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MATH168-DIFFERENTIALEQUATIONSI
solutions to first order odes - exact differential equations 1 1 This is the first part of lecture four
Lecturer: Dr. Peter Amoako-Yirenkyi - A series of lecture notes primarily
taken from the text book (Elementary
2 differential Equation and boundary
Recommended Textbook:Elementary differential Equation value problems by Boyce and Diprima.
The laws of the universe are written largely in the language of math- After going through this lecture notes,
ematics. Algebra is sufficient to solve many static problems, but the you should be able to:
most interesting naturally phenomena involve change and are best de- • Youshouldbeable to understand
scribed by equations that relate changing quantities. Many important the principle behind Exactness of
and significant problems in engineering, the physical sciences, and the differential Equations
social sciences such as economics and business when formulated in • Find the solution to exact differen-
mathematical terms require the determination of a function satisfying tial equations
an equation containing the derivatives of unknown function. Such • Solve non-exact differential equa-
equations are called differential equation. tions using method of integrating
factors
2 W.W. Boyce & R.C. DiPrima. Elemen-
Exact Differential Equation tary Differential Equations and Boundary
Value Problems. John Wiley & Sons,
Inc., tenth edition, March 2012. ISBN
Through out the previous weeks we have been discussing first order 978-0-470-45832-7
differential equations and a number of methods that are applicable to various classes of problems. The
most important of these had been separable differential equations and linear coefficients of the differential
form of some differential equations(both homogeneous and non-homogeneous). Although these methods
can solve a number of differential equations, first order equations that can be solved by these methods are
rather special since most first order differential equations cannot be solved in this way. The next class of
equations we need to consider are known as exact differential equations which employs the idea of total
differential.
Given a function f(x,y), its total differential, df , is defined as:
df = ∂f dx+ ∂f dy (1)
∂x ∂y
This shows that the family of curves (or general solution) f(x,y) = c
satisfies the differential equation
df =0 (2)
=⇒ ∂fdx+∂fdy =0 (3)
∂x ∂y
So if there exists a function f(x,y) such that:
M(x,y) = ∂f and (4)
∂x
N(x,y) = ∂f (5)
∂y
3 3
then M(x,y)dx+N(x,y)dy is called an exact differential , and the In order to establish that a differential
equation: equation is exact, recall that in partial
M(x,y)dx+N(x,y)dy = 0 (6) derivatives if f(x,y) has continuous
second order partial derivatives, then:
is said to be an exact equation, whose solution is the family of func- ∂2f ∂2f
tions f(x,y) = c . ∂x∂y = ∂y∂x
(that is, the order of differentiation does
not matter). So if ∂f = M and ∂f = N
∂x ∂y
then: ∂M = ∂N (which is ∂2f = ∂2f )
∂y ∂x ∂x∂y ∂y∂x
Not only is this condition necessary for
exactness, it also determines exactness.
math168 - solutions to odes - exact differential equations and integration factors 2
Example 1. Solve the following differential equation:
(1−2xy)dx+(4y3−x2)dy =0
Solution
With (M(x,y) = 1−2xy) and N(x,y) = 4y3 −x2, we notice that:
∂M = −2x = ∂N So the given functions M(x,y) and N(x,y) passed
∂y ∂x 4
the test for exactness . All that is left to do is to find the function 4
f (x, y) such that: ∂f = M and ∂f = N. The way to accomplish this
∂x ∂y Example 2. Solve the differential equation
is to integrate M with respect to x, integrate N with respect to y, and dy
then merge the results: (3x2 +2xy)+(x+y2) =0
dx
Solution
Z Z Here M = 2x, N = 1,thegiven
2 x x
M(x,y)dx = (1−2xy)dx = x−x y+h(y) (10) equation is not exact. To see that it
Z Z cannot be solved by the procedure
M(x,y)dx = (4y3 −x2)dy = y4 −x2y+g(x) (11) above, let us suppose that there is a
function f(x,y) such that:
These calculations imply that a function f(x,y) which satisfies both: fx = 3x2 +2xy, fy = x+y2 (7)
Integrating the first of (7) gives:
∂f = M =1−2xy and ∂f = N =4y3−x2 is f(x,y) = x2y+y4 f (x, y) = x3 + x2y + ψ(y) (8)
∂x ∂y
where ψ(y) is an arbitrary function of y
Therefore, the exact equation we were given is satisfied by the family only. To try to satisfy the second of (7)
of curves: x − x2y+y4 = c wecompute fy from (8), obtaining:
f = x2+dψ =x+y2 or
Example 3. Solve the differential equation (ycosx + 2xey) + (sinx + y dy
x2ey +2)dy = 0 dψ 2 2
dx dy = x+y −x (9)
Solution Since the right-hand side of (9) depends
M((x,y) = ycosx+2xey), N(x,y) = sinx+x2ey+2 on x as well as y, it is impossible to
∂M = cosx+2xey, ∂N = cosx+2xey Therefore, the differential solve (9) for ψ(y). Thus there is no
∂y ∂x f (x, y) satisfying both of (7). So in the
equation is exact. We know that ∂f = M(x,y); ∂f = N(x,y). Thus next section we will try to find a factor
∂x ∂y to make the equation exact and have a
there is an f(x,y) such that function satisfying both (7) - the factor
∂f ∂f is called integrating factor.
=ycosx+2xey and =sinx+x2ey+2 (12)
∂x ∂x
5 Integrating the first of these equations: R dy = R (ycosx +2xey)dx, 5 Note that either of equations (10)
keeping y constant gives: having h(y) as a function of y alone or
(11) with g(x) as a function of x alone
f (x, y) = ysinx + x2ey +ψ(y) (13) maybechosenconveniently
and hence differentiating the new function w.r.t y keeping x constant:
∂f = sinx+x2eydψ = sinx+x2ey +2 i.e. the differential is equal to
∂y dy R R
the second equation in (12). Thus dψ = 2 =⇒ dψ = 2dy and
dy
ψ = 2y the constant of integration can be omitted; we do not require
the most general one. Substituting for ψ(y) in (13) gives f(x,y) =
ysinx+x2ey+2y. Before f(x.y) = c, constant. Hence c = ysinx+
x2ey +2y
math168 - solutions to odes - exact differential equations and integration factors 3
Using Integrating factors for Non-Exact Equations
(I f )M(x,y)dx + N(x,y)dy = 0 (14)
is not exact, sometimes we can turn it into 6an exact differential
equation by multiplying the whole equation by an appropriate factor,
called an integrating factor, let say µ = µ(x,y), which is a function of
x, y . For which we obtain: 6 Remark: Finding the integrating
factor for a given equation can be very
µM(x,y)dx+µN(x,y)dy =0 (15) difficult. Some of the important rules
and procedures follow.
and expect it to be exact. Then for (15) to be exact then: Theorem1. 1. Assume µ = µ(x) =⇒
∂µ = µy = 0, then: µx =
∂y
∂(µM) ∂(µN) dµ M −N
= = y x,whichisafunctionof
∂y ∂x dx N
x alone, call this function ξ(x). Then
Applying product rule we get: µ(x) = eR ξ(x)dx is an integrating factor
of (14)
2. Assume µ = µ(y) =⇒ ∂µ =
∂M ∂µ ∂N ∂µ ∂x
dµ M −N
µ +M =µ +N 0, then: µy = = y x,whichis
∂y ∂y ∂x ∂x dy −M
a function of y alone, call this function
∂µ ∂µ ψ(y). Then µ(y) = eR ψ(y)dy is an
=⇒ M −N =µ(∂N∂x−∂M∂y) (16) integrating factor of (14)
∂y ∂x 3. If ∂M + ∂N 6= 0, then: µ(x,y) =
∂x ∂y
1 , is the integrating factor of (14)
Simply put Mµ −Nµ =µ(N −M ), µ(M −N ) ∂M ∂N
y x x y y x ∂x + ∂y
Sometimes an integrating factor may be found by inspection, after 4. If M(x,y)dx + N(x,y)dy = 0 can
grouping the terms in the equation, and recognizing a certain group be written in the form yg(x,y)dx +
7 xh(x,y)dy = 0 where h(x,y) 6=
as being a part of an exact differential. The following tabulates g(x,y), then µ(x,y) = 1
possible terms and their respective Integrating factors and the exact xM−yN
equations they produce: 7 The following observations are often
helpful in finding integrating factors:
1. If a first-order differential equation
contains the combination xdy +
GroupofTerms Integrating factor Exact differential df(x,y) ydy = 1d(x2 +y2) try some function
2 22
ydx−xdy −1 xdy−ydx = d y x +y asamultiplier.
x2 x2 x 2. If a first-order differential equation
ydx−xdy 1 ydx−xdy = d x contains the combination xdy +
y2 y2 y ydx = d(xy), try some function xy as
ydx−xdy −1 xdy−ydx = d ln y a multiplier.
xy xy x
ydx−xdy − 1 xdy−ydx = d arctan y 3. If a first-order differential equation
x2+y2 x2+y2 x
1 ydx+xdy contains the combination xdy−ydx,
ydx+xdy xy xy =d(ln(xy)) try 1 or 1 as a multiplier. If
h i x2 y2
ydx+xdy 1 , n > 1 ydx+xdy = d −1 neither of these works, try 1 or
(xy)n (xy)n h(n−1)(xy)n−1 i xy
ydy+xdx 1 or some function of these
ydx+xdy 1 =d 1ln(x2+y2) x2+y2
x2+y2 x2+y2 2 expressions, as an integrating factor,
ydx+xdy 1 , n > 1 ydy+xdx = dh −1 i remember that dtan−1 y =
(x2+y2)n (x2+y2)n 2(n−1)(x2+y2)n−1 xdy−ydx y xxdy−ydx
aydx+bxdy xa−1yb−1 xa−1yb−1(aydx+bxdy) = d(xayb) xy and tan−1 x = x2+y2
NOTE:Ifanon-exact equation has a solution, then an integrating
factor is guaranteed to exist, but that does not mean it is easy to find.
math168 - solutions to odes - exact differential equations and integration factors 4
Example 4. Solve the differential equation 8 8
dy Example 5. Solve (y2 −y)dx+xdy = 0
(3xy+y2)+(x2+xy) =0
dx Solution
Solution
M(x,y) = 3xy+y2, N(x,y) = x2+xy M(x,y) = y2 −y, N(x,y) = x
M =3x+2yandN = 2x+yClearly, ∂M 6= ∂N,thereforethe M =2y−1, N =1
y x ∂y ∂x y x
differential equation is not exact. Involve (16): Mµ − Nµ = µ(N − Hence the differential equal is not
y x x exact and no integrating factor is
My). Let µ be a function of x alone, that is, µ = µ(x). Applying 1 immediately apparent. Note, however,
M −N 3x+2y−(2x+y) 1 that if terms are strategically regrouped,
ξ(x) = y x = = the differential equation can be written
N x2 +xy x as:
R ξ(x)dx R 1dx lnx −(ydx−xdy)+ydx (17)
µ(x) = e =e x =e =x The group of terms in parenthesis
has many integrating factors in table
. Multiplying µ(x) = x through the differential equation we obtain: (above). Trying each integrating factor
separately, we find that the only one
(3x2y+xy2)+(x3+x2y)dy =0 that makes the entire equating exact is:
dx µ(x,y) = 1 ,
y2
which is a new equation in same form i.e.: M(x,y)+ N(x,y)y′ = 0 Using this integrating factor we can
M(x,y) = (3x2y+xy2); N(x,y) = (x3+x2y) rewrite (17) as:
−ydx−xdy +1dx=0 (18)
M =3x2+2xy; N =3x2+2xy y2
y x
which is exact. Let f = 3x2y+xy2. Integrating: Since (18) is exact, it can now be solved.
x Alternatively, we note from table(above)
Z that (18) can be rewritten as −d(x/y)+
f (x, y) = (3x2y+xy2)dx = x3y+ 1x2y2+ψ(y) 1dx = 0 or as dx = 1dx. Integrating,
2 y
weobtain the solution:
Find the derivative w.r.t y and equating with N(x,y) we obtain x = x+c or y = x
y x+c
=⇒ fy = x3+x2yψ′(y) = x3+x2y
ψ′(y) = 0 =⇒ ψ(y)=c.
Finally x3y+ 1x2y2 = c
2
Linear Equations and those Reducible to the form
Afirst-order linear equation is defined as a differential of the form:
A(x)y′ +B(x)y = C(x) (19)
where A(x) 6= 0 such that the derivative of the dependent variable
exist otherwise it is not a differential equation. Then (16) in the stan-
dard form is written as
y′ + p(x)y = q(x) (20)
p(x) = B(x) and q(x) = C(x)
A(x) A(x)
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