DIFFERENTIAL EQUATIONS I
(MATH 251)
Fredrick Asenso Wireko
Department Of Mathematics
Kwame Nkrumah University Of Science and Technology
March 16, 2022
Fredrick Asenso Wireko Department Of Mathematics
DIFFERENTIAL
Kwame
EQUATIONS
Nkrumah University
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March 16, 2022
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FIFTH LECTURE
Fredrick Asenso Wireko Department Of Mathematics
DIFFERENTIAL
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EQUATIONS
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Integrating Factors for Non - Exact Equations
Let
M(x, y )dx + N(x, y )dy = 0
(1)
be non-exact. We need to ensure that the differential equation is exact.
To do this, we multiply (41) by an appropriate factor
µ = µ(x, y )
(2)
called an Integrating factor, which is a function of x and y . To obtain :
µM(x, y )dx + µN(x, y )dy = 0
(3)
and expect it to be exact. Then for (3) to be exact then :
δ(µN)
δ(µM)
=
δy
δx
Fredrick Asenso Wireko Department Of Mathematics
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Cont...
Applying the Product rule,
δµ
δN
δµ
δM
+M
=µ
+N
δy
δy
δx
δx
δµ
δµ
δN
δM
=⇒ M
−N
=µ
−
δy
δx
δx
δy
µ
(5)
(6)
Simply put
Mµy − Nµx = µ(Nx − My )
Sometimes an integrating factor may be found by inspection, and
recognizing a certain group as being part of an exact differential. The ff
tabulates possible terms and their respective integrating factors and the
exact equations they produce :
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Cont...
Group of Terms
Integrating Factor
Exact differential
xdy −ydx
x2
= d( yx )
ydx − xdy
- x12
ydx − xdy
1
y2
ydx−xdy
y2
= d( yx )
ydx − xdy
1
xy
xdy −ydx
xy
= d(ln yx )
ydx − xdy
1
- x 2 +y
2
ydx + xdy
1
xy
ydx + xdy
1
(xy )n ,
aydx + bxdy
x a−1 y b−1
xdy −ydx
xy
= d(arctan yx )
ydx+xdy
xy
n>1
ydx+xdy
(xy )n
= d(ln(x, y ))
−1
= d (n−1)(xy )n−1
x a−1 y b−1 (aydx + bxdy ) = d(x a y b )
Fredrick Asenso Wireko Department Of Mathematics
DIFFERENTIAL
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EQUATIONS
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Some Important Rules For Integrating Factor
1
2
M −N
dµ
y
x
Assume µ = µ(x) =⇒ δµ
,
δy = µy = 0, then µx = dx =
N
which
is
a
function
of
x
alone,
call
this
function
ξ(x).
Then
µ(x)
=
R
ξ(x)dx
e
is an integrating factor of (1)
Assume µ = µ(y ) =⇒
δµ
δx
= 0, then µy =
dµ
dy
=
My −Nx
−M ,
which is a
function of y alone, call this function ψ(y ). Then ψ(y ) = e
an integrating factor of (1)
3
If
δM
δx
+
δN
δy
6= 0, then µ(x, y ) =
1
δM
+ δN
δx
δy
R
ψ(y )dy
is
, is the integrating factor of
(1)
4
If M(x, y )dx + N(x, y )dy = 0 can be written in the form yg (x, y )dx
1
+ xh(x, y )dy = 0 where h(x, y ) 6= g (x, y ), then µ(x, y ) = xM−yN
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Some Observations For Finding Integrating Factors
1
If a first-order differential equation contains the combination
xdy + ydy = 21 d(x 2 + y 2 ) try some function x 2 + y 2 as a multiplier.
2
If a first-order differential equation contains the combination
xdy + ydx = d(x, y ), try some function xy as a multiplier.
3
If a first-order differential equation contains the combination
xdy − ydx, try x12 or y12 as a multiplier. If neither of these works, then
1
1
xy or x 2 +y 2 or some function of these expressions, as an integrating
factor, remember that dtan−1 ( yx ) =
xdy −ydx
xy
and tan−1 ( yx ) =
xdy −ydx
x 2 +y 2
Fredrick Asenso Wireko Department Of Mathematics
DIFFERENTIAL
Kwame
EQUATIONS
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Example
Solve the differential equation
(3xy + y 2 ) + (x 2 + xy )
dy
=0
dx
Solution : M(x, y ) = 3xy + y 2 , N(x, y ) = x 2 + xy
My = 3x + 2y and Nx = 2x + y , Clearly, δM
δy 6=
differential equation is not exact. Involving
δN
δx ,
therefore the
Mµy − Nµx = µ(Nx − My ).
Let µ be a function of x alone, that is µ = µ(x).
ξ(x) =
M y − Nx
3x + 2y − (2x + y )
1
=
=
2
N
x + xy
x
Fredrick Asenso Wireko Department Of Mathematics
DIFFERENTIAL
Kwame
EQUATIONS
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Cont...
µ(x) = e
R
ξ(x)dx
=e
R
1
dx
x
= e lnx = x
Multiplying µ(x) = x through the differential equation we obtain:
(3x 2 y + xy 2 ) + (x 3 + x 2 y )
dy
=0
dx
which is a new equation in the same form i.e M(x, y ) + N(x, y )y’ = 0
M(x, y ) = (3x 2 y + xy 2 ); N(x, y ) = (x 3 + x 2 y )
My = 3x 2 + 2xy ; Nx = 3x 2 + 2xy
which is exact. Let fx = 3x 2 y + xy 2 . Integrating:
Z
1
f (x, y ) = (3x 2 y + xy 2 )dx = x 3 y + x 2 y 2 + ψ(y )
2
Fredrick Asenso Wireko Department Of Mathematics
DIFFERENTIAL
Kwame
EQUATIONS
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Cont...
Find the derivative w.r.t y and equating with N(x, y ) we obtain
=⇒ fy = x 3 + x 2 y ψ 0 (y ) = x 3 + x 2 y
ψ 0 (y ) = 0 =⇒ ψ(y ) = c
Finally
1
x 3y + x 2y 2 = c
2
Fredrick Asenso Wireko Department Of Mathematics
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EQUATIONS
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Equations Reducible to Linear Form
Other first order equations in first degree which are not linear may be
reduced to the linear form by means of appropriate transformations. No
general rule can be stated; in each instance, the proper transformation is
suggested by the form of the equation. Examples of such equations are
Bernoulli’s equation, Ricatti’s equation and the one considered below.
Consider
{xs(y ) + t(y )}y 0 = µ(y )
(7)
in which the co-efficient of y’ is a linear function of x and no other x
appears anywhere again.
Thus (7) can be expressed as a linear differential equation with x as the
dependent variable and y the independent variable. To be able to do this ,
dx
dx
dx
to obtain: µ(y ) dy
= xs(y ) + t(y ) or µ(y) dy
multiply (7) through by dy
s(y )x = t(y ).
Thus this is a linear differential equation of x as the function of the
independent variable y
Fredrick Asenso Wireko Department Of Mathematics
DIFFERENTIAL
Kwame
EQUATIONS
Nkrumah University
I (MATHOf251)
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Example
Solve the differential equation
(3x − 4y 3 )y 0 + y = 0
Solution : Clearly the co-efficient of y 0 is linear in x and further more x
dx
to obtain :
appears no where again. Therefore we multiply through by dy
(3x − 4y 3 ) + y
whose standard form is :
dx
dx
= 0 =⇒ y
+ 3x = 4y 3
dy
dy
3
dx
+ x = 4y 2
dy
y
Fredrick Asenso Wireko Department Of Mathematics
DIFFERENTIAL
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EQUATIONS
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Cont...
Integrating factor:
µ(y ) = e
Z
3
R
dy
y
3
= e lny = y 3 =⇒ y 3
d(xy 3 ) =
Z
dx
+ 3y 2 x = 4y 5
dy
4y 5 =⇒ xy 3 = (4/6)y 6 + c
=⇒ 3xy 3 = 2y 6 + c
Fredrick Asenso Wireko Department Of Mathematics
DIFFERENTIAL
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EQUATIONS
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Bernoulli’s Equation
A Bernoulli equation is a first order differential of the form
y 0 + p(x)y = q(x)y n
(8)
where p(x) and q(x) are continuous functions on the interval of discussion
and n is a real number. This equation differs from a standardized linear
differential equation in first degree at the right side of the equation. Here
there is a multiplication of y n , n 6= 0, n 6= 1, at the R.H.S. Then divide
(8) through by y n and substitute z = y 1−n .
y 0 y −n + p(x)y 1−n = q(x)
0
z
= y n y 0 . The equation
Let z = y 1−n =⇒ z 0 = (1 − n)y −n y 0 =⇒ 1−n
becomes:
z 0 + (1 − n)p(x)z = (1 − n)q(x).
The solution is obtained using first order linear equation in first degree
solution techniques.
Fredrick Asenso Wireko Department Of Mathematics
DIFFERENTIAL
Kwame
EQUATIONS
Nkrumah University
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Example
Solve the differential equation
4
y 0 + y = x 3y 2
x
y (2) = −1,
x >0
Solution : Write it in the form y 0 + p(x)y = q(x)y n that is :
4
y 0 + y = x 3y 2
x
(9)
Divide through by y −2
4
y −2 y 0 + y −1 = x 3
x
Let z = y −1 and z 0 = −y −2 y 0 which is linear first order in first degree
=⇒ µ(x) = e
R
− x4 dx
= e −4ln|x| = x −4
Fredrick Asenso Wireko Department Of Mathematics
DIFFERENTIAL
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Cont...
Therefore,
0
Z
Z −4
z.x =
x v dx = −x −1 dx
4
x −4 v = −ln|x| + c =⇒ z(x) = cx 4 − x 4 lnx
but
z = y −1 =⇒ y −1 = x 4 (c − lnx)
Applying the initial conditions,
(−1)−1 = c24 − 24 ln2 =⇒ c = ln2 −
y (x) =
x 4 (ln2
1
16
1
−16
= 4
1
x (1 + 16ln x2 )
− 16 − lnx)
Fredrick Asenso Wireko Department Of Mathematics
DIFFERENTIAL
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Cont...
Solve the differential equations
x 2y − x 3
dy
= y 4 cosx
dx
Solution : Write it in the form y 0 + p(x)y = q(x)y n that is :
y0 −
y
cosx
= − 3 y4
x
x
(10)
Divide through by y 3 :
y −4 y 0 −
y −3
cosx
=− 3
x
x
, Let z = y −3 and z 0 = −3y −4 y 0 , which is linear first order in first degree
=⇒ µ(x) = e
R
p(x)dx
=e
R
3
dx
x
= e 3lnx = x 3
Fredrick Asenso Wireko Department Of Mathematics
DIFFERENTIAL
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Cont...
Therefore,
3
z.x =
Z
3cosx 3
x dx =
x
Z
3cosxdx =⇒ zx 3 = sinx + c
But
z = y −3 =⇒
Hence,
y3 =
x3
= 3sinx + c
y3
x3
3sinx + c
Fredrick Asenso Wireko Department Of Mathematics
DIFFERENTIAL
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EQUATIONS
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Cont...
Exercise
Solve the following IVP
y 0 = 5y + e −2x y 2
y (0) = 2
y √
− y =0
x
y (1) = 0
and
y0 +
Fredrick Asenso Wireko Department Of Mathematics
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