Differential Equations

Chapter 5 Differential equations 5.1 Ordinary and partial differential equations A differential equation is a relation between an unknown function and its derivatives. Such equations are extremely important in all branches of science; mathematics, physics, chemistry, biochemistry, economics,. Typical example are • Newton’s law of cooling which states that the rate of change of temperature is proportional to the temperature difference between it and that of its surroundings. This is formulated in mathematical terms as the differential equation dT = k(T − T ), dt 0 where T (t) is the temperature of the body at time t, T0 the temperature of the surroundings (a constant) and k a constant of proportionality, • the wave equation, ∂2u ∂2u = c2 , ∂t2 ∂x2 where u(x, t) is the displacement (from a rest position) of the point x at time t and c is the wave speed. The first example has unknown function T depending on one variable t and the relation involves the first dT order (ordinary) derivative . This is a ordinary differential equation, abbreviated to ODE. dt The second example has unknown function u depending on two variables x and t and the relation involves ∂2u ∂2u the second order partial derivatives and . This is a partial differential equation, abbreviated to PDE. ∂x2 ∂t2 The order of a differential equation is the order of the highest derivative that appears in the relation. The unknown function is called the dependent variable and the variable or variables on which it depend are the independent variables. A solution of a differential equation is an expression for the dependent variable in terms of the independent one(s) which satisfies the relation. The general solution includes all possible solutions and typically includes arbitrary constants (in the case of an ODE) or arbitrary functions (in the case of a PDE.) A solution without arbitrary constants/functions is called a particular solution. Often we find a particular solution to a differential equation by giving extra conditions in the form of initial or boundary conditions. 49 Example 5.1 Show that cos ct and sin ct are solutions of the second order ODE u¨ + c2u = 0, where c is a constant. Deduce that A cos ct + B sin ct is also a solution for arbitrary constants A, B. Remark It is conventional to useu ˙ to denote the derivative (of u) with respect to t andu ¨ the second derivative with respect to t. In a similar way we will use u0 and u00 to denotes derivatives with respect to x. Solution : ¤ Remarks 1. A differential equation which contains no products of terms involving the dependent variable is said to be linear. For example, d2y + x2 = 0, u + u = 0, dx t x are linear but d2y + y2 = 0, u + uu = 0, dx2 t x are nonlinear. The ODE in the above example is also linear. As illustrated, any linear combination (A cos ct+B sin ct) of given solutions (cos ct, sin ct) is also a solution. This is true for all linear differential equations and makes them much easier to solve. It is not true of nonlinear differential equations. 2. The solution u = A cos ct + B sin ct contains two arbitrary constants and is the general solution of the second order ODEu ¨ + c2u = 0. This illustrates the fact that the general solution of an nth order ODE contains n arbitrary constants. 50 5.2 First order ODEs We will study methods for solving first order ODEs which have one of three special forms. Separable type1 Consider first, for example, the ODE dy = x. dx This is precisely the same as writing Z 1 2 y = x dx = 2 x + C, where C is an arbitrary constant—the constant of integration. This is the general solution of the ODE. More generally, ODEs of the form dy = f(x)g(y), dx are called separable and can be solved in a similar way. Dividing by g(y) and integrating both sides with respect to x we get Z Z 1 dy dx = f(x) dx. g(y) dx Recalling the formula for integration by change of variables, we see that the integral on the left is equal to Z 1 dy. g(y) Hence the separable ODE is equivalent to the relationship between integrals Z Z 1 dy = f(x) dx. g(y) Assuming that these integrals may be evaluated, we get G(y) = F (x) + C, where C is an arbitrary constant. This gives the general solution. Example 5.2 Find the general solution of y0 = ex+4y, and the particular solution for which y(0) = 0. Solution : 1This type of ODE is studied in level-1 modules. 51 Answer: The general solution, 1 x 1 x y = − 4 log(−4(e + C)) = − 4 log(C − 4e ), where we have relabelled the arbitrary constant C (= −4 × old value of C). The particular solution is 1 x y = − 4 log(5 − 4e ). ¤ Remark The same techniques may also be used to solve PDEs which are separable in the same sense. For example, the PDE ∂u y = u2, ∂x only has a derivative with respect to x and so we regard y as fixed and rewrite it as Z Z du y 2 = dx. y fixed u y fixed Hence 1 −y −y = x + A(y), i.e., u = . u x + A(y) where the “constant of integration” is the arbitrary function A(y). This must, in general, depend on y since this variable was fixed during the integration. Exact type An ODE of the form d φ(x, y) = 0, dx is said to be exact and obviously has the general solution φ(x, y) = C, where C is an arbitrary constant. Being able to recognise that an ODE can be expressed in this form is more difficult however. Using the chain rule for functions of two variables, we have d ∂φ dy ∂φ φ(x, y) = + , dx ∂x dx ∂y (see Example ??). Hence an ODE of the form dy P (x, y) + Q(x, y) = 0, dx is exact provided there exists some φ(x, y) such that ∂φ ∂φ P = and Q = . ∂x ∂y For example, the ODE dy y + x = 0, dx is exact since we can take φ = xy giving ∂φ ∂φ = y and = x. ∂x ∂y In general, exact ODEs are characterised by the following theorem. 52 Theorem The ODE dy P (x, y) + Q(x, y) = 0, dx is exact if and only if ∂P ∂Q = . ∂y ∂x Proof If the ODE is exact then there exists φ such that ∂φ ∂φ P = and Q = , ∂x ∂y and hence it is necessary that ∂P ∂2φ ∂2φ ∂Q = = = . ∂y ∂y∂x ∂x∂y ∂x The proof that this condition is enough to guarantee the existence of φ is omitted. ¤ Example 5.3 Show that the ODE ¡ ¢ dy y + cos(x + y) + x − y + cos(x + y) = 0, dx is exact an find its general solution. Solution : 53 Answer: The general solution is 1 2 xy + sin(x + y) − 2 y = C, where C is an arbitrary constant. ¤ Integrating factors The differential equation dy P + Q = 0, (1) dx has the same solutions as the one obtained by multiplying through by a factor µ(x, y) dy (µP ) + (µQ) = 0. (2) dx This opens up the possibility that, by multiplying by an integrating factor µ, we may convert a non-exact ODE (1), for which ∂P ∂Q 6= , ∂y ∂x into an exact one (2), for which ∂ ∂ (µP ) = (µQ). ∂y ∂x An important special case of this is covered at level-1. Recall that the linear first order ODE dy + a(x)y = b(x), dx has an integrating factor µZ ¶ µ = exp a dx . The ODE may be written as d ¡ ¢ µy = µb, dx and the general solution obtained by integration. Example 5.4 Find the general solution of dy x − 2y = 2x5. dx Solution : 54 Answer: The general solution is Z 2 2 2 2 3 2 5 2 y = x 2x dx = x ( 3 x + C) = 3 x + Cx . ¤ Example 5.5 Show that the ODE dy 2y + 2x2y2 + (x + x3y) = 0, dx is not exact. Find an integrating factor xn (for some n, to be determined) and hence find its general solution. Solution : 55 Answer: The general solution is 2 1 4 2 φ = x y + 2 x y = C. ¤ Example 5.6 Solve the ODE dy 4xy + 5y3 = , dx 4x2 − 2y3 by finding an integrating factor depending only on y. Solution : Answer: The general solution is 2x2 φ = + 5x + 2y = C. y2 ¤ 5.3 Second order ODEs We will only consider linear second order ODEs, those of the form d2y dy a(x) + b(x) + c(x)y = f(x). (1) dx2 dx 56 If the right hand side is identically zero, we say that the ODE is homogeneous. More generally, there is a homogeneous ODE associated with each ODE of the form (1), obtained by replacing f(x) by 0, d2y dy a(x) + b(x) + c(x)y = 0. (2) dx2 dx If the general solution of the homogeneous form (2) is known, the general solution of the full ODE (1) may be found using the following result. Theorem The general solution of the ODE d2y dy a(x) + b(x) + c(x)y = f(x), dx2 dx is y = CF + PI, where CF is the general solution of homogenous form d2y dy a(x) + b(x) + c(x)y = 0, dx2 dx called the complementary function and PI is any solution of the full ODE, called a particular integral.

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