Solving Differential Equations for Soil Water Flow

The article explains how to solve differential equations for soil water flow, with a focus on the steady flow rate of water through soil and hydraulic conductivity constant. It also includes initial conditions and step-by-step solutions.

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