Module 12 - Differential Equations 2 (self Study)

SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self Study Course

MODULE 12

DIFFERENTIAL EQUATIONS II

Module Topics 1. First order ordinary differential equations of the type

x dx =f dt t

2. Exact first order ordinary differential equations 3. Linear first order ordinary differential equations 4. Applications

A:

Work Scheme based on JAMES (THIRD EDITION)

1. You should remember that differential equations were considered in Module 6. A number of basic definitions were introduced and then some first order ordinary differential equations of separable type and second order homogeneous ordinary differential equations with constant coefficients were solved. Remind yourself of the contents of that Module before moving on. 2. Some other types of first order ordinary differential equations which can be solved analytically are studied in this Module. Turn to the top of p.692 and study the solution method for equations of type 10.16, in x dx depends only on the ratio . The equation is reduced to separable type using the which the derivative dt t x substitution y = . t N.B. The notation used in J. is different from that in the Data Book. Do make sure you can apply the formulae in the Data Book to equations which have x and t (and not y and x) as the dependent and independent variables respectively. Work through Example 10.14. The key point to note is that the original equation in the variables x and t is replaced by a new equation in the variables y and t. This is achieved by introducing the new variable dx dy dy x =t + 1.y = t + y, which y = , i.e. x = ty. Differentiating the latter with respect to t then gives t dt dt dt is substituted into the original ordinary differential equation, as shown in Example 10.14. An analytical solution exists when the resulting equation for y(t) can be solved. ***Do Exercises 13(b), 17(d), 18(b) on pp.693 and 694*** 3. You must next study exact first order ordinary differential equations. Before reading the relevant section in J. we must reconsider partial derivatives, which were introduced at the end of Module 4. Much more on partial derivatives will appear in Module 19, but two topics require consideration here in order that you might understand the theory on solving exact first order ordinary differential equations. First order partial derivatives were defined in Module 4, and second and higher order partial derivatives can be obtained with an obvious generalisation of the procedures for ordinary derivatives. For example, given f (x, y) = x2 y 2 , it follows that ∂f ∂f = 2xy 2 , = 2x2 y. ∂x ∂y –1–

Higher derivatives are then determined, using an obvious notation, by ∂ ∂2f = 2 ∂x ∂x ∂ ∂2f = ∂x∂y ∂x ∂ ∂2f = ∂y∂x ∂y



∂f ∂x

 



∂f ∂y ∂f ∂x

=

∂ (2xy 2 ) = 2y 2 , ∂x

=

∂ (2x2 y) = 4xy, ∂x

=

∂ (2xy 2 ) = 4xy. ∂y

 

∂2f ∂2f = . It can be ∂x∂y ∂y∂x proved that these mixed second order partial derivatives are always equal for all reasonable functions f . For the particular function f chosen here, f (x, y) = x2 y 2 , it is shown above that

The second topic needed for our current discussion concerns the chain rule. You have considered the latter applied to differentiation of functions of a single variable, but the chain rule can be applied to functions of more than one variable. It is sufficient here to note that for a function h satisfying h(x, t) = c, where c is a constant, which links the dependent variable x to the independent variable t, the appropriate expression for the time derivative of h, using the chain rule for functions of two variables (to be considered more fully in Module 19), is ∂h dx ∂h dh = + . dt ∂x dt ∂t 4. Turn now to J. p.694 and study section 10.5.5 up to the beginning of Example 10.16, noting that condition 10.19 must hold for the equation 10.17 to be exact. Work through Example 10.16. For this Example p(x, t) = ln(sin t) − 3x2 , q(x, t) = x cos t + 4t, and hence the consistency condition 10.19 can be checked. After confirming the latter is satisfied equations 10.18 are written down, for solution. Unfortunately, solving partial differential equations, however simple they are, is more complicated than solving comparable ordinary differential equations. Consider, for instance, the elementary partial differential equation ∂h = x, ∂x which has the general solution x2 + c(t). h= 2 The usual constant of integration must now become an arbitrary function of the other variable(s), since when differentiating partially with respect to x the other variable(s), here only t, is held fixed and hence c(t) is constant. The partial differential with respect to x of this term is therefore zero, and one recovers the original ∂h ∂h differential equation. In Example 10.16 the more complicated expressions for and are integrated ∂x ∂y in a similar way, giving two forms for the function h, each containing a function to be determined. These two expressions for h are different representations of the same function and, therefore, must be consistent, leading to the final solution stated. ***Do Exercises 23(e), 25(b), 24(d) on p.697*** 5. First order linear ordinary differential equations have the general form of Equation 10.20 on p.698. This type of equation occurs in a number of different physical situations. Read through the theory in section 10.5.7 up to the bottom of p.699. This determination of the solution procedure is quite complicated, but in order to solve particular problems you only need to remember the four important steps in the process.

–2–

dx is 1. Observe the form of p(t). (a) Write the equation in the general form 10.20 where the coefficient of dt R (b) Calculate g(t) = e p(t)dt ; g(t) is called the integrating factor. (c) Multiply the equation (in the form 10.20) by g(t). The resulting equation can then always be rewritten d (xg(t)) = g(t) r(t). dt (d) Integrate this equation with respect to t to determine x. Work through Example 10.18 observing the four steps above. First the equation is in the standard form 10.20, so it is possible to calculate g(t) immediately, exp(t2 /2). Multiplying throughout by this integrating
d x exp(t2 /2) . Finally, in this case both factor it is apparent that the left-hand-side can be expressed dt t2 in the integral on the right-hand-side, if you need sides can be integrated with respect to t (using u = 2 to introduce a substitution). ***Do Exercises 31(c), 33(c), 32(c) on p.701*** 6. First order ordinary differential equations of separable type were considered in Module 6, and three other types of first order equations have been solved in the sections above. Some first order equations fall into more than one category, and you should investigate the types of first order equation in the following order: separable,

x dx =f , dt t

linear,

exact,

solving any given equation using the method appropriate to the first category in which it appears - this will usually give you the easiest method of solution. dx + x = 0. dt Z Z dx dt This equation is separable since it can be written =− ; x t

For illustrative purposes, consider the equation t

and linear with p(t) =

1 , r(t) = 0. t dx x = − , with the right-hand-side depending only dt t

However, the equation can also be rewritten in the form on the ratio x/t; and it is exact since p(x, t) = t and q(x, t) = x imply

∂ ∂ (t) = 1 = (x). ∂t ∂x

dx Thus the equation t + x = 0 falls into all four categories. The simplest method of solution is to treat the dt equation as separable, but it would be good for you to find the solution using all four methods. You should get the same answer in all cases, of course! Some special methods exist for other first order ordinary differential equations. However, there remain a large number of equations which cannot be solved analytically, and for these equations numerical methods must be used. Such methods are not considered in this course. 7. As mentioned earlier first order ordinary differential equation arise in many engineering situations and to conclude this Module it is useful to look at some of these applications. At the beginning of Module 6 you read section 10.2. Four applications were discussed - two of which resulted in first order ordinary differential –3–

equations, so read again sections 10.2.2 and 10.2.3 on pp.673–675. You should now be able to classify Equations 10.6 and 10.7! Finally, turn to J. p.710 and read through section 10.7. Hopefully you have found these applications of interest, but don’t worry if you have found it difficult to understand the modelling involved in setting up the differential equations. In the mathematics examinations this year you will only be asked questions on the solution methods.

B:

Work Scheme based on STROUD (FIFTH EDITION)

A considerable part of the material in this Module can be found in Programme 24 of S. Turn to p.825 and work through frames 27–64. You should note that S. develops the theory using the variables y and x whereas J. uses x and t. Since James is the principal text you should expect questions to be written in terms of the latter pair of variables. Exact first order ordinary differential equations are not considered in S. For this topic you should study sections 3 and 4 of A: Work scheme based on JAMES (THIRD EDITION) above.

–4–

Specimen Test 12 1.

Classify, but do not solve, the equation dx + x = t3 dt

t under as many as possible of the headings:

2.

Separable,

x dx =f , dt t

Find the solution of the equation dx 2 + x=t dt t 1 which satisfies the boundary condition x = when t = 1. 4

3.

Using the substitution x = yt find the general solution of the equation t2

4.

dx = tx + x2 . dt

Write down the condition for the equation p(x, t)

dx + q(x, t) = 0 to be exact. dt

Verify that the following equation is exact and find its general solution: (t + ex )

dx + 2 cos t + x = 0 . dt

–5–

Exact,

Linear.