Understanding Engineering Mathematics

This can now be solved by direct integration:

Iy=

∫

IQdx+C

and we finally get the solution

y=

1

I

(∫

IQdx+C

)

=

1

I

∫

IQdx+

C
I
Thus, the purpose of multiplying by the integrating factor is to convert the left-hand
side to the derivative of a product so that we can integrate the resulting equation directly.
But all this depends on findingIfrom the equation

dI

dx

=IP

Thisequation is separable:

dI

I

=Pdx

so ∫
dI
I

=

∫

P(x)dx

Note that we needn’t bother to introduce an arbitrary constant here – you can if you like,
but it will simply cancel out in the end.
Thus, forIwe obtain

lnI=

∫

P(x)dx

So the integrating factor is given by

I=e

∫

P(x)dx

I don’t encourage you to remember the above results, either forI or the solution for
y. Rather, you should try to remember how to reproduce the above arguments to derive
solutions directly yourself. This may seem daunting, but with plenty of practice it is not
too bad, and is a very useful skill. To help with this the following problems work through
the procedure step by step. It also illustrates that you can often short cut the process of
finding the integrating factor by rearranging the left-hand side of the original equation to
be the derivative of a product ‘by inspection’.

Problem 15.9

Convert the DE

xy′Yy=x^2

to standard linear form.