Sample Solutions of Assignment 2 for MAT3270B: 2.6-2wei/odeas2sol.pdf · 9 and the solutions of original equation are given by e3xx2y + 1 3 e 3xy = c (b). Answer: My¡Nx ¡M = 1 y

Sample Solutions of Assignment 2 for MAT3270B: 2.6-2.4

Note: We amend 6(b),10.(c) and 10.(f) in this Sample solutions.

1. Determine if the following equation is exact. If it is exact, find out

the solution.

(a). (x4 + 4y) + (4x− 3y8)y′= 0

(b). (x + y) + (2x− y)y′= 0

(c). (2xy2 + 2y) + (2x2y + 2x)y′= 0

(d). (x log x + xy)dx + (y log x + xy)dy = 0

(e). x

(x2+y2)32dx + ydy

(x2+y2)32

= 0

(a). Answer: The original ODE can be rewritten as

(x4 + 4y)dx + (4x− 3y8)dy = 0

∂M(x, y)

∂y= 4

∂N(x, y)

∂x= 4

So this ODE is exact for ∂M(x,y)∂y

= ∂N(x,y)∂x

. Thus there is a ψ(x, y) such

that

∂ψ

∂x= M(x, y) = x4 + 4y

Integrating above equation, we obtain

ψ(x, y) =1

5x5 + 4xy + f(y)

Setting ψy = N gives

f′(y) = −3y8

1

2

then we get

f(y) =−1

3y9

Hence,

ψ(x, y) =1

5x5 + 4xy − 1

3y9

and the solutions of original equation are given by

1

5x5 + 4xy − 1

3y9 = c

(b). Answer: The original ODE can be rewritten as

(x + y)dx + (2x− y)dy = 0

∂M(x, y)

∂y= 1

∂N(x, y)

∂x= 2

So this ODE is not exact for ∂M(x,y)∂y

6= ∂N(x,y)∂x

.

(c). Answer: The original ODE can be rewritten as

(2xy2 + 2y)dx + (2x2y + 2x)dy = 0

∂M(x, y)

∂y= 4xy + 2

∂N(x, y)

∂x= 4xy + 2


= ∂N(x,y)∂x


that∂ψ

∂x= M(x, y) = 2xy2 + 2y


ψ(x, y) = x2y2 + 2xy + f(y)


f′(y) = 0

then we get

f(y) = constant

3

Hence,

ψ(x, y) = x2y2 + 2xy + constant


x2y2 + 2xy = c

(d). Answer: The original ODE can be rewritten as

(x log x + xy)dx + (y log x + xy)dy = 0

∂M(x, y)

∂y= x

∂N(x, y)

∂x=

y

x+ y

So this ODE is not exact for ∂M(x,y)∂y

6= ∂N(x,y)∂x

.

(e). Answer:∂M(x, y)

∂y=

−3xy

(x2 + y2)52

∂N(x, y)

∂x=

−3xy

(x2 + y2)52


= ∂N(x,y)∂x


that∂ψ

∂x= M(x, y) =

x

(x2 + y2)32


ψ(x, y) =−1

(x2 + y2)12

+ f(y)


f′(y) = 0

then we get

f(y) = constant

Hence,

ψ(x, y) =−1

(x2 + y2)12

+ constant

4


−1

(x2 + y2)12

= c

2. Using the integrating factors given below to solve the corresponding

ODEs.

(a). x2y3 + x(1 + y2)y′= 0, µ(x, y) = 1

xy3

(b). ( sin yy− 2e−x sin x)dx + cos y+2e−x cos x

ydy = 0, µ(x, y) = yex

(c). (3x + 6y)dx + (x2

y+ 2 y

x)dy = 0, µ(x, y) = xy

(a). Answer: Multiplying the original ODE by µ, we get the fol-

lowing equation

−xdx =1 + y2

y3dy

Integrating both sides of the above equation, and the solutions of the

original equation are given by

log |y| − 1

2y2+

x2

2= c

(b). Answer: Multiplying the original ODE by µ = yex, we get the

following equation

(ex sin y − 2y sin x)dx + (ex cos y + 2 cos x)dy = 0

∂M(x, y)

∂y= ex cos y − 2 sin x

∂N(x, y)

∂x= ex cos y − 2 sin x


= ∂N(x,y)∂x


that∂ψ

∂x= M(x, y) = ex sin y − 2y sin x

5


ψ(x, y) = ex sin y + 2y cos x + f(y)


f′(y) = 0

then we get

f(y) = constant

Hence,

ψ(x, y) = ex sin y + 2y cos x + constant


ex sin y + 2y cos x = c

(c). Answer: Multiplying the original ODE by µ = xy, we get the

following equation

(3x2y + 6x)dx + (x3 + 2y2)dy = 0

∂M(x, y)

∂y= 3x2

∂N(x, y)

∂x= 3x2


= ∂N(x,y)∂x


that∂ψ

∂x= M(x, y) = 3x2y + 6x


ψ(x, y) = x3y + 3x2 + f(y)


f′(y) = 2y2

then we get

f(y) =2

3y3

6

Hence,

ψ(x, y) = x3y + 3x2 +2

3y3


x3y + 3x2 +2

3y3 = c

3. Show that if Nx−My

xM−yN= R, where R depends on the quality xy

only, then the different equation M(x, y)dx + N(x, y)dy = 0 has an

integrating factor of the form µ(xy). Using this to solve the following

ODE.

(3x +6

y) + (

x2

y+ 3

y

x)dy

dx= 0

Answer: Let ω(z) = e∫

R(Z)dz, which is well defined for R depends on

the quality xy only. Then setting µ(x, y) = ω(xy) and multiplying the

original ODE by µ(x, y), we get the following equation

M(x, y)µ(x, y)dx + N(x, y)µ(x, y)dy = 0

Set M = Mµ, N = Nµ

∂M(x, y)

∂y= e

∫R(Z)dz(My + MR(Z)x), Z = xy

∂N(x, y)

∂x= e

∫R(Z)dz(Nx + NR(Z)y), Z = xy

If Nx−My

xM−yN= R, where R depends on the quality xy only, then ∂M(x,y)

∂y=

∂N(x,y)∂x

. So the original ODE has an integrating factor of the form of

µ(xy).

Now, we solve the special equation with method above. It is easy to

get R(xy) = 1xy

and ω(z) = e∫

1z = z

7

Multiplying the original ODE by µ = ω(xy) = xy, we get the follow-

ing equation

(3x2y + 6x)dx + (x3 + 3y2)dy = 0

∂M(x, y)

∂y= 3x2

∂N(x, y)

∂x= 3x2


= ∂N(x,y)∂x


that∂ψ

∂x= M(x, y) = 3x2y + 6x


ψ(x, y) = x3y + 3x2 + f(y)


f′(y) = 3y2

then we get

f(y) = y3

Hence,

ψ(x, y) = x3y + 3x2 + y3


x3y + 3x2 + y3 = c

4. Finding out the integrating factors of the following ODEs and solve

the corresponding ODEs.

(a). (3x2y + 2xy + y3)dx + (x2 + y2)dy = 0

(b). dx + (xy− sin y)dy = 0

(c). ydx + (2xy − e−2y)dy = 0

8

(d). exdx + (ex cot y + 2y csc y)dy = 0

(a). Answer:∂M(x, y)

∂y= 3x2 + 2x + 3y2

∂N(x, y)

∂x= 2x

My−Nx

N= 3 is a function of x only, then there is an integrating factor µ

that also depends on x only and satisfies the following equation

dµ

dx= 3µ

Integrating above equation, we get µ = e3x. Multiplying the original

ODE by µ = e3x, we get the following equation

(3e3xx2y + 2e3xxy + e3xy3)dx + (e3xx2 + e3xy2)dy

∂M(x, y)

∂y= 3e3xx2 + 2e3xx + 3e3xy2

∂N(x, y)

∂x= 3e3xx2 + 2e3xx + 3e3xy2

So the resulting ODE is exact for ∂M(x,y)∂y

= ∂N(x,y)∂x

. Thus there is a

ψ(x, y) such that

∂ψ

∂x= M(x, y) = 3e3xx2y + 2e3xxy + e3xy3


ψ(x, y) = e3xx2y +1

3e3xy3 + f(y)


f′(y) = 0

then we get

f(y) = constant

Hence,

ψ(x, y) = e3xx2y +1

3e3xy3

9


e3xx2y +1

3e3xy3 = c

(b). Answer: My−Nx

−M= 1

yis a function of y only, then there is

an integrating factor µ that also depends on y only and satisfies the

following equationdµ

dy=

1

yµ

Integrating above equation, we get µ = y. Multiplying the original

ODE by µ = y, we get the following equation

ydx + (x− y sin y)dy = 0

∂M(x, y)

∂y= 1

∂N(x, y)

∂x= 1


= ∂N(x,y)∂x

. Thus there is a

ψ(x, y) such that∂ψ

∂x= M(x, y) = y


ψ(x, y) = xy + f(y)


f′(y) = −y sin y

then we get

f(y) = y cos y − sin y

Hence,

ψ(x, y) = xy + y cos y − sin y


xy + y cos y − sin y = c

10

(c). Answer: My−Nx

−M= 1−2y

−yis a function of y only, then there is



dy=

1− 2y

−yµ

Integrating above equation, we get µ = 1ye2y. Multiplying the original

ODE by µ, we get the following equation

e2ydx + (2xe2y − 1

y)dy = 0

∂M(x, y)

∂y= 2e2y

∂N(x, y)

∂x= 2e2y


= ∂N(x,y)∂x

. Thus there is a


∂x= M(x, y) = e2y


ψ(x, y) = xe2y + f(y)


f′(y) =

1

−y

then we get

f(y) = − log y

Hence,

ψ(x, y) = xe2y − log y


xe2y − log y = c

(d). Answer: My−Nx

−M= cot y is a function of y only, then there is


11


dy= cot yµ

Integrating above equation, we get µ = sin y. Multiplying the original


sin yexdx + (ex cos y + 2y)dy = 0

∂M(x, y)

∂y= ex cos y

∂N(x, y)

∂x= ex cos y


= ∂N(x,y)∂x

. Thus there is a


∂x= M(x, y) = sin yex


ψ(x, y) = sin yex + f(y)


f′(y) = 2y

then we get

f(y) = y2

Hence,

ψ(x, y) = sin yex + y2


sin yex + y2 = c

5. State the region in the ty-plane where the (existence and uniqueness)

hypotheses of Theorem 2.4.2 are satisfied.

12

(a). dydt

= log |ty|1−t2+y2

(b). dydt

= 1+t2

3y−y2

(a). Answer: Setting f(t, y) = log |ty|1−t2+y2

If ty > 0 and 1− t2 + y2 6= 0 then

∂f

∂y=

1y(1− t2 + y2)− 2y log ty

(1− t2 + y2)2

Obviously f and ∂f∂y

are continuous if ty > 0 and 1− t2 + y2 6= 0.

If ty < 0 and 1− t2 + y2 6= 0 then

∂f

∂y=

1y(1− t2 + y2)− 2y log−ty

(1− t2 + y2)2


are continuous if ty > 0 and 1− t2 + y2 6= 0.

Hence, if t 6= 0, y 6= 0 and 1− t2 + y2 6= 0, the hypotheses of Theorem

2.4.2 are satisfied.

(b). Answer: Setting f(t, y) = 1+t2

3y−y2

If 3y − y2 6= 0 then

∂f

∂y=

(−3 + 2y)(1 + t2)

(3y − y2)2


are continuous if 3y − y2 6= 0.

Hence, if 3y − y2 6= 0, the hypotheses of Theorem 2.4.2 are satisfied.

6. Verify if the following functions satisfy the Lipschitz condition

|f(t, y1)− f(t, y2)| ≤ L|y1 − y2|, (t, y1) ∈ R, (t, y2) ∈ R

at the given domain R.

(a). f(t, y) = t√

y, R = [−2, 2]× [1, 10],

(b). f(t, y) = t√

y, R = [−2, 2]× [0, 10],

(c). f(t, y) = (sin t)(sin y), R = (−∞, +∞)× (−∞, +∞).

13

(a). Answer:

|f(t, y1)− f(t, y2)| = |t√y1 − t√

y2|

=|t|√

y1 + y2

|y1 − y2|, if y1 + y2 6= 0

In R = [−2, 2]×[1, 10], we have y1+y2 6= 0,√

y1 + y2 ≥√

2, and |t| ≤ 2.

So let L =√

2, we can get

|f(t, y1)− f(t, y2)| ≤ L|y1 − y2|,

(t, y1) ∈ [−2, 2]× [1, 10], (t, y2) ∈ [−2, 2]× [1, 10].

(b). Answer: Assume that there exists L, which satisfy the condi-

tion

|f(t, y1)− f(t, y2)| = |t√y1 − t√

y2| ≤ L|y1 − y2|,

(t, y1) ∈ R = [−2, 2]× [0, 10], (t, y2) ∈ R = [−2, 2]× [0, 10]

Obviously L > 0. Let t = 32, y1 = 1

4L2 , y2 = 116L2 , then

|t√y1 − t√

y2| = 3

8L

L|y1 − y2| = 3

16L

So we have

|t√y1 − t√

y2| > L|y1 − y2|This is a contradiction. So f does not satisfy the Lipschitz condition

in R.

(c). Answer: |f(t, y1)− f(t, y2)| = | sin t|| sin y1 − siny2|

= 2| sin t|| sin y1 − y2

2|| cos

y1 + y2

2| ≤ 2| sin y1 − y2

2| ≤ |y1 − y2|

for | sin x| ≤ |x|. So f satisfies the Lipschitz condition in R.

14

7. Solve the following ODE and state the interval of Existence in terms

of y0.

(a). y′ = 2ty2, y(0) = y0

(b). y′= t2

y(1+t2), y(0) = y0


dy

y2= 2tdt, y 6= 0

Integrating the above equation, we get

y(t) =−1

t2 + c

From y(0) = y0, then c = −1y0

and y(t) = y0

1−y0t2

Since y(t) 6= 0, then y0 6= 0

1− y0t2 6= 0, then t 6= ± 1√

y0

Hence, the interval of existence is (− 1√y0

, 1√y0

)


ydy =t2dt

1 + t2, y 6= 0

Integrating the above equation, we get

1

2y2 = t− tan−1 t + c

From y(0) = y0, then c = 12y0

2. Substituting to the equation,

y2 = 2t− 2 tan−1 t + y02

i.e.

y =√

2t− 2 tan−1 t + y02

Let f(t) = 2t− 2 tan−1 t + y02, then

f′(t) = (

2t2

1 + t2) > 0, if t 6= 0;

Obviously, f′(0) = 0. y(0) = y2

0 ≥ 0.

Then there exists t0 ∈ (−∞, 0) such that f(t0) = 0. Therefore the

15

interval of existence is (t0, +∞).

8. Show that

y(t) =

{(t− C)2 for t ≥ C ≥ 00 for t ≤ C

(1)

satisfies y′= 2y

12 , y(0) = 0 for any constant C ≥ 0. So we don’t have

uniqueness, why?

Answer: Obviously,

y′=

{2(t− C) for t ≥ C ≥ 00 for t < C

(2)

It is easy to check y′= 2y

12 and y(0) = 0. The function f(t, y) = 2y

12 is

continuous everywhere, but ∂f∂y

= y−12 is not continuous when y = 0. So

Theorem 2.4.2 does not apply to this problem and we can not assure

the uniqueness of the solution.

9. Solve the following Bernoulli equations and state the Interval of

Existence.

(a). t2y′ + 2ty − y3 = 0, t > 0, y(1) = 1

(b). y′= 2y − y2, y(0) = 1

(a). Answer: (a) The original ODE can be rewritten as

t2

y3y′+

2t

y2− 1 = 0

Let v = 1y2 , then

dv

dt=−2

y3

dy

dt

16

1

y3y′=−1

2

dv

dtSubstituting to the equation, we get the following resulting equation

dv

dt− 4

tv = − 2

t2

v(t) = e∫

4tdt[−

∫ 2

t2(e

∫ −4t

dt)dt]

=2

5t+ ct4

Hence, the solutions are

y2(t) =1

25t

+ ct4

From y(1) = 1, then c = 35

and y2(t) = 5t3t5+2

Since y 6= 0, then t 6= 0 and y(t) =√

5t3t5+2

. Therefore the interval of

existence is (0, +∞).

Answer: (b). Let v = 1y, then

dv

dt=−1

y2

dy

dt

dy

dt=−1

v2

dv

dtThe original ODE can be rewritten as

dv

−2v + 1= dt

Integrating above equation, we get

y(t) =2

e2(t+c) − 1

Since y(0) = 1, then c = 12log 3. Substituting to the equation, the

solution is

y(t) =2

3e2t − 1From 3e2t − 1 6= 0, then t 6= 1

2log 3. Hence the interval of existence is

(12log 1

3, +∞).

17

10. Solve the following ODEs

(a). y′= 2x+y

3+3y2−x, y(0) = 0

(b). y′= y

t−y

(c). xy′+ y − y2e2x = 0, y(1) = 1

(d). y′ = ex+y

(e). xdy − ydx = 2x2y2dy, y(1) = −2

(f). dydx

= −2xy+1x2+2y

(g). y′= y3

1−2xy2 , y(0) = 1

(h). (2y + 1)dx + x2−yx

dy = 0


(2x + y)dx + (−3− 3y2 + x)dy = 0

∂M(x, y)

∂y= 1

∂N(x, y)

∂x= 1


= ∂N(x,y)∂x


that∂ψ

∂x= M(x, y) = 2x + y


ψ(x, y) = x2 + xy + f(y)


f′(y) = −3− 3y2

then we get

f(y) = −3y − y3

18

Hence,

ψ(x, y) = x2 + xy − 3y − y3


x2 + xy − 3y − y3 = c

To satisfy the initial condition we must choose c = 0, so

x2 + xy − 3y − y3 = 0

is the solution of the given initial value problem.


ydt + (−t + y)dy = 0

My−Nt

−M= −2

yis a function of y only, then there is an integrating factor

µ that also depends on y only and satisfies the following equation

dµ

dy=−2

yµ

Integrating above equation, we get µ = 1t2

. Multiplying the original


y

t2dt + (

−1

t+

1

t2y)dy = 0

∂M(t, y)

∂y=

1

t2

∂N(t, y)

∂x=

1

t2

So the resulting ODE is exact for ∂M(t,y)∂y

= ∂N(t,y)∂t

. Thus there is a

ψ(t, y) such that∂ψ

∂t= M(t, y) =

y

t2


ψ(t, y) =−y

t+ f(y)


f′(y) =

y

t2

19

then we get

f(y) =1

2

y2

t2

Hence,

ψ(t, y) =−y

t+

1

2

y2

t2


−y

t+

1

2

y2

t2= c

(c). Answer: The original ODE can be rewritten as

x

y2

dy

dx+ (

1

y− e2x) = 0

Let v = −1y

, then

dv

dx=

1

y2

dy

dxdy

dx=

1

v2

dv

dxSo we get the following resulting equation

xdv

dx= v − e2x

It can be written as

(v − e2x)dx + xdv = 0

∂M(x, y)

∂v= 1

∂N(x, y)

∂x= 1

So this ODE is exact for ∂M(x,v)∂v

= ∂N(x,v)∂x

. Thus there is a ψ(x, v) such

that∂ψ

∂x= M(x, v) = v − e2x


ψ(x, y) = vx− 1

2e2x + f(y)


f′(y) = 0

20

then we get

f(y) = constant

Hence,

ψ(x, y) = vx− 1

2e2x

and the solutions of resulting equation are given by

vx− 1

2e2x = c.

Obviously, the solutions of the original equations are given by

−x

y− 1

2e2x = c.

To satisfy the initial condition we must choose c = −1− 12e2, so

x−x

y− 1

2e2x + 1 +

1

2e2 = 0


(d). Answer: The original ODE can be rewritten as

exdx− e−ydy = 0

Integrating above equation, the solutions of original equation are given

by

ex + e−y = 0

(e). Answer: The original ODE can be rewritten as

(−y)dx + (x− 2x2y2)dy = 0

∂M(x, y)

∂y= −1

∂N(x, y)

∂x= 1− 4xy2

My−Nx

N= −2

xis a function of x only, then there is an integrating factor

µ that also depends on x only and satisfies the following equation

dµ

dx=−2

xµ

21

Integrating above equation, we get µ = 1x2 . Multiplying the original


−y

x2dx + (

1

x− 2y2)dy = 0

∂M(x, y)

∂y=−1

x2

∂N(x, y)

∂x=−1

x2


= ∂N(x,y)∂x

. Thus there is a


∂x= M(x, y) =

−y

x2


ψ(x, y) =y

x+ f(y)


f′(y) = −2y2

then we get

f(y) =−2

3y3

Hence,

ψ(x, y) =y

x− 2

3y3


y

x− 2

3y3 = c.


y

x− 2

3y3 − 10

3= 0


(f). Answer: The original ODE can be rewritten as

(2xy + 1)dx + (x2 + 2y)dy = 0

∂M(x, y)

∂y= 2x

22

∂N(x, y)

∂x= 2x


= ∂N(x,y)∂x


that∂ψ

∂x= M(x, y) = 2xy + 1


ψ(x, y) = x2y + x + f(y)


f′(y) = 2y

then we get

f(y) = y2

Hence,

ψ(x, y) = x2y + x + y2


x2y + x + y2 = c

(g). Answer: The original ODE can be rewritten as

(y3)dx + (2xy2 − 1)dy = 0

∂M(x, y)

∂y= 3y2

∂N(x, y)

∂x= 2y2

My−Nx

−M= −1

yis a function of y only, then there is an integrating factor

µ that also depends on y only and satisfies the following equation

dµ

dy=−1

yµ

Integrating above equation, we get µ = 1y. Multiplying the original


y2dx + (2xy − 1

y)dy = 0

23

∂M(x, y)

∂y= 2y

∂N(x, y)

∂x= 2y


= ∂N(x,y)∂x

. Thus there is a


∂x= M(x, y) = y2


ψ(x, y) = xy2 + f(y)


f′(y) =

−1

ythen we get

f(y) = − log y

Hence,

ψ(x, y) = xy2 − log y


xy2 − log y = c.


xy2 − log y = 0


(h). Answer:∂M(x, y)

∂y= 2

∂N(x, y)

∂x= 1 +

y

x2

My−Nx

N= 1

xis a function of x only, then there is an integrating factor

µ that also depends on x only and satisfies the following equation

dµ

dx=

1

xµ

24

Integrating above equation, we get µ = x. Multiplying the original


(2xy + x)dx + (x2 − y)dy = 0

∂M(x, y)

∂y= 2x

∂N(x, y)

∂x= 2x


= ∂N(x,y)∂x

. Thus there is a


∂x= M(x, y) = 2xy + x


ψ(x, y) = x2y +1

2x2 + f(y)


f′(y) = −y

then we get

f(y) =−1

2y2

Hence,

ψ(x, y) = x2y +1

2x2 − 1

2y2


x2y +1

2x2 − 1

2y2 = c.

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Sample Solutions of Assignment 2 for MAT3270B: 2.6-2wei/odeas2sol.pdf · 9 and the solutions of original equation are given by e3xx2y + 1 3 e 3xy = c (b). Answer: My¡Nx ¡M = 1 y