MAT1CDE Assignment 4: Differential Equations and Applications, 2018

This document contains solutions to several differential equations problems, likely part of the MAT1CDE Assignment 4. The problems cover topics such as separable differential equations, finding particular solutions given initial conditions, and applications involving steady flow and hydraulic conductivity. Various integration techniques are applied to solve these equations, and the solutions are presented with detailed steps. The solutions also include determination of integration constants and domains. One problem involves determining exponential and constant values. This resource is valuable for students seeking assistance with differential equations and related mathematical concepts.

Q1)
a)
dy
dx = ( y−2 )2 sec2 ( x−1 )
Y = 2 when x = 3
dy
( y−2 ) 2 =sec2 ( x−1 ) dx
∫ ( y −2 )−2 dy=∫ sec2 (x) dx
( y−2 )−1
−1 =tan ( x ) +C
1
y−2 =−tan ( x ) +C
y- 2 = 1
−tan ( x ) +C
y = 1
−tan ( x ) +C +2
at y = 2 when x = 3
substituting to equation
1
y−2 =−tan ( x ) +C
1
2−2 =−tan ( 3 ) +C
C = -0.05241
The equation will then be;
y = 1
−tan ( x )−0.05241 +2
for domain
-tan(x) ≠ 0

X = 3
Domain
X∈ R−{3 }
b)
dy
dx = y2
4 x2
dy
y2 = dx
4 x2
∫ 1
y2 dy =∫ 1
4 x2 dx
∫ y−2 dy =∫ ( 4 x )−2 dx
y1
−1 = 4 x−1
−1 +C
y−1= ( 4 x )−1 +C
1
y = 1
4 x +C
Substituting y = 1 at x = 1
1
1= 1
4 +C
C = ¾
The equation will be
y−1= ( 4 x )−1 + 3
4
Domain
{X : 4x ≥ 0 }={4 }

Q2)
I = ∫ 1
y3 ∗e
1
y dy
Let 1/y = u
Then −1
y2 dy=du
I =∫ 1
y ∗e
1
y∗
( 1
y2 )dy
∫u∗eu∗(−du)
−∫u∗eu∗(du)
− ( u eu−eu ) +C
eu−u eu +C
I = e
1
y − 1
y e
1
y +C
b)

dx
dy − x
y2 = m
y3 , x = 1 when y = 1
If ¿ e∫ −1
y2 dy
=e
1
y
x e
1
y =∫ m
y2 ∗e
1
y dy+C
= m [e
1
y − 1
y e
1
y ¿+ C
When, y = 1, x = 1
C = e
x e
1
y =m[e
1
y − 1
y e
1
y ]+e
c)
dx
dy = y2
x
∫ dy
y2 =∫ dx
x
−1
y =lnx+C
Y = 1 when x = 1
-1 = 0 + C
C = -1
−1
y =lnx−1
d) part b the value of C is equivalent to exponential value e1, whereas in part b the value of C is
equivalent to -1

Q3)
a)
Given that
3
( 1
a∗da
dt )
2
=ρ ………………………….1
Where ρ= ᴧ [is eistetain constant ]
From equation 1 we get
1
a
da
dt = √ ρ
3 = √ ᴧ
3
1
a
da
dt = λ , assume √ ᴧ
3 =λ
da
dt =λdt

Integrating on both sides we get
∫ da
a = λ∫ dt
ln(a) = λt +ln (c), where c = constant of integration
a ( t )=C∗eλt ………………….2
The initial condition
a(0) = 1, that is
a = 1 when t = 0
From equation 2 we get
a(0) = Ce0=C
C = 1
Substituting this value of C in equation 2
a(t) = 1 * e λt
a(t) = e λt
a(t) = e √ ᴧ
3 t
b) here ρ= 3 m
4 π a3
from equation 1
3 ( 1
a
da
dt )2
= 3 m
4 π a3
( 1
a
da
dt )
2
= m
4 π a3

1
a ( da
dt ) = √ m
4 π ∗1
a
2
a
√ a∗da
dt = √ m
4 π =x
√a∗da=kdt
Integrating on both side
∫ a
1
2 da=k ∫dt
a
3
2
3
2
=kt +σ
2
3 a
3
2 = ( kt +σ ) ………………………………..3
Initial condition a(0) = 1, that is a = 1, when t = 0
From equation 3 we get
2
3∗a ( 0 )
2
3 =k∗0 σ
2
3∗1
3
2 =σ
σ = 2
3
By substituting the value of σ ∈¿ equation 3
2
3∗a
3
2 =kt+ 2
3
2
3 (a
3
2 −1)=kt
a3/2 – 1 = 3
2 kt

a3/2 = 3
2 kt + 1
a(t) = (1 + 3
2 kt)2/3
a(t) = (1 + 3
2 √ m
4 π t)2/3
Q4)
The steady flow rate of water through soil is given by keah
(1− dh
dz )=−e
Where, h is the capillary potential
A is a constant
K > 0 is the hydraulic conductivity constant
The initial condition is h = 0 at z = l (depth of water table)
a)
keah
(1− dh
dz )=−R
( 1−dh
dz )=−R
k e−ah
dh
dz =1+ R
k e−ah ………………………………………..1
b)
given that

u = eah
differentiate with respect to z
du
dz = ( eah ) a dh
dz
du
dz = 1
a ( eah ) a dh
dz
Comparing with equation 1
(1 + R
k e−ah ¿= 1
a ( e−ah ) du
dz
u = eah
now
(1+ R
ku )= u
a
du
dz
∫ a ( 1
u + R
k ( 1
u2 ) )du=∫dz
alnu− R
ku = z+C
aln eah− R
k e−ah=z +C
h = 0, z = L
then,
aln e0 − R
k e0=L+C
a = −R
k −L=C

aln eah− R
k e−ah=z +a− R
k −L
Z = aln eah− R
k e−ah−a+ R
k + L