Calculus: Analytic Geometry and Calculus, with Vectors

4.1 Indefinite integrals 209

example, find the solutions of the following differential equations satisfying the
given boundary conditions.

(a) Lx=0,y =1when x=0 11ns.:y = 1

(b)±x=1,y=2when x=3 Ans.:y=x-1

dy

(c) =cos 2x, y = 0 when x = 0 lIns.: y = sin 2x

dx

(d)ay=ea=,y=1when x=0

dx

Ans.:y=-Tear+s

9 A body moves to and fro on a line in such a way that its scalar velocity

v at time t is given by the formula

v=t2-8t+15.

During what interval of time is the scalar velocity negative, and how far does the

body move during that time? Hint: Ifs is the coordinate of the body at time t,

then ds/dt = v and hence

to

3 -4t2+15t+c,

where c is a constant that is 0 if we choose the origin such that s = 0 when t = 0.

11n$.: g units.

10 If y is a function of x satisfying the differential equation

(1)

and if we know that y > 0, then we can divide by y to obtain the first and then

the rest of the formulas

(2) y dx =

k, log y = kx + c, y = ekx+o, y = ekxe°,

where c is a constant that depends upon the particular function y with which

we started. But ec is a constant that we can call .1, so

(3) y = Aekx.

If we know that y satisfies the boundary condition y = yo when x = 0, then we

can put x = 0 in (3) to find that A = yo and hence

(4) y = yoekx

Remark: Without assuming that y 5-6 0, we can solve (1) with the aid of a trick.

Transposing a term in (1) and multiplying bye kx give the first and hence the

second of the formulas

ekx(dx-ky)=0, de kxy=0.