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Ordinary Differential Equations (ODE)
SK/EUM111/ODE/09 1
Ordinary Differential Equations (ODE)
LEARNING OBJECTIVES
The objectives of this chapter are to:
1. Impart to students the knowledge of ordinary differential equations and their classifications.2. Impart to students the concept of solutions of differential equations.3. Equip students with the knowledge of solving first order ordinary differential equations using
the method of separating the variables.
4. Equip students with the knowledge of solving linear first order ordinary differential equationsusing integrating factor.
5. Equip students with the knowledge of solving homogenous first order ordinary differentialequations.
6. Equip students with the knowledge of solving linear second order ordinary differentialequations with constant coefficients.
Differential equations form the language in which the basic laws of science are expressed. The science
tells us how the system at hand changes "from one instant to the next." The challenge addressed by thetheory of differential equations is to take this short-term information and obtain information about
long-term overall behavior. So the art and practice of differential equations involves the followingsequence of steps: one "models" a system (physical, chemical, biological, economic, or even
mathematical) by means of a differential equation; one then attempts to gain information aboutsolutions of this equation; and one then translates this mathematical information back into the
scientific context.
Differential Equation:
Short terminformation
Solving Behaviour over time
Physical World
InterpretationModel
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A basic example is given by Newton's law, F = ma. a = acceleration, the second derivative of x =position. Forces don't effect x directly, but only through its derivatives. This is a second order ODE,
and we will study second order ODEs extensively later in the course.
Definition: A differential equation is an equation containing derivatives of an unknown function. A
differential equation (DE) contains one or more terms involving derivatives of one variable (thedependent variable,y) with respect to another single independent variable (the independent variable,x) is said to be anordinary differential equation.
In the DE 3+4x+7x=dx
dy 2 y=f(x) is the unknown function.dx
dyis the derivative of the
unknown function.
For example, xdx
dy wherey is the dependentvariable and
x is the independentvariable
txdt
dx wherex is the dependentvariable and
tis the independentvariable
The DE is an ordinary DE (ODE) if the unknown function depends on only one independent variable.
The following are ODE
5x=y+)dx
dy3(+)
dx
yd((d)
1=)dx
dy2(+
dx
dyec)(
t
1+x=
dt
dx(b)
4-2x=dx
dya)
263
2
2
3y
2
An equation involving thepartial derivatives of one or more terms involving derivatives of onevariable oftwo or more independent variables is called apartial differential equation.
(Partial differential equations)2
2
u u
t x
(heat equation)
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2 2
2 20
u u
x y
(Laplaces equation )
Order: The order of a differential equation is the order of the highest derivative that appears in the DE
equation.
First-Order : dy/dx = 3x2+10x+3 , y = 3x
3+4x+2
Second-Order: d2y/dx2 = 4x + 5, y = 5x4+3
Third-Order:y = 6x +2
Forth-Order:y(4) = 4x2
+ 4x+ 7
A general nth-order, ordinary differential equation is often represented by the
symbolism
, , ,...,n
n
dy d yF x y
dx dx
Thus first-order differential equations, contain onlyy' and may containy and functions ofx. Hence it is
written asF(x,y,y) =0
second-order differential equations, contain onlyyand may containy,y and functions ofx. Hence itis written as
F(x,y,y,y) =0
Degree: The degree of a DE is the highest power of the highest derivative (when the derivatives arecleared of radicals and fractions).
Linear: A DE. is said to be linear if the dependent variable and its derivative are of degree one andthere are no product involving the dependent variable ,the derivative and in transcendental functions
such as sine, exponential.
A DE is linear if it can be written in the form :1
1 1 0n 1
d y d y dy(x) +a (x) +...+a (x) +a (x) y = f(x)
dx dx dx
n n
n n n
a
. (1)
Two properties:
i) The dependent variables y and all its derivative are of first degree (power of each term isinvolving only 1)
ii) Each coefficient depends on only the independent variable x. iii) Not involving the dependent variable in transcendental functions such as sine, exponential
.For example
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Differential Equation Linearity
1.tx
dt
dx
Linear, ordinary and order is 1
2.t
dt
dx
dt
xd 3
2
2
Linear, ordinary and order is 2
3.y
dx
dy2)(
non-linear (y)2
2.tx
dt
dxx non-linear (
dxx
dt)
4. y x y y x4 2 cos is linear, ordinary and order is 2
5. y y y y x4 2 cos is nonlinear y y
6.
2
2
u
x
v
tu v u sin
is linear in v but nonlinear in u sinu . Theequation is nonlinear
7. d x
dt
dy
d tx y t
2
2 sin
is linear in each of the dependent variables x and
y. But it is nonlinear in the set of {x, y}. The
equation is nonlinear
Homogeneous or NonHomogeneous Equation
From equation (1), it can be written as
1
1 1 0n 1
d y d y dy(x) +a (x) +...+a (x) +a (x) y = f(x)
dx dx dx
n n
n n na
.
1
1 01
n 1
a (x) a (x)a (x)d y d y dy f(x)+ +...+ + y =
dx (x) dx (x) dx (x) (x)
n n
n
n
n n n na a a a
1
1 01
n 1
a (x) a (x)a (x)d y d y dy f(x)+ +...+ + y =
dx (x) dx (x) dx (x) (x)
n n
n
n
n n n na a a a
( ) ( 1) '
1 1 0( ) ... ( ) ( ) ( )n n
ny p x y p x y p x y r x
(2)
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where 1 01a (x) a (x)a (x) f(x)
,..., , and(x) (x) (x) (x)
n
n n n na a a a
are represented with1 1 0( ),..., ( ), ( ) ( )np x p x p x and r x
respectively.
When1 1 0( ),..., ( ) ( )np x p x and p x are constants, we say that equation (2) has constant
coefficients; otherwise it has variable coefficients. Ifr(x) = 0, equation (2) becomes
( ) ( 1) '
1 1 0( ) ... ( ) ( ) 0n n
ny p x y p x y p x y
(3)
and is calledhomogeneous.
Ifr(x) is not identically zero, the equation is callednonhomogeneous.
Constant-coefficient linear equations
A linear differential equation has constant coefficients if the coefficients of the dependent variable
and its derivatives are constants.
For example, 1) tdt
dx
dt
xd 3
2
2
constant-coeff linear ODE
2) 32
2
745 xydx
dy
dx
yd constant-coeff linear ODE
3) teydt
dyt 5 linear ODE but not constant-coeff
Exercise :State the linear or nonlinear, and homogeneous or nonhomgeneous of the following
differential equations.
i. x5 y''' + x2 y' + 8y = 0 - the homogeneous linear equationii. y'' + 2y = 6e-xsinx - nonhomogeneous linear differential equation
iii. (2 x2)y'' 5xy' + 6y = 0 - homogeneous linear differential equationiv. x(y''y + y'2) + 2y'y = 0 - homogeneous nonlinear differential equation
Solution of an ODE
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A solution is an expression which satisfies the differential equation. A solution of a DE which
involves the
independent variable x and dependent variable y is the finding of the equation f(x,y)=0 which satisfies
the DE.
General Solution: A solution is a set of all possible solutions of the DE.A general solution of a
differential equation represents an infinite number of solutions, where the arbitrary constant C may be
any real or complex number. For example, the general solution ofy = 2x yieldsy = x2 + C.
Exercise. 1 : Show thaty = e2x is a solution of y
dx
dy2
Exercise 2: Show thaty =5e2x is a solution of y
dx
dy2
There are many different expressions which can satisfy a differential equation, that is, there are
many solutions. A solution from which many solutions can be found is called the general solution. It
contains the same number of arbitrary constants as the order of the equation.
The general solution of ydx
dy2 isy = Ce2x where C is any constant. C is called an arbitrary
constant.
Particular Solution: Particular Solution is any one solution of the DE.A particular solution is created
when the constant C has a specific value, which occurs when initial conditions are given and the
correspondence of a particular value of the independent variablex gives a specific valuey.
Example 1 - the particular solution ofy = 2x with the initial condition ofy(0) = 2, yieldsy = x2 + 2.
Initial Conditions and Boundary Conditions: These two conditions determine the value(s) of the
constant(s) in the general solution and form the particular solution.
Example 2 - the general solution of ydx
dy
2 isy = Ce2x
If y(0) = 4
then y = 4e2x which is a particular solution
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If the initial conditions (e.g. at t=0 or at initial displacement) are given, the problem is called an
initial value problem. If the conditions are given at different times, then the problem is called a
boundary value problem.
Explicit and Implicit Solutions
A solution of an ordinary differential equation that can be expressed in the formy =f(x) is said to be an explicit solution.
A relation G(x,y) =0 is said to be an implicit solution of an ordinary differentialequation on an intervalIprovided it defines one or more explicit solutions onI.
Example : For -1
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Exercise 2: Show that xx BeAxey is the general solution of 022
2
ydx
dy
dx
yd
Hence, determine the particular solution satisfying 1)0(,0)0( dx
dyy
Exercise 3: Obtain the solution of
a) xdx
dycos
b) tdt
xd
2
2
with conditions 7)2(and3)0( xx
Exercise 4: Determine the values of the constants C1 and C2 in y=C1sin x + C2cos x given
(a) the initial conditions y(0)=1, y(0)=2
(b) the boundary conditions 1=)(y'1,=y(0)
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First-Order Ordinary Differential Equations
First-order differential equation: ( , , ) 0F x y y
-- involves a first but no higher derivatives
Example: ( ) 2cos 4y x x
yfunction of x and xindependent variable
( , ( ), ( )) 0F x x x ( )x : solution
Whenfis independent of the variabley, that isf (x,y) =g(x) ,
The differential equation
( )dy g xdx
can be solved by direct integration.
Direction Field as a geometrical representation of first order DE
-- A set of line segments tangent to a curve
-- Give a rough outline of the shape of the curve
( , , ) 0F x y y
Solve forslope
( , )direction field
y f x y
SLOPE FIELDS
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Slope fields (also called vector fields or direction fields) are a tool to graphically obtain the solutions
to a first order differential equation.
Example 1: Draw a slop field for dy/dx = y
This is the FIELD in which we plot the SLOPES.
So we first tried it with the expression ( dy/dx = y ). This
means that the derivative of any point in the function is the
value of the function at that point (y-coordinate).
Let's say we want to plot the point (1,1). The derivative
(SLOPE) of the function at that point is 1, so we draw a
small line with slope 1 at that point.
Next, take the point (3,2). The derivative at that point is 2,
so there's another small line, but with a slope 2 at the point
(3,2).).
Now, you understand how they are constructed, this is how
the SLOPE FIELD of the function(s) that have ( dy/dx = y)
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From here, you can solve for the point (1,1).
To do this, draw a curve that follows the slope of the line in
either direction until the next whole number is reached. At
that point, change the curve's slope to that of the new point
and continue until the next whole number and so on...
WHAT FUNCTION DOES THE CURVE LOOK LIKE?It
is easy to see that the function that has a derivative equal tothe value of the function and passes through the point (1,1)
is y=ex.
However, you can also see that there are many other functions with a similar shape that could have
been drawn, depending on the starting point.
This is because the slope fields shows the entire family of functions [ (x)dx + C ].So, by solving for a point, you can find the exact one function from that family.
Let say the slope fields for dy/dx = 2x and dy/dx = -x/y are
Can you guess by observation what functions they are?
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Example 2:
y= -2xy
The slope,y'(x), of the solutionsy(x), is determined once we know the values forx andy , e.g., ifx=1
andy=-1, then the slope of the solutiony(x) passing through the point (1,-1) will be (-2).1.(-1)=2. Ifwe graphy(x) in thex-y plane, it will have slope 2, givenx=1 andy=-1. We indicate this graphicallyby inserting a small line segment at the point (1,-1) of slope 2.
Thus, the solution of the differential equation with the initial conditiony(1)=-1 will look similar to thisline segment as long as we stay close tox=-1.
Of course, doing this at just one point does not give much information about the solutions. We want to
do this simultaneously at many points in thex-y plane.
We can get an idea as to the form of the differential equation's solutions by " connecting the dots." So
far, we have graphed little pieces of the tangent lines of our solutions. The " true" solutions should notdiffer very much from those tangent line pieces!
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Example 3:2
y y The slope:2
y
General Solution:1
yx k
Direction field fory'=y and integral curves through (0,1), (0,2), (0,3), - 1, (0, - 2), and (0, - 3).
Example 4: Let's consider the following differential equation:
y = e-y
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Here, the right-hand side of the differential equation depends only on the dependent variabley, not onthe independent variablex. Such a differential equation is called autonomous. Autonomous
differential equations are always separable.
Autonomous differential equations have a very special property; their slope fields are horizontal-shift-invariant, i.e. along a horizontal line the slope does not vary.
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Directly Integrable Equations
Equation type:y = g(x)
Integrate on both sides of the equation will directly yield the solution.
( )y g x Solve by integrating both sides directly.
Example 1:y = 2x + 5 yieldsy = x2 + 5x + C
Example 2:y = 6e3x + 2/x yields y = 2e3x + 2ln x + C
Separation of Variables
( ) ( ) ( ) ( ) 0F x G y dx f x g y dy
This type of DE is called separable because it can be written in the form (variables can be separated)
0dy)y(Ndx)x(M
The first equation is usually not exact but multiplying it by the appropriate integrating factor will make
it exact, but use of an integrating factor may eliminate solutions or may lead to extraneous solutions.
After multiplying by the integrating factor)y(G)x(f
1the equation becomes:
0dy)y(G
)y(gdx
)x(f
)x(F
where)x(f
)x(F)x(M and
( )( )
( )
g yN y
G y .
Solutions are of the form cdy)y(Ndx)x(M where 0)y(G0)x(f & .
Solve by separation of variables intody dy f(x)
= f(x)/g(y) ( or y' = = )dx dx g(y)
or Equation type:f(y)
dy/dx = g(x)
Solve by separation of variables intof(y) dy = g(x) dx, and integrate each side.
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( ) ( )dy
g y dx f x dx cdx
=> ( ) ( )g y dy f x dx c
Example 1:
1. ( 2 )' x yy xe is separable2. y = 3x-y is not separable
Example 2:y = xy2
(1/y2) dy = x dx
1/y = x2/2 + C
Example 3:y = y sin x with initial conditionsy(0) = 1
General Solution: (1/y) dy = sin x dx
ln y = - cos x + C
Particular Solution: Solve to find C = 1 (ln 1 = -cos 0 + C => 0 = -1 + C)
ln y = - cos x + 1
Example 4:y = 4x (y-2)
1/(y-2) dy = 4x dx
ln (y-2) = 2x2
+ C which is reducible toy = 2 + eC => y = 2 + C1
First Order DE. Linear
A first order differential equation is said to be linear if:
(1) the dependent variable and its derivatives occur to the first order only,
(2) there are no products involving the dependent variable with its derivatives, and
(3) there are no non-linear functions of the dependent variable such as sine, exponential, etc.
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Afirst order linear DE may be written in the standard formdy
+ P(x) y =Q(x).dx
Exercise : What are P(x) and Q(x) referred to the standard form?
2
2 3
dy(a) - 3x y - 5x = 0dx
dy(b) x = 5y
dx
dy(c) x + 3x y + 2x = 5y.
dx
Linear First Order Homogeneous Differential Equation
If the DE given is a linear 1st
order DE., i.e.,dy
+ P(x) y =0dx
, the general solution of the DE can be
solved as follows.
The general solution of the DE can be found if the integral ( )P x dx exists.Example : y'+3y=0 with initial conditiony(1)=1
Solution : Since3( ) 3 3 3 e xP x dx x y C
and3 3[1 ](1) 1 e ( ) e xy C y x
First Order Linear Non-homogeneous DE
By definition, a linear first-order differential equation in y cannot contain products, powers or
other nonlinear combinations of y or y'. Have its most general form is
( )
( ) 0
( )
ln ( )
eP x dx
dyP x y
dx
dyP x dx
y
y P x dx C
y C
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( ) ( )dy
P x y Q xdx
withf(x)0, we can obtain the general solution of above equation as follows:
Multiply both sides by ])(exp[ dxxa :
[ ' ( ) ]exp[ ( ) ] ( )exp[ ( ) ]
exp[ ( ) ] ( )exp[ ( ) ]
exp ( ) ] ( )exp[ ( ) ]
y P x y P x dx Q x P x dx
dy P x dx Q x P x dx
dx
y P x dx Q x P x dx C
or exp ( ) ( )exp ( )y P x dx Q x P x dx dx C Here ])(exp[ dxxa is called the integration factor. Using integration factor to find the solution of DE.is a standard way in other form of DE. as well.
Integrating Factor
The linear DE in standard form has the integrating factor
P(x)dxe=x)(
Exercise : Find the integrating factor of the DE
dy(a) = 5y
dx
dy(b) x + 2y = 2x
dx
dy(c) + y tan x = sin x
dx
General Solution
The general solution of the DEdy
+ P(x)y = Q(x)dx
is
(x) y = (x) Q(x) dx + C
Example 1 Solve the differential equation
1x
y
dx
dy
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Solution
Here we have
xP
1
and 1Q
The integrating factor,
xeex
dxx ln1
We integrate with respect tox:
Cxxdxxy2
2
1
x
Cxy
2
1
We muptiply through the given equation with x :
dy
x y xdx
This is an exact equation because it can be written as
xxydx
d)(
Exercise :Find the general solution of the DE.
4 x
dy(a) = 5y
dx
dy(b) x + 2y = 2x
dx
dy(c) + y tan x = sin x
dx
dy(d) x -3y = x e
dx
EXACT EQUATIONS AND INTEGRATING FACTORS
First-order Differential Equations for which we can find exact solutions
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Study the patterns carefully. The first step of any solution is correct identification of the type of
differential equation.
A Linear First Order DE with Variable Coefficients can be written in the following form.
)x(Qy)x(Pdxdy
This equation is exact only when P(x) = 0.
Proof:
Write DE in the form 0dy)y,x(Ndx)y,x(M
)x(Qy)x(Pdx
dy
dx)x(Qydx)x(Pdy
0dydx)x(Qydx)x(P
0dy1dx)x(Qy)x(P
But ( , ) ( ) ( ) ( )M x y P x y Q x P xy y
0)1(x
)y,x(Nx
P(x) must equal 0 for the DE to be exact.
The equation 0dy1dx)x(Qy)x(P can be solved though by finding a proper integrating factor.The factor depends only on x, so call it )x( .
The new equation 0dy)x(dx)x(Q)x(y)x(P)x( is exact so
since )x(P)x()x(Q)x(y)x(P)x(y
)x(dxd)x(
x
( ) ( ) ( )d
x P x xdx
ordx
d)x(P
d1
dx)x(P
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d1
dx)x(P
lndx)x(P
dx)x(P
e)x( assuming 0)x(
The solution of the DE is therefore of the form
cdx)x(Qeey
dx)x(Pdx)x(P
Total Differential of a Function F(x,y)
F is a function of two variables which has continuous partial derivatives over a domain D. The total
differential of F is defined as :
dF(x,y) = dyy
)y,x(Fdx
x
)y,x(F
Dyx ).(
Exact Differential
M(x,y)dx + N(x,y)dy is an exact differential over D if )y,x(F such that
)y,x(dFdy)y,x(Ndx)y,x(M
dyy
)y,x(Fdx
x
)y,x(F
= dF(x,y)
In other words, M(x,y) =x
)y,x(F
N(x,y) =y
)y,x(F
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M(x,y)dx + N(x,y)dy = 0 is an exact differential equation if
M(x,y)dx + N(x,y)dy is an exact differential.
Then
i.e. dF x, y 0.
Integrating, we have F x, y c arbitrary constant
Given a DE that can be written in the form M(x,y)dx + N(x,y)dy = 0,
find F(x,y) such that .dy)y,x(Ndx)y,x(M)y,x(dF
The following theorem supplies a method for determining whether or not a DE is exact.
Theorem
M(x,y) and N(x,y) are functions with continuous first partial derivatives in the domain D.
M(x,y) and N(x,y) form the DE
M(x,y)dx + N(x,y)dy = 0
The above DE is exact in D if and only if
)y,x(Nx
)y,x(My
Dyx ),(
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Solution Method for Exact Equations
1. Showy
dx)y,x(M =
x
dy)y,x(N
Set )y,x(Mx
)y,x(F
. Integrate with respect to x to get F ),( yx
F ),( yx = dx)y,x(M + )y( 2. Differentiate with respect to y to get )y,x(N
y
[ dx)y,x(M + )y( ] = )y,x(N 3. Solve for )y( 4. The answer will be of the form : F ),( yx = g(x,y) + )(y , if no boundary value
is given.
Example : Determine whether the following equations are exact and solve the ones that are exact.
0)2( dyyxedxe yy
Solution:
(a) ,yeM yxeN y 2
so thatx
Ne
y
M y
Hence the given equation is exact and
yeyxMx
f ),( (1)
and yxeyxNy
f y2),(
(2)
We integrate (1) with respect to x:
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dxeyxf y),(
)(ygxey
Differentiating with respect toy:
yxeyxNdy
dgxe
y
f yy 2),(
ydy
dg2
Integrating, we have
1
2)( cyyg
The desired solution
21
2),( ccyxeyxf y
cyxey 2
Exercise 1 : Which of these equations are exact ?
a. dxy )3(2 + ( dyxy )42 = 0
b. dxyx )23(2 dyyx )( 3 = 0
c. 0sin2cos)1(2 rdrdr
d. 0dy)yy
x(dx)x
y
x(
3
2
2
e. dxx
y)
12(
2/1
2/3 + (3x
2/1y dy)1
2/1 =0
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Exercise 2: Solve
a. 0dy)1xy3yx2(dx)x2yyx3(23322 given 12y
b. 0dyxy2edxye2ye x2xx
c. Find A so that this equation is exact. 0dyxy4xdxy2yAx 322
Modified Exact First Order DE
Sometimes, a non-exact differential equation M x y dx N x y dy( , ) ( , ) 0 can be turned into an
exact differential equation by multiplying the whole equation by an appropriate factor, called an
integrating factor.
What ifx
)y,x(N
y
)y,x(M
We may be able to rewrite the equation so that it is exact.
If 0dy)y,x(Ndx)y,x(M is not exact, but 0)y,x(N)y,x(dx)y,x(M)y,x( is exact, then
)y,x( is called an integrating factor.
Example : Show that 0dyxdxxy2y 22 is not exact, then find n such that ynis an integrating factor.
i. x2y2xy2yy
2
x2)x(x
2
therefore the DE is not exact.
ii. Multiply the DE by yn, then solve.
0dyxydx)xy2y(y2n2n
n1n1n2n xy1n2y2nxy2yy
must equal
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xy2xyx
n2n
which means 1ny2n must equal 0 and
nxy1n2 must equal nxy2 for this to be so n must equal -2
iii. Now, solve the equation.
0dy)x(ydxxy2yy 2222
Solution:
1 2 21 2 0xy dx x y dy
2 1
( , )F x y x x y c
Theorem (a) If1
N
M
y
N
x
is a function ofx only, sayf(x), then exp ( )f x dx is an integrating
factor of the equation M x y dx N x y dy( , ) ( , ) 0
(b) if
y
M
x
N
M
1is a function ofy only, say g(y), then exp ( )g y dy is an integrating factor
of the equation M x y dx N x y dy( , ) ( , ) 0
Proof
(a) By the hypothesis, if ( )x is the integrating factor which depends on variablex only. We see
that
( ) ( , ) ( ) ( , )x M x y dx x N x y dy 0
is the exact differential. From theorem, we have the necessary condition
yM
xN
M
y
N
xN
d
d x
( ) ( )
Eventually, we obtain
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d
d x
M
y
N
x
Nf
which has a general solution ( ) exp ( )x f x dx
The proof for (b) is similar and it is left for Exercise.
Example
Find the integration factor of the following ODE and solve the corresponding equation
3 2 02 3 2 2x y x y y dx x y dy
Solution
For this ODE,
M x y x y x y y
M
yx x y
M
xx y y
( , )
3 2
3 2 3 6 2
2 3
2 2
and
N x y x yN
xx
N
yy( , ) 2 2 2 2
and
We can see that1
3N
M
y
N
x
. Therefore the integration factor is given by
exp 3 3d x e x and the ODE is reduced into
e x y x y y dx e x y dyx x3 2 3 3 2 23 2 0
The differential function is f x y( , )
fx
e x y x y yx 3 2 33 2
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f
ye x y f x y e x y
yC x
f
xe xy e x y
yC x
f
xe xy x y y C x
x x
x x
x
3 2 2 3 23
3 3 23
3 2 3
3
2 33
2 3
( , ) ( )
' ( )
' ( )
By comparing the above equations, we have
constantthenand0 )()(' xCxC
Therefore the general solution of the ODE is
33 2( , )
3
x yf x y e x y k
Homogeneous DE
If 0dy)y,x(Ndx)y,x(M can be written in the form )y,x(fdx
dy where
x
yg)y,x(f then
the DE is homogeneous or is homogeneous if the functionf(x,y) is homogeneous, that is-
( , ) ( , )f tx ty f x y . If ( , ) ( , )nf tx ty t f x y then f is homogeneous of degree n.
Check that the functions
Example 1:2 2
( , )xy
f x yx y
2 2
2 2 2 2 2 2( , ) ( , )
txty xyf tx ty t t f x y
t x t y x y
is homogeneous degree 2.
Example 2:
2
3 3 2( , ) ( ) 4
x
yf x y x y e xy
2
3 3 2
2
3 3 3 3 2
3
( , ) (( ) ( ) ) 4 ( )
(( ) ( ) ) (4 )
( , )
tx
ty
x
y
f tx ty tx ty e tx ty
t x y e t xy
t f x y
is homogeneous degree 3.
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A homogeneous equation can be transformed to a separable equation by a change of variable vxy .
The new equation is separable in v and x, so it can be solved.
A homogeneous equation 0dy)y,x(Ndx)y,x(M , is always transformed into a separable one by
letting vxy .
Thendy dv
v xdx dx
.
Since the given DE is homogeneous, we know it can be written in the form
x
yg
dx
dy. Since vxy
,
vgx
vxg
x
yg
.
And since
x
yg
dx
dy
( )dv
v x g vdx
.
Separating variables dx)v(gxdvvdx : 0xdvdx)]v(gv[ : 0dv)v(gv
1dx
x
1
To solve, integrate: cdv)x(g(v 1dxx1
c)v(Fxln or cx
yFxln
Let us summarize the steps to follow:
i) Recognize that your equation is an homogeneous equation; that is, you need to check thatf(tx,ty)=f(x,y), meaning thatf(tx,ty) is independent of the variable t;
ii) Write out the substitution v=y/x;iii)Through easy differentiation, find the new equation satisfied by the new functionz.
You may want to remember the form of the new equation:
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dx
dvxv
dx
dy
iv)Solve the new equation (which is always separable) to find v;v) Go back to the old functiony through the substitutiony =x v;vi) If you have an IVP, use the initial condition to find the particular solution.vii)Since you have to solve a separable equation, you must be particularly careful about the
constant solutions.
Example 3: Find all the solutions of2 5
2
dy x y
dx x y
Follow these steps:
i) It is easy to check that 2 5( , )2
x yf x y
x y
is homogeneous;
ii) Consider v=y/x ;iii) We have
2 5 2 5'
2 2
x xv zxv v
x yv z
which can be rewritten as
1 2 5' .
2
vv v
x v
This is a separable equation. If you don't get a separable equation at this point, then yourequation is not homogeneous, or something went wrong along the way.
iv) All solutions are given implicitly by 4 ln | 2 | 3ln(| 1|) ln(| |)
1
2
v v x C
v
v
v) Back to the functiony, we get
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4 ln | 2 | 3ln(| |)
2
y x y x C
y x
y x
Note that the implicit equation can be rewritten as
3 4
1( ) ( )y x C y x
Example 4:2
yxy y
x
22
2 ( ) (A)
y y y y
y x x xx
Lety
vx
, y vx , 'y v x v
(A) 2'v x v v v 2'v x v
2
1 1dv dx
v x 1 ln | |x c
v
1ln | | yv x c x ln | |xy x c
Non-Homogeneous DE Reducible to Homogeneous Form
Consider the non-homogeneous equation
1 1 1
2 2 2
a x b y cdy
dx a x b y c
(1)
where a1, b1,c1,... c2 are all constants.
Case1:
If
1 11 1
2 22 2
, . ., 0a ba b
i ea ba b
then the transformation (shift of origin)
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reduces the non-homogeneous equation (1) to the homogeneous equation of the form
1 1 1 1
2 1 2 1
a x b ydy
dx a x b y
(2)
here the unknown constants h,k are determined by solving the pair of equations
1 1 1
2 2 2
0
0
a h b k c
a h b k c
Now the equation (3) is homogenous in the new variables x1 and y1, it can be solve homogenous
equation method
.
Case 2:
If
1 11 1
2 22 2
, . ., 0a ba b
i ea ba b
then the transformation
reduces the non-homogeneous equation (1) in the variables x and z which can be solved using
homogenous equation method
Example 1: Solve the non-homogeneous equation10 2 2
3 9
dy x y
dx x y
1 1,x x h y y h
1 1 1z a x b y
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Example 2: Solve the non-homogeneous equation7 3 7
7 3 3
dy x y
dx y x
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Bernoulli Equations as First Order Linear Non-homogeneous differential equation.
The Bernoulli differential equation has the form
d y
d xP x y Q x y
n ( ) ( )
n=0 ( ) ( )y P x y Q x linear
n=1 ( ) ( )y P x y Q x y separable
( ( ) ( )) ( )
1 1( ) ( )
y Q x P x y R x y
dyR x dy R x dx
y dx y
If n 1 then Bernoulli equations can be transformed to linear equations by changing the dependent
variablen1
yv
.
Proof:
ny)x(Qy)x(Pdx
dy
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(y-n
) multiply => )x(Qy)x(Pdx
dyy n1n
letn1
yv
dxdyyn1
dxdv n => )x(Qv)x(P
dxdy
dxdv
dydx
)n1(1
dxdv
dy
dx
n1
1y n
(1-n) multiply => )x(Q)n1(v)x(Pn1dx
dv
let )x(P)n1()x(P1 => )x(Qv)x(Pdx
dv11
which is linear in v!
)x(Q)n1()x(Q1
Example : y3xy4dx
dy1x 2
change to Bernoulli y1x
3y
1x
x4
dx
dy22
where1x
x4)x(P
2 &
1x
3)x(Q
2
since yn
= y1
y-1
multiply =>1x
3
1x
x4
dx
dyy
22
1
Let2)1(1n1
yyyv solving for 2/1vy
dx
dyy2
dx
dv solving for
dx
dv
y2
1
dx
dy
1x
3
1x
x4
dx
dvv
2
1v
22
2/12/1
1x
x43
dx
dv
v2
12
separate and integrate
dx1x
x43dv
v2
12
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ylnvlndvv1
2
1 2/1=
dx
1x
x432
= dx1x1
31x
x22
22
yln = xarctan31xln2 2
solving for y
3arctan
22 1
xe
yx
Solving Second Order Differential Equation by Reduction of Order
Let us consider a second-order differential equation in the form
d y
d xF y y
2
2 ,
where F y y, is a continuous function. To solve the differential equation of this type, we can at leastreduce its order by one
Let pd y
d x , then
d y
d x
d p
d x
d p
d y
d y
d xpd p
d y
2
2
Therefore, the original differential equation Eq.(6.6) is reduced to a first-order differential equation
pyFdy
dpp ,
The above method is known as the reduction of order.
Example
Solve the differential equationsd y
dxy
dy
dx
2
2 by reduction of order:
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Solution
Let pd y
d x , we have
dy
dpp
dx
yd
2
2
and the ODE becomes
pd pd y
y p d p ydy d p y dy
p yC d y
d xy C
d y
C yd x
C
y
Cx C
y c c x c
1
2 2
1
2
1
2
1 1
2
2
22
2 2
2 2
1
1 1 2
tan '
tan
Riccati Equation
2( ) ( ) ( )y P x y Q x y R x
( ) 0P x linear
Let S(x) be a solution and let ( ) 1/ y S x z
The Riccati equation is transformed into linear
Example: 21 1 2 (A)y y yx x x
( ) 1 / P x x , ( ) 1 / Q x x , ( ) 2 / R x x
By inspection, ( ) 1y S x is a solution of (A)
Let1 1
( ) 1y S xz z
2
1y z
z
2
21 1 1 1 1 2(A) (1 ) (1 )
1 1 1 2(1 )(1 1)
1 1 1 2(1 )(2 )
zz x z x z x
x z z x
x z z x
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2 2
2 2
2
2 2
1 1 2(1 )(2 )
3 1 2(2 )
2 3 1 2 3 1
z zz
x z z x
z z
x z z x
z z z zx x x x x x
3 1z zx x
(linear)
Integrating factor
3( )
3dx
xe x
3 2 3 23 ( )x z x z x z x
Integrate 3 313
x z x c and3
1
3
cz
x
Solution:3
3 3
1 1 3 21 1
1/ 3 / 3
c xy
z c x c x
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Application Problems
Exponential decay or increaseFor radioactive decay, the decay rate is proportional to the number of the mother nuclei left. For
example, if the decay rate constant is and the initial no. of mother nuclei is N0,then we have,
dN
dtN and the initial condition is N(0)=N0.
( )t
dNN
dt
N t Ce
Initial condition implies C=N0 and thus N(t)=N0t
e
Note : Although the above two examples are very simple, it should be pointed this is not the usualcase. The remained sections in this chapter will try to illustrate some methods to solve some of the
standard form of DE.
Another example:
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Solution of this problem is:
Population Dynamics
The easiest mathematical model offered to govern the population dynamics of a certain species is
called the exponential model, that is, the rate of change of the population is proportional to the
existing population. In other words, ifP(t) measures the population, we have
dPkP
dt ,
where the rate kis constant. It is fairly easy to see that ifk> 0, we have growth, and ifk
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P(t)=P0kt
e
,
where P0 is the initial population, i.e. P(0)=P0. Therefore, we conclude the following:
ifk>0, then the population grows and continues to expand to infinity, that is, lim P t
x
ifk0, is not adequate and the model can be dropped. The main argument for this
has to do with environmental limitations. The complication is that population growth is eventually
limited by some factor, usually one from among many essential resources. When a population is far
from its limits of growth it can grow exponentially. However, when nearing its limits the populationsize can fluctuate, even chaotically. Another model was proposed to remedy this flaw in the
exponential model. It is called the logistic model (also called Verhulst-Pearl model). The differential
equation for this model is
(1 )dP P
kPdt M
,
whereMis a limiting size for the population (also called the carrying capacity). Clearly, when P issmall compared toM, the equation reduces to the exponential one. In order to solve this equation we
recognize a nonlinear equation which is separable. The constant solutions are P=0 and P=M. The non-constant solutions may obtained by separating the variables
(1 )
dPkdt
PP
M
and integration
(1 )
dPkdt
PP M
The partial fraction techniques gives
11
(1 ) (1 )
dP M dPP PP
PM M
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which gives 1P
In P In kt cM
|
Easy algebraic manipulations give
(1 )
ktP CeP
M
where Cis a constant. Solving for P, we get
If we consider the initial condition P(0)=P0 (assuming that P0 is not equal to both 0 orM), we get
,
which, once substituted into the expression for P(t) and simplified, we find
It is easy to see that
lim P tx
However, this is still not satisfactory because this model does not tell us when a population is facing
extinction since it never implies that. Even starting with a small population it will always tend to thecarrying capacityM.