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Exam (elaborations)

APM 2611- DIFFERENTIAL EQUATIONS-2019

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APM 2611- DIFFERENTIAL EQUATIONS-2019

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  • September 23, 2022
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APM2611/201/1/2019




APM 2611- DIFFERENTIAL EQUATIONS-2019




Tutorial letter 201/1/2019


DIFFERENTIAL EQUATIONS
APM2611

Semester 1


Department of Mathematical Sciences


This tutorial letter contains solutions for assignment 01.




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Downloaded by Anna Maina (annamurugijoe@gmail.com)

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Question 1


(1.1) Find an interval of definition for the equation

dx
x2 = 3x3t2.
dt
Since the equation has the solution x(t) = 0 for all t, any interval may be taken as an interval
of definition. The largest interval of definition is (−∞, ∞).

(1.2) Solve the initial value problem
dx
x2 = 3x3t2, x(0) = 1
dt
and give an interval of definition.
Notice that x = 0 is a solution for the equation, but does not satisfy the initial value problem.
The equation is separable, separating variables yields
dx
= 3t2 dt.
x
Integrating both sides yields
ln |x| = t3 + c
where c is the constant of integration. Since x(0) = 1

ln |1| = 03 + c

we have c = 0. Thus
3
ln |x| = t3 ⇒ |x| = et .

Since x(0) = 1 (which is positive), and by continuity of x(t)
3
x = et .

Since the solution is valid for all t, an interval of definition is (−∞ , ∞ ) or any other interval
which contains the initial value 0.




2

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