Differential equations of the first order and first degree. Methods of solution. Separation of variables. Homogeneous, exact and linear equations.
Integrating factors. Bernoulli’s equation.Website owner: James MillerDifferential equations of the first order and firstdegree. Methods of solution. Separation of variables.Homogeneous, exact and linear equations. Bernoulli’s equation.Differential equations of the first order and first degree. Any differential equationof the first order and first degree can be written in the formExample. The differential equationcan also be written as(x - 3y)dx + (x - 2y)dy = 0Existence of a solution.
The general solution of the equation dy/dx = g(x, y), if it exists, hasthe form f(x, y, C) = 0, where C is an arbitrary constant. Under what circumstances does ageneral solution exist? We have the following theorem.Theorem 1. The direction at each point of R is that of thetangent to that curve of the family f(x, y, C) = 0that passes through the point.A region R in which a direction is associated witheach point is called a direction field. 1 isshown the direction field and integral curves forthe differential equation dy/dx = 2x.
The generalsolution of this equation is y = x 2 + C. Theintegral curves are parabolas.Methods of solving differentialequations of the first order andfirst degreeI Separation of variables.
If an equationM(x, y) dx + N(x, y) dy = 0can be brought into the formP(x) dx + Q(y) dy = 0,the variables are referred to as having been separated. The general solution is then given byExample. Solve(1 + x 2)dy - xy dx = 0Solution. Dividing by y(1 + x 2) and transposing we getIntegrating both sides, we getln y = ½ ln (1 + x 2) + ln Corln y = ln C(1 + x 2) ½Taking exponentialsy = C(1 + x 2) ½The arbitrary constant was added in the form “ln C” to facilitate the final representation.Homogeneous polynomials, functions and equationsDef. Homogeneous polynomial. A polynomial whose terms are all of the same degreewith respect to all the variables taken together. Thusx 2 + 2xy - 2y 2 is homogeneous of degree 22x 3y + 3 x 2y 2 + 5y 4 is homogeneous of degree 42x + 5y is homogeneous of degree 1The concept of homogeneity is extended to general functions in the following way:Def.
Homogeneous function. A function such that if each of the variables is replaced by ktimes the variable, k can be completely factored out of the function whenever k ≠ 0. The powerof k which can be factored out of the function is the degree of homogeneity of the function.Thus2x 2 ln x/y + 4y 2 is homogeneous of degree 2x 2y + y 3 sin y/x is homogeneous of degree 3II Homogeneous differential equations. A differential equation of the formM(x, y) dx + N(x, y) dy = 0is said to be homogeneous if M(x, y) and N(x, y) are homogeneous functions of the samedegree. Such an equation can be transformed into an equation in which the variables areseparated by the substitution y = vx (or x = vy), where v is a new variable.Note.
Differentiating y = vx gives dy = v dx + x dv, a quantity that must be substituted for dywhen vx is substituted for y.Example. Solve(x 2 - y 2)dx + 2xy dy = 0Solution. We cannot separate the variables, but M(x, y) and N(x, y) are homogeneous functionsof degree 2. Substitutingy = vx and dy = v dx + x dvwe get(1 - v 2)dx + 2v(v dx + x dv) = 0Separating the variables givesIntegrating we getln(v 2 + 1) = - ln x + ln CTaking exponentials we obtainx(v 2 + 1) = CFinally, since v = y/x, this becomesx 2 + y 2 = CxReason why the substitution y = vx transforms the equation into one in which thevariables are separable. The reason the substitution y = vx transforms the equation into onein which the variables are separable can be seen when the given equation is written in the formIf M(x, y) and N(x, y) are homogeneous functions of the same degree and one substitutes vx for yone finds that the x’s all cancel out on the right side of 2) and the right side becomes a functionin v alone i.e.
The equation takes the form3) dy/dx = g(v)Substituting dy = v dx + x dv then gives4) v dx + x dv = g(v) dxwhere the variables can be separated asIII Exact differential equationsDef. Exact differential equation. A differential equation which is obtained by setting thetotal differential of some function equal to zero.The total differential of a function u(x, y) is, by definition,and the exact differential equation associated with the function u(x, y) isThe primitive of 5) is u(x, y) = C.If in a given differential equation6) M(x, y) dx + N(x, y) dy = 0the quantityM(x, y) dx + N(x, y) dyhappens to be exactly the total differential of some function u(x, y) i.e. If there exists somefunction u(x, y) such thatthen 6) is an exact differential equation and its primitive is u(x, y) = C.Since ordinarily one cannot determine by inspection whether or not a given equation is exact, atest for exactness is necessary.
That test is given by the following theorem.Theorem 2. Cotherm type tse manual meat. Let M(x, y), N(x, y), ∂M/∂y and ∂N/∂x be continuous functions of x and y. Thena necessary and sufficient condition that the differential equationM(x, y) dx + N(x, y) dy = 0be exact isIf the differential equation is exact the next step is to produce the function u(x, y) of which M(x,y) dx + N(x, y) dy is the total differential. Sometimes it can be determined from inspection.More often it cannot. The following example illustrates the usual method of solution.Example. Test the following equation for exactness and find the solution if it is exact.7) (3x 2y - y)dx + (x 3 - x + 2y)dy = 0Solution.M = 3x 2 - y N = x 3 - x + 2y∂M/∂y = 3x 2 - 1 ∂N/∂x = 3x 2 - 1Since ∂M/∂y = ∂N/∂x, the equation is exact i.e.
There is a function u(x, y) of which the left-handside of 7) is exactly the total differential. To find this function we integrate ∂u/∂x = M = 3x 2 - ywith respect to x, holding y constant. We obtain8) u(x, y) = ∫M ∂x = ∫(3x 2 - y)∂x = x 3y - yx + φ(y)where φ(y) consists of terms that are free from x ( ∫M ∂x denotes integration with respect to x,holding y constant). In the same way, we integrate ∂u/∂y = N = x 3 - x + 2y with respect to y,holding x constant. We obtain9) u(x, y) = ∫N ∂y = ∫( x 3 - x + 2y)∂y = x 3y - xy + y 2 + ψ(x)where ψ(x) consists of terms that are free from y ( i.e.
Terms containing x only or constants).Comparing 8) and 9) we see that the general solution isx 3y - xy + y 2 = CExample from Middlemiss. Differential and Integral Calculus. 443There is a formula that can be used to obtain the solution. It is:Formula for general solution of exact equation M dx + N dy = 0.
The general solution isgiven by∫ M ∂x + ∫f(y)dy = Cwhere f(y) is composed of all the terms in N which are free from x (i.e. All terms not containing x— terms containing y only or constants) and ∫M ∂x denotes integration with respect to x,keeping y constant.Alternate formula:∫N ∂y + ∫f(x)dx = Cwhere f(x) is composed of all the terms in M which are free from y.The above formulas will give the correct result in the vast majority of cases but they are notinfallible and in exceptional cases may give an incorrect result. Consequently the solution shouldalways be checked by substituting it into the original equation.There is another more complicated formula that also gives the general solution:Formula for the general solution.where ∫M ∂x denotes integration with respect to x, keeping y constant.IV Integrating factors. If a differential equation is not exact it may be possible to make itexact by multiplying it through by some function. For example, the equation11) y dx + 2x dy = 0is not exact since ∂M/∂y ≠ ∂N/∂x. However, multiplying it by y gives the exact equationy 2dx + 2xy dy = 0in which the left-hand side is exactly the differential of xy 2.
A function which, when multipliedinto a differential equation, makes it exact is called an integrating factor. In this example y is anintegrating factor for 11).Theorem 3. Let the equation12) M(x, y) dx + N(x, y) dy = 0possess a solution f(x, y, C) = 0, where C is an arbitrary constant. Then if equation 12) is notexact, it can always be made exact by multiplying it through by some proper function of x and yi.e. There exists some integrating factor μ(x, y) that will make it exact. Moreover, if μ(x, y) is anintegrating factor for 12) then a · μ(x, y) is also an integrating factor, where a is an arbitraryconstant.Thus if 12) is solvable, it is either exact or can be made exact by some integrating factor.
Thereis however no general rule for finding that integrating factor. The theorem merely assures us thatan exact version the equation exists.The conditions for 12) to be solvable are not very severe. They are stated in Theorem 1 above.There are many equations which are not integrable as they stand but which become integrablewhen multiplied by the right factor. There is no general method for finding the right factor butin many simple cases one can be found by inspection. The ability to do this depends largely uponrecognition of certain common exact differentials and upon experience.
One employs ingenuityand trial and error to, through manipulation, possibly transformation of variables, and somemultiplying factor, transform the equation into one that can be integrated. For example, if onenotices the groups of terms “x dy - y dx” or “x dy + y dx” in an equation he may be able tomanipulate the equation into an equation that can be integrated using a multiplying factor thatcreates one of the following exact differentials:Example. Solve the equation(x 2 + y 2 + y)dx - x dy = 0Solution. Let us write the equation as(x 2 + y 2)dx + y dx - x dy = 0Remembering the formulawe decide to multiply the equation through by the factor 1/(x 2 + y 2) to obtainorWe then integrate to obtain the solutionTheorem 4. A differential equation of the formcan be reduced to the formby means of the integrating factorIts primitive isExample. Solve the equation(3x 2y - xy) dx + (2x 3y 2 + x 3y 4) dy = 0Solution. We rewrite the equation asy(3x 2 - x) dx + x 3(2y 2 + y 4) dy = 0and multiply by the factor 1/yx 3 which givesIntegrating, we get the primitiveTheorem 5.a) A necessary and sufficient condition for μ(x, y) to be an integrating factor for the equation13) M(x, y) dx + N(x, y) dy = 0is that it satisfy the equationb) If the quantityis a function of x alone, i.e.thenis an integrating factor for the equationM dx + N dy = 0.c) If the quantityis a function of y alone, i.e.thenis an integrating factor for the equationM dx + N dy = 0.Example.
Solve y dx + (3 + 3x - y) dy = 0Solution.M = y N = 3 + 3x - y∂M/∂y = 1 ∂N/∂x = 3Nowis not a function of x alone, butis a function of y alone. Hence an integrating factor is given byIf one multiplies the equation by y 2, one can confirm that it becomes exact and that its solution isxy 3 + y 3 - y 4/4 = CV Linear equations of first order.
A linear equation of first order is an equation of typeThis equation hasas an integrating factor. The general solution is given byExample. Tony christie rapidshare premiums. SolveSolution.P = 2/xMultiplying both sides of the equation by this factor and integrating we haveorx 2y = x 6 + CBernoulli’s equation.
Differential Equations Pdf Textbook
The equationis known as Bernoulli’s equation. It can be transformed into a linear equation by thetransformation14) y - n+1 = vwhere v is a new variable.Let us divide 13) by y n to obtain the equivalent equationIf we now take the derivative of 14) with respect to x, we obtainSubstituting 14) and 16) into 15) yieldsorwhich is a linear equation in the variable v. We can now solve this equation by the method forlinear equations.The general solution for Bernoulli’s equation isReferences1. Differential and Integral Calculus. Mathematics Dictionary.3.
Applied Differential Equations.4. Differential Equations and Applications.5. Differential Equations (Schaum).6. Handbook of Engineering Fundamentals.7. Earl Rainville. Elementary Differential Equations.More from SolitaryRoad.com:.
The Garland Science website is no longer available to access and you have been automatically redirected to CRCPress.com. INSTRUCTORSAll instructor resources (.see Exceptions) are now available on our. SummaryA Contemporary Approach to Teaching Differential EquationsApplied Differential Equations: An Introduction presents a contemporary treatment of ordinary differential equations (ODEs) and an introduction to partial differential equations (PDEs), including their applications in engineering and the sciences. Designed for a two-semester undergraduate course, the text offers a true alternative to books published for past generations of students. It enables students majoring in a range of fields to obtain a solid foundation in differential equations.The text covers traditional material, along with novel approaches to mathematical modeling that harness the capabilities of numerical algorithms and popular computer software packages.
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It contains practical techniques for solving the equations as well as corresponding codes for numerical solvers. Many examples and exercises help students master effective solution techniques, including reliable numerical approximations.This book describes differential equations in the context of applications and presents the main techniques needed for modeling and systems analysis. It teaches students how to formulate a mathematical model, solve differential equations analytically and numerically, analyze them qualitatively, and interpret the results. Reviews'The author covers traditional material along with modern approaches for analyzing, solving, and visualizing ODEs. The topics are accompanied by mathematical software codes for some of the most popular packages.
The text contains a large number of examples from different areas. It also includes advanced material for students who want to obtain a deeper knowledge on this subject. The textbook contains a great number of exercises. In addition, each chapter ends with summary and review questions, making the text well-suited for self-study as well.' — Zentralblatt MATH 1326'Two notable aspects of the book are its comprehensiveness and tight integration with applications. There is much to like about this book — lucid writing, clear development of the basic ideas, and a very large number of exercises with a good range of difficulty. A very attractive text.'
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