544 CHAPTER 29. FIRST ORDER SCALAR ODE

Let the initial position be at (0,0) and let the coordinates of the point be (x(t) ,y(t)) .What is the initial velocity? It is

(30√

3,30). Then the acceleration is given by(

x′′ (t) ,y′′ (t))=−32(0,1)− .2

(x′ (t) ,y′ (t)

)Thus y′′+ .2y′ =−32. Let’s solve for y′.(

e.2ty′)′= (−32)e.2t

e.2ty′ (t) =−32.2

e.2t +C, so y′ (t) =−160+Ce−.2t

So what is C? When t = 0, we get C−160 = 30 and so C = 190. Hence

y′ (t) = 190e−.2t −160

y(t) =−160t−950e−0.2t +D

What is D? When t = 0 we want y(0) = 0 and so D = 950. Thus

y(t) =−160t−950e−0.2t +950

As to x,x′′+ .2x′ = 0 so

(x′e.2t)′ = 0

and so x′ (t) =Ce−.2t . To satisfy the initial condition, x′ (t) = 30√

3e−.2t . Then

x(t) =30√

3−(1/5)

e−.2t +D

What is D? to satisfy the initial condition for the position, D = 150√

3 and so

x(t) =−150√

3e−.2t +150√

3

The position of the pumpkin is

(x(t) ,y(t)) =(−150

√3e−.2t +150

√3,−160t−950e−0.2t +950

)The following is a summary of the above discussion.

PROCEDURE 29.1.15 To solve the first order linear differential equation

y′+a(t)y = f (t) ,

do the following:

1. Find A(t) ∈∫

a(t)dt. That is, find A(t) such that A′ (t) = a(t).

2. Multiply both sides by the integrating factor eA(t).

3. The above step yields (eA(t)y(t)

)′= eA(t) f (t)

4. Do∫

dt to both sides. Then choose the arbitrary constant to satisfy a given initialcondition.