Understanding Engineering Mathematics

giving the required DE

d^2 y

dx^2

+ 4 y= 0

(b) Similarly:
d^2 y
dx^2

=−12 cos 2x=− 4 y

so

d^2 y

dx^2

+ 4 y= 0

i.e. this function is also a solution.

(c)

d^2 y

dx^2

=−8sin2x+4cos2x

=− 4 (2sin2x−cos 2x)

=− 4 y

and again therefore

d^2 y

dx^2

+ 4 y= 0

You may again realise that since both sin 2xand cos 2xare solutions of the equation,

andsince it is a linearequation, then any function of the form

y=Acos 2x+Bsin 2x

whereAandBare arbitrary constants isalsoa solution.

If we tryy=2sinxwe have

d^2 y

dx^2

=−2sinx=−y

so this satisfies

d^2 y

dx^2

+y= 0

which isnotthe required equation.

Problem 15.5

Find the general solution of Problem 15.2(i) and find particular solutions

satisfying

(a) y=1atx=0(b)y=0atx= 1

In Problem 15.3 we were asked to check a given solution to the equation

dy

dx

=x^2 .Here

we wish to actuallyfindthe most general possible solution. We can in fact directly integrate

the equation in this case, remembering to add an arbitrary constant.