MAT1512
ASSIGNMENT 4
2025
, QUESTION 1
a).
dp
= t 2 p − 5p + t 2 − 5
dt
dp
= (t 2 − 5)p + (t 2 − 5)
dt
dp
= (t 2 − 5)(p + 1)
dt
1
dp = (t 2 − 5)dt
(p + 1)
1
∫ dt = ∫(t 2 − 5) dt
(p + 1)
t3
ln(p + 1) = − 5t + C
3
t3
( −5t+C)
p+1= e 3
t3
( −5t) C
p+1= e 3 e ∴ Let ∶ A = eC
t3
( −5t)
p + 1 = Ae 3
t3
( −5t)
p(t) = Ae 3 −1
b).
dy
cos(x) sec 2 (y) = csc 2 (y) sec(x)
dx
1 1 1
∴ sec(x) = , sec(y) = and csc(y) =
cos(x) cos(y) sin(y)
dy
cos(x) sec 2 (y) = csc 2 (y) sec(x)
dx
ASSIGNMENT 4
2025
, QUESTION 1
a).
dp
= t 2 p − 5p + t 2 − 5
dt
dp
= (t 2 − 5)p + (t 2 − 5)
dt
dp
= (t 2 − 5)(p + 1)
dt
1
dp = (t 2 − 5)dt
(p + 1)
1
∫ dt = ∫(t 2 − 5) dt
(p + 1)
t3
ln(p + 1) = − 5t + C
3
t3
( −5t+C)
p+1= e 3
t3
( −5t) C
p+1= e 3 e ∴ Let ∶ A = eC
t3
( −5t)
p + 1 = Ae 3
t3
( −5t)
p(t) = Ae 3 −1
b).
dy
cos(x) sec 2 (y) = csc 2 (y) sec(x)
dx
1 1 1
∴ sec(x) = , sec(y) = and csc(y) =
cos(x) cos(y) sin(y)
dy
cos(x) sec 2 (y) = csc 2 (y) sec(x)
dx
