1550078481-Ordinary_Differential_Equations__Roberts_

150 Ordinary Differential Equations

Since t he different ial equation in (3a) involves the single variable q 1 and is

linear, we can solve the init ial value problem (3a) expli citly and substitute the

result into (3b). The resulting initial value problem will involve the variable

q2 only. We will then use SOLVEIVP to solve this initial value problem

numerically. (In chapter 7, we will show how to solve the system of two, first-

order, initial value problems consisting of (3a) and (3b) simultaneously- that

is, we will show how to find q 1 and q 2 without first solving for q 1 explicitly

and substituting into (3b).)

In chapter 2 , we proved that the general solution of the nonhomogeneous

li near first-order differential equation y' = a( t )y + b( t) where a( t) and b( t) are

continuous on the interval I is

(4) y(t) = Y1(t)(K + v(t))

where K is an arbitrary constant and

Y1(t) = ef'a(x)dx and

f

t b(x)

v(t) = Yi(x)dx.

Equation (3a) is a nonhomogeneous linear differential equation in which y(t) =

q1(t), a(t) = -1/30, and b(t) = .5. Hence,

Yi(t) = e-f'dx/3 0 = e-t/3o+c,

where C 1 is a constant of integration. Since we need only one constant in the

general solution- which is the constant K - we set C 1 = 0. Substituting t he

resulting expression for y 1 (t) into the equation for v(t), yields

f

v(t) = t --db(x) x = f t -·-^5 dx = .5 f t exl^30 dx = 15 et/^3 o + C

Y1(x) e -x/3 0 2

where C2 is a constant of integration. Again since we need only one constant
in the general solution , we set C 2 = 0. Substituting our expressions for y 1 (t)

and v(t) with C1 = C2 = 0 into equation (4), we find the general solution for

equation (3a) to be

q1(t) = Y1(t)(K + v(t)) = e-t/^30 (K + 15et/^3 o) = Ke-t/^3 o + 15.

Since q1(0) = 90 = K + 1 5, we have K = 90 - 1 5 = 75. Hence,

q1(t) = 7 5e-t/^3 o + 15.

Substituting this expression for q 1 (t) into the differential equation of (3b), we
find q2 must satisfy the init ial value problem

(5) dq2 = 2 5 -t/30 + 5 - q2.

dt · e · 15'