MAT1512 Assignment 4 2025 - Due August 2025

MAT1512
ASSIGNMENT 4 2025
DUE: AUGUST 2025 (MEMO)

, Assignment 4, 2025 - MAT1512
July 2, 2025

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Question 1

Solve the following differential equations:

dp
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(a) = t2 p − 5p + t2 − 5.
dt
dy
(b) cos(x) sec2 (y) = csc2 (y) sec(x).
dx
(c) (ey − 5)y ′ = 2 + cos x.




1. Differential equations ​




dp
(a) = t2 p − 5p + t2 − 5.
dt
First rewrite the equation by factoring:

dp
= (t2 − 5)p + (t2 − 5) = (t2 − 5)(p + 1).
dt
This is a separable first-order ODE. We separate variables as follows

dp
= (t2 − 5) dt.
p+1

Integrating both sides gives

dp 2 t3
∫ = ∫ (t − 5) dt ⟹ ln ∣p + 1∣ = − 5t + C.
p+1 3
3
Exponentiating yields p + 1 = C ′ e t /3−5t where C ′ = eC . Hence the general solution is
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3
p(t) = C ′ e t /3−5t − 1.

We can rename the constant C ′ as C . Thus,

t3
p(t) = Ce 3 −5t − 1,
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where C is an arbitrary constant.