082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

548 Chapter 19 Ordinary differential equations I
Consider again the differential equation of the previous example:
d
dx (x2 y) =cosx
Using the product rulefordifferentiation we can expand the l.h.s. as follows:
d
dx (x2 y) =x 2dy
dx +2xy
Doing this, the differential equation can bewritten inthe equivalent form
x 2dy
dx +2xy=cosx
Suppose we had posed the question in this form. A method of solving this equation
would be to recognize that the equation is exact and that the l.h.s. could be written as
d
dx (x2 y).
Itiseasy torecognize an exact equation because itwill always take the form
µ dy
dx +µ′ y=f(x)
where µ is some function ofx. That is, the coefficient ofyis the derivative of the coefficient
of dy
dx . When thisisthe case the l.h.s. can be written d dx (µy).
Example19.13 The following equations are exact. Note in each case that the coefficient of y is the
derivative ofthe coefficient of dy . Solve them.
dx
(a) x 3dy
dx +3x2 y =e 2x
(b) cosx dy − (sinx)y =1
dx
Solution (a) The equation can be written
d
dx (x3 y) =e 2x
and so, upon integrating,
∫
x 3 y = e 2x dx
so that
= e2x
2 +c
y = e2x
2x 3 + c x 3
(b) The equation can be written
d
((cosx)y) =1
dx