Solving Differential Equations for Soil Water Flow

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