Understanding Engineering Mathematics

(back where we started, but bear with me on this – we’ll come back to it later). We know
that the left-hand side must now be the derivative of a product, namelyIy, because that
is precisely why we chose the integrating factor:

d

dx

(Iy)=

d

dx

(xy)=x^2

Integrating both sides finally gives

xy=

x^3

3

+C

(Nowwe introduce the arbitrary constant), so the general solution is

y=

x^2

3

+

C

x

Problem 15.13

Repeat the solution but dispense with the integrating factor and solve the

original equation directly.

We ended up, in Problem 15.12, after finding the integrating factor, with the original
equation:

x

dy

dx

+y=x^2

In fact, in this case, the IF is really using a sledgehammer to crack a nut. Simply notice
that by reversing the product rule

x

dy

dx

+y≡

d

dx

(xy)

and we have

d

dx

(xy)=x^2

which is where we got to after finding the integrating factor. We are therefore led directly
to the result

xy=

x^3

3

+C

In general of course it may not be quite so easy to spot the required derivative of a
product, and we may need to go through the full procedure of finding the IF, but try to
avoid it when you can. You can in any case see that the better developed your skills of
differentiation, the easier it will be to spot such short cuts.

Exercise on 15.4

Find integrating factors for the following equations and hence obtain the general solution

(i) xy′+y=x (ii) xy′− 2 y=x^3 + 2

Can you dispense with the integrating factor, by finding a derivative of a product?