V . Ramesh I Mech-B 100111137064
Types of Solutions of a Differential equation
A differential equation is a mathematical equation for an unknown function of one or several variables that relates the values of the function itself and its derivatives of various orders. Differential equations play a prominent role in engineering engineering,, physics physics,, economics economics,, and other disciplines. Differential equations are mathematically studied from several different perspectives, mostly concerned with their soluti so lutions²the ons²the set of o f functions that satisfy the equation. Only the simplest differential equations admit solutions given by explicit formulas; however, some properties of solutions of a given differential equation may be determined without finding their exact form. If a self-contained formula for the solution is not available, the solution may be numerically approximated using computers. The theory of dynamical of dynamical systems puts emphasis on qualitative analysis of systems described by differential equations, while many numerical methods have been developed to determine solutions with a given degree of accuracy. A relation y = (x) is said to be a solution or integral of the differential equation f (x, y, y', . . . , equat ion, it gives a result that y{sup (n)}) = 0 in the range a x b if, when substituted into the equation, is identically zero in that range (see 169 169). ). It is frequently difficult or undesirable to express a solution y explicitly as a function of x, but instead to have an implicit implicit relation re lation F (x, y) = 0 between the solution y and the independent variable x. Such a relation is a solution if, when solved explicitly for y in terms of x, it yields a solution in the way described above. For example, if both sides of a polynomial equation in x and y are differentiated with respect to x, then the function y(x) determined by this implicit relation will be a solution of the differential equation thereby obtained (see 170 170). ). The simplest of all ordinary or dinary differential equations is y' = g (x), in which g is an elementary function. An example is g (x) = 2x. The problem of solving so lving this differential equation is equivalent to finding a function y(x) the derivative of which is 2x; this leads to the solution y = x + c in which c is an arbitrary constant. The process of integration has led to the occurrence of an arbitrary constant in the solution. It can be seen intuitively (and it can be proved rigorously) that finding the solution of a differential equation of the nth order will somehow be equivalent to performing n integrations and that the final solution y(x) will contain n arbitrary constants of integration. Any solution of an nth order equation of the form y = (x, c , c , . . . , c ) in which c , . . . , c are arbitrary constants is called a general solution of the equation. equat ion. Any solution that may be obtained from the general solution of an equation by assigning particular values to the constants is called a particular solution of that equation. It sometimes happens that a nonlinear differential equation has a solution that cannot be obtained by assigning specific values to the arbitrary constants in the general solution. Such a solution is called a singular singular solution solution of the differential equation.
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For example, a particular nonlinear differential equation (see 171 171)) has a general genera l solution that is linear in the independent variable (see 172 172)) with an arbitrary constant c. A particular solution may be obtained from (172 (172)) by taking c = 1 (see 173 173). ). On the other hand, a function proportional to x (see 174 174)) also satisfies the differential equation. Because this solution cannot be obtained from the general solution by assigning a particular value to c, it is a singular solution of the differential equation (171 (171). ). In this instance the graphs of the functions defined by (172 (172)) are straight lines, and the graph of the function defined by (174 (174)) is a parabola that is the envelope of that family of straight lines (cf. Figure 18). 18). Similarly, a nonlinear equation with the cube of the derivative proportional to y (see 175 175)) has a general solution with y proportional to (x - c) (see 176 176)) and the singular solution y = 0. Here the general solution is represented by a family of semicubical parabolas, and the singular solution is the line that passes through the cusp of each member of the family (cf. Figure 19). 19). Such a line is called a cusp locus. The differential equations are very much helpful in many areas of science. But most of interesting real life problems involve more than one unknown function. Therefore, the use of system of differential equations is very useful. Without loss of generality, we will concentrate on systems of two differential equations
As a motivation let us consider an island with two type of species: Rabbits and Fox. Clearly one plays the role of predator while the other one the role of a prey. If we are interested to model the populations growths of both species, then we have to keep in mind that if, for example, the population of the Fox increases, then the Rabbit population will be affected. So the rate of change of the population of one type will depend on the actual population of the other type. For example, in the absence of the Rabbit population, the Fox population will decrease (and fast) to face a certain extinction. Something that most of us would like to avoid. A model for this Predator-Prey problem was developed by Lotka (in 1925) and Volterra (in 1926) and is known as the Lotka-Volterra system
where R(t ) measures the Rabbit population, popu lation, F (t ) measures the Fox population, and all the involved constant
are positive numbers. Note that a and b are the growth rate of the 2
prey, and the death rate of the predator. and are measures measures of the effect of the interaction interaction between the Rabbits and The Fox. Note that in the Lotka-Volterra system, the variable t is missing. This kind of system is called autonomous system and are written
Phase Plane
Let us go back to the general case
A solution to this system is the couple of functions (x(t ), ),y(t )) )) which satisfy both differential equations of the system. When we change the variable t , then we get a set of points on the xyplane which, in physics, we usually call a trajectory. The moving object has the coordinates (x(t ), ),y(t )) )) at time t . The velocity to the trajectory at time t is given by
Note that we do not need to know the solution (x(t ), ),y(t ) to determine the velocity vector vecto r at time t . Indeed, we have
as long as we know x, y, and t . In particular, we can draw all the velocity vectors everywhere on the plane for the t he autonomous systems. This is known as the vector vect or field. The vector field should be understood as the analogue of the direction field for differential equations. Example.
Draw the vector field for the predator-prey problem
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A differential equation (or "DE") contains derivatives or differentials. Recall from the Differential section in the Integration chapter, that a differential can be thought of as a derivative where "d y/ y/ d dx " is not written in fraction form.
Examples
of Differentials
d x
(this means "an infinitely small change in x")
d
(this means "an infinitely small change in ")
t (this d t
means "an infinitely small change in t ") ")
Examples Example
of Differential Equations
1.
We saw the following example in the Introduction to this chapter. It involves a derivative (d y/ y/ d dx ):
As we did before, we would integrate it to produce the required solution. This will be a general solution (involving K , a constant of integration). So we proceed as follows:
and this gives 4
But where did that d y go from the d y/ y/ d dx ? Why did it seem to disappear? In this example, we appear to be integrating the x part only (on the right), but in fact we have integrated with respect to y as well (on the left). Differential equations are like that - you need to integrate with respect to two (sometimes more) different variables, one at a time. We could have written wr itten our question only using differentials : d y
2
= (x
í 3)d x
(All I did was to multiply both sides of the original d y/ y/ d dx in the question by d x.) Now we integrate both sides, the left side with respect to y (that's why we use "d y") and the right side with respect to x (that's why we use "d x") : d y
=
(x2 í 3)d x
Then the answer is the same as before, but this time we have arrived at it considering the d y part more carefully:
On the left hand side, we have integrated
d y
= 1 d y to give us y.
We have integrated both sides, but there's a constant of integration on the right side only. What happened to the one on the left? The answer is quite straightforward.
Note about the constant:
We do actually get a constant on both sides, but we can combine them into one constant ( K ) which we write on the right hand side. Example
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This example also involves differentials: 2 d = sin(t + 0.2) d t t We have: A function of with d on the left side, and A function of t with d t t on the right side.
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To solve this, we would integrate both sides, one at a time, as follows: 2 d = sin( t + 0.2) d t t 3/3 = ícos(t + 0.2) + K We have integrated with respect to on the left and with respect to t on the right. Solving a differential equation
From the above examples, we can see that solving a differential equation means finding an equation with no derivatives that satisfies the given differential equation. Solving a differential equation always involves one or more integration steps. It is important to be able to identify the type of differential equation we are dealing with before we attempt to solve it.
Definitions First order DE: Contains
only first derivatives
Second order DE: Contains
second derivatives (and possibly first derivatives also)
Degree: The highest power of the highest derivative which occurs in the
Examples
differential equation.
- Order and Degree
1) This DE has order 2 (the highest derivative appearing is the second derivative) and degree 1 (the power of the highest derivative is 1.)
2) This DE has order 1 (the highest derivative appearing is the first derivative) and degree 5 (the power of the highest derivative is 5.) 6
3) ( y " ) 4 + 2( y ' ) 7 í 5y = 3 This DE has order 2 (the highest derivative appearing is the second derivative) and degree 4 (the power of the highest derivative is 4.)
General and Particular Solutions
When we first performed integrations, we obtained a general solution (involving a constant, constant, K ). ). We obtained a particular solution by substituting known values for x and y. These known conditions are called boundary conditions (or initial conditions). It is the same concept when solving differential equations - find general solution first, then substitute given numbers to find particular solutions. Let's see some examples of finding solutions of first order, first degree DEs. Example
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a. Find the general solution for the differential equation d y
+ 7x d x = 0.
b. Find the particular solution solution given that y(0) = 3. Answer a. We simply need to subtract 7x d x from both sides, then insert integral signs and integrate:
NOTE 1: We could have written it in a more familiar way as: 7
Then
So
NOTE 2:
means
. We also have:
and so on.
b. We now use the information y(0) = 3 to find K . The information means that at x = 0, y = 3. We substitute these values into the equation that we found in part (a), to find the particular solution.
gives K = 3.
So the particular solution is: Here is the graph of our solution.
Example
2
Find the particular solution of y' = 5
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given that when x = 0, y = 2. Answer We can write y' = 5
as a differential equation: d y
= 5 d x
Integrating both sides gives: y = 5x + K
Applying the boundary conditions: x = 0, y = 2, we have K = 2 so: y = 5x + 2
Example
3
Find the particular solution of y''' = 0
given that: y(0) = 3, y' (1) (1) = 4, y'' (2) (2) = 6
Answer Since y''' = 0, when we integrate once we get: y'' = A (A is a constant)
Integrating again gives: y' = Ax + B (A, B are constants)
Once more: y = Ax
2
/2 + Bx + C (A, B and C are constants)
The boundary conditions are:
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(1) = 4, y'' (2) (2) = 6 y(0) = 3, y' (1) We need to substitute these values into our expressions for y'' and y' and our general solution, y = Ax2/2 + Bx + C.
Now y(0) = 3 gives C = 3
and y'' (2) (2) = 6 gives A = 6 (actually, y'' = 6 for any value of x in this problem since there is no x term)
Finally, (1) = 4 gives B = -2 y' (1) So the particular solution for this question is: y = 3x2 í 2x + 3
Checking the solution by differentiating and substituting initial conditions: y' = 6x í 2 y' (1) (1) = 6(1) í 2 = 4 y'' = 6 y''' = 0
Our solution is correct.
Example
4
After solving the differential equation,
(we will see how to solve this DE in the next section Separation of Variables), Variables), we obtain the result y = c ln x
Did we get the correct general solution? 10
Answer
Now, if y = c ln x, then
[See Derivative of the Logarithmic Function if you are rusty on this.) So
We conclude that we have the t he correct solution.
Second Order DEs
We include two more examples here to give you an idea of second order DEs. We will see later in this chapter how to solve such Second Order Linear DEs. DEs. Example
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The general solution of the second order DE y'' + a 2y = 0
is y = A cos ax + B
sin ax
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Example
6
The general solution of the second order DE y'' í 3y' + 2y = 0
is 2x
y = Ae
+ Bex
If we have the following boundary conditions: (0) = 5 y(0) = 4, y' (0) then the particular soluti so lution on is given by: y=e
2x
+ 3ex
Now we do some examples using second order DEs. We are given a solution and we need to check if it is the correct co rrect solution. Example
7 - Secon d Order DE
Show that y = c1 sin 2x + 3 cos 2x
is a general solution for the differential equation
Answer We have a second order differential equation and we have been given the general solution. Our job is to show that the solution so lution is correct. We do this by substituting the answer into the original 2nd order differential equation. We need to find the second derivative of y: y = c1 sin 2x + 3 cos 2x
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First derivative:
Second derivative:
Now for the check step:
Example
8 - Secon d Order DE
has a solution of y = c1 + c2e2x
Show that
Answer Since 2x
y = c1 + c2e
, then:
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and
It is obvious that
Differential equations arise in many areas of science and technology, specifically whenever a deterministic relation involving some continuously varying quantities (modeled by functions) and their rates of change in space and/or time (expressed as derivatives) is known or postulated. This is illustrated in classical mechanics, mechanics, where the motion of a body is described by its position and velocity as the time t ime varies. Newton's laws allow one to relate the t he position, velocity, acceleration and various forces acting on the body and state this relation as a differential equation for the unknown position of the body as a function of time. In some cases, this differential equation (called an equation of motion) motion) may be solved so lved explicitly. An example of modelling a real world problem using differential equations is determination of the velocity of a ball falling through the air, considering only gravity and air resistance. The ball's acceleration towards the ground is the acceleration due to gravity minus the deceleration due to air resistance. Gravity is constant but air resistance may be modelled as proportional to the ball's velocity. This means the ball's acceleration, accelerat ion, which is the derivative of its velocity, depends on the velocity. Finding the velocity as a function of time involves solving a differential equation.
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