lecture 4c_substitution methods

7/29/2019 Lecture 4C_Substitution Methods

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Copyright Ren Barrientos Page

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MAP2302 Lecture 4C 2011-3

SUBSTITUTION METHODS

The equation 1 2is neither separable nor linear. However, we may obtain a solution by means of a substitution. This sectionexplores substitution methods that can be used to solve certain types of differential equations. The methodsillustrated are by no means exhaustive and the goal here is simply to demonstrate the general idea behind asubstitution in the context of a limited set of first and higher order equations.

Substitutions Suggested by the Form of the Equation

Sometimes the variables in an equation occur in certain combinations that that suggest a substitution. For examplethe variables and might appear as the combination throughout the equation. We call thiscombination bilinear in and and we say that a function, is a bilinear function1 of and if For example, the function sin2 3 1is bilinear in and. On the other hand, 3is not.

We now study differential equations of the form 1

Example 1 Solve the equation .Solution

Observe that, . This suggests that we let . We first compute/: 1 Substituting in the differential equation we obtain the separable equation 1 1 1 1

Simplifying,

1Sometimes we also say that the function has abilinear argument.

1

Bilinear Substitution:

The Substitution converts equation (1) into a separable equation.Procedure: Let and replace the original differential equation by one in the variables and :Then equation(1) becomes the separable equation:


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1This is a separable equation: Thus,

2 since , 2 or

where 2.Example 2 Solve the equation 2 2 4 .Solution

Let 2 4. Then 2or 2

Substituting in 2 2 4,2 2

or Separating the variables,

Integrating, /

or2/ But . Hence 2 2 4/

or

where 2 4 0.Example 3 Solve the equation sin ; 0 /4 .Solution

Let . Then 1 1Substituting in the differential equation,


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1 sin sin 1which is separable: sin 1 Multiplying numerator and denominator by

sin 1:

sin 1sin 1 sin 1 sin 1sin 1 1sincos Dividing each term in the numerator by cos ,sec cos sin Integrating,

tan cos

ortan sec But . Therefore, tan sec Finally, applying the initial condition0 /4,

tan/4 0 sec/4 0 0 1 2Thus, is an implicit solution.

The previous example illustrates a point: when using transformations, one can avoid unnecessary steps bytransforming the initial conditions as well. Consider again the original transformation

which takes us from the t-xcoordinate system to the t-u coordinate system (the variable t is being retained as theindependent variable). We need to find what u is when 0. Since 0 /4, the point though which theintegral curve passes in the t-x is 0,/4). Thus, /4 when 0 as well and the corresponding initialcondition in the-u plane is 0 4Warning: this is just a coincidence. In general the initial condition after the transformation will be different than the original one.In this example, we are making the substitution . Hence, the initial condition remains the same only because itis specified at 0.

The original IVP thus becomes

sin 1 ; 0 4Now we may use the definite integral: sec cos sin

tan 1cos


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tan 1costan/4 1cos/4 Finally, substituting foru, An Application to Wave Motion

Consider a wave that moves from left to right with speed. At any instant it has a definite shape , , thatis, if we fix , what we see is a shape frozen in time:

But now we wish to make this wave move as time evolves. How do we describe thistraveling wave?

Let us make a simplifying assumption that the shape of the wave does not change, that is, it does notdisperseas ittravels its course. Let us also assume that when 0 the wave looks like this:

Suppose that the wave move to the right with speed

. If this wave is shifted

units to the right then it can be

described by which is just another snapshot at a later time. Setting , the traveling wave can berepresented by a function of the form This function is bilinear in and and satisfies theone-dimensional wave equation: 1 This is an example of a linear partial differential equation and one which you will encounter often in your futurestudies.

Exercise: show that sin whereis an arbitrary constant satisfies the one-dimensional wave equation.Missing Intermediate Terms

Consider the second order equation 2 This linear differential equation which may be replaced by a first order equation via the substitution For then, 2 which has an integrating factor . Using the method of integrating factors we have

x

,

x

Wave frozen at time


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Integrating, 14 2 1 or

14 2 1 Since , 14 2 1 Integrating again to obtain: 14 2 1 Observe that we have two arbitrary constants and . This is not surprising since we originally had a second orderequation.

As we shall see, it is common practice to use subscripts to denote arbitrary constants when there is more than one ofthem. The above solution may be written as

where and are arbitrary constants.Example 4 find a solution of the equation 4Solution

Let . Then 4or 2/

Let us solve 2/. By separation of variables,/ 2Integrating, 43/ 2

or

32 /

where We Integrate again to reproduce: 32 /

23/ where 32

Thus,

/


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Differential Forms with Homogeneous Coefficients

We now study differential equations of the form , , 2or their associated differential form

, , 0

We first define the concept of homogeneous of function.

Example 5 The function, 5 is homogeneous of degree 2 because, 5 5 5

, Example 6 The function, / is homogeneous of degree 0 because, / / / ,Example 7 The function, is not homogeneous because

, and no factoring will allow us to write this as, for any value of.When equation (2) involves homogeneous functions it is possible to use another substitution which converts it to aseparable equation

Remarks: do not memorize equation (3). Go through the procedure and perform the algebra as you go along. Also, thesubstitution may be used instead the procedure is the same except that the separable equation involves the variables and.

Definition

A function, is homogeneous of degreeif, , for any real numbersuch that,is in the domain of.

, , 0

1 1, 1, 1, ; 0 3

If both coefficients, and, in (2)are homogeneous of the same degree, then substitution converts the differential equation into a separable equation.

Procedure: Set ,Substituting,

By the homogeneity of the functions and, this equation is separable and


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Example 8 Find a solution of the equation 2 0.Solution

This equation is neither linear nor separable. However, notice that both coefficients arehomogeneous ofdegree2.

Let . Then . Substituting and , 2 0Using homogeneity and factoring:1 2 0

Cancelling: 1 2 0Collecting like terms: 13 2 0

This is a separable equation: 13 2Thus, 1

2

13

Integrating, || ||Note: we introduce ln|| as the constant of integration. This can be done without loss of generalitybecause any real number may be written as the log of a positive number. Introducingln|| instead of justallows us to combine all terms and simplify our final answer.

Multiplying by3: 3 ln|| ln1 3 3ln||or

ln|| ln

13

Renaming to: ln|| ln 13Since theln function is monotone, this equation holds if and only if

|| 13or 1 3

where is an arbitrary constant.Finally, since

,

Hence, is an implicit solution of the differential equation.

Example 8 Find a solution of the equation


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SolutionThe functions, and, are homogeneous of degree 3 [verify]. Let .Then by the product rule, Substituting:

1or 1

which gives us the separable equation 1Combining like terms,

1 1 Separating,

1 1 or

1 1 Integrating, ln|| 13 ln || ln ||

or

||

Exercise: show that the last solution can be expressed as

/ ; /Example 9 (an application) An airplane leaves its port located at,0 and travels with a constant airspeed (that isspeed relative to the air) of. Its intended destination is located at 0,0, however, there is a northerly windwith wind speed of magnitude and therefore the planes course must be adjusted. Find the equation of the curvetraced by the airplanes flight path.

SolutionWe assume that the pilot adjusts the aircrafts flight direction to compensate for the wind speed bymaintaining a heading that is always directed toward the origin, as shown in the figure below:

,0

,0 ,

y

x


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Let , be a point on the trajectory. From the figure on the right, cos ; 0 sin ; 0 0Hence,

sin cos

Also from the figure,sin / and cos / . Hence / /

Simplifying, This equation has homogeneous functions of like degree, as can be seen if we write it thus:

1

Let / so that . Then and substitution into the equation above gives us

1 or

1

Separating variables, 1 Hence, 1 1

The integral on the left is found in any table of integrals:

ln 1 ln|| Substituting /,

ln 1

ln|| Applying the condition 0,ln 0 1 0 ln||

which gives us ln||.


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Hence, the trajectory is given by

ln 1 ln|| ln||or

ln 1 ln Since all arguments are nonnegative,

Example 10 (an application) A light source located at the origin reflects its light off a mirrored surface in such a waythat the reflected light beams are parallel to the x-axis (as shown in the figure below). Find the shape of the surface.

SolutionWhen light impinges on a mirror, theLaw of Reflection states that the angle of incidence is equal to the

angle of reflection. Thus, in the figure below, .

From these figures it is clear thattan tan and that2 2 so that tan Sincetan / we have 21

This is a quadratic equation in: 2 0Solving for:

2 4 42 Hence,

In differential form, 0The coefficients, and , are homogeneous of degree1. Therefore, thesubstitutions or may be used to solve this equation and, as an exercise, you should proceedwith the substitution .

2 90 and also2 90. Therefore, 2

,

x

y

,


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However, we are going to take a different direction because we recognize something in this equation that is

familiar. If we implicitly differentiate /, we obtain

This expression appears in the equation

because if we rearrange its terms, we obtain 1The left-hand side is exactly/ /. Thus,

1Integrating, where is an arbitrary constant. Simplifying,

or 2 we immediately recognize this to be the equation of a family of parabolas with vertex on the x-axis.performing some more algebra give us the following final result: 4 which corresponds to a parabola with vertex at,0 and focus at0,0.

lecture 4c_substitution methods

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