) = 1x, f ′(x) = −1

x2 , f ′′(x) = 2x3

|ET | ≤K(b − a)3

2(3− 1)3

|ET | ≤ 1310−4.

I 1310−4 ≥ 4

3n2 → n2 ≥ 4 · 104 → n ≥ 2 · 102 = 200.

Question 13

13. Find the family of orthogonal trajectories to the family of curves given by

y = kx2.

I For the family of curves described above dydx

= 2kx .

I Also y = kx2 giving k = yx2 .

I Therefore the given family satisfy the differential equationdydx

= 2 yx2 x = 2 y

I The family of orthogonal trajectories must satisfy the differential equationdydx

= −x2y

I Separating the variables, we get 2ydy = −xdx

Zydy = −

Zxdx , or y 2 =

2+ C .

I Hence our family of orthogonal trajectories is a family of curves of theform

y 2 +x2

2= C ,

a family of ellipses.

Question 13

y = kx2.

= 2kx .

= 2 yx2 x = 2 y

= −x2y

Zydy = −

Zxdx , or y 2 =

2+ C .

y 2 +x2

2= C ,

Question 13

y = kx2.

= 2kx .

= 2 yx2 x = 2 y

= −x2y

Zydy = −

Zxdx , or y 2 =

2+ C .

y 2 +x2

2= C ,

Question 13

y = kx2.

= 2kx .

= 2 yx2 x = 2 y

= −x2y

Zydy = −

Zxdx , or y 2 =

2+ C .

y 2 +x2

2= C ,

Question 13

y = kx2.

= 2kx .

= 2 yx2 x = 2 y

= −x2y

Zydy = −

Zxdx , or y 2 =

2+ C .

y 2 +x2

2= C ,

Question 13

y = kx2.

= 2kx .

= 2 yx2 x = 2 y

= −x2y

Zydy = −

Zxdx , or y 2 =

2+ C .

y 2 +x2

2= C ,

Question 13

y = kx2.

= 2kx .

= 2 yx2 x = 2 y

= −x2y

Zydy = −

Zxdx , or y 2 =

2+ C .

y 2 +x2

2= C ,

Question 13

y = kx2.

= 2kx .

= 2 yx2 x = 2 y

= −x2y

Zydy = −

Zxdx , or y 2 =

2+ C .

y 2 +x2

2= C ,

Question 14

Solve the initial value problem

8><>:x2y ′ + 2xy = 1,

y(1) = 2.

I We put the equation in standard form: dydx

+ 2xy = 1

I The integrating factor is given by I (x) = eR

(2/x)dx = e2 ln |x| = x2.

I Multiplying by the integrating factor, we get x2 dydx

+ 2xy = 1.

I Therefore ddx

x2y = 1 and x2y =R

1dx = x + C

I Dividing both sides by x2, we get y = x+Cx2 .

I The initial condition gives y(1) = 2 or 1 + C = 2 and C = 1. Hencey = x+1

Question 14

8><>:x2y ′ + 2xy = 1,

y(1) = 2.

+ 2xy = 1

(2/x)dx = e2 ln |x| = x2.

+ 2xy = 1.

I Therefore ddx

x2y = 1 and x2y =R

1dx = x + C

Question 14

8><>:x2y ′ + 2xy = 1,

y(1) = 2.

+ 2xy = 1

(2/x)dx = e2 ln |x| = x2.

+ 2xy = 1.

I Therefore ddx

x2y = 1 and x2y =R

1dx = x + C

Question 14

8><>:x2y ′ + 2xy = 1,

y(1) = 2.

+ 2xy = 1

(2/x)dx = e2 ln |x| = x2.

+ 2xy = 1.

I Therefore ddx

x2y = 1 and x2y =R

1dx = x + C

Question 14

8><>:x2y ′ + 2xy = 1,

y(1) = 2.

+ 2xy = 1

(2/x)dx = e2 ln |x| = x2.

+ 2xy = 1.

I Therefore ddx

x2y = 1 and x2y =R

1dx = x + C

Question 14

8><>:x2y ′ + 2xy = 1,

y(1) = 2.

+ 2xy = 1

(2/x)dx = e2 ln |x| = x2.

+ 2xy = 1.

I Therefore ddx

x2y = 1 and x2y =R

1dx = x + C

Question 14

8><>:x2y ′ + 2xy = 1,

y(1) = 2.

+ 2xy = 1

(2/x)dx = e2 ln |x| = x2.

+ 2xy = 1.

I Therefore ddx

x2y = 1 and x2y =R

1dx = x + C

question 1 - university of notre dameapilking/math10560/work/old exams... · 2020. 3. 16. ·...

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