“main”2007/2/16page 67iiiiiiii1.8 Change of Variables 6718. Consider the special case of the RLC circuit in whichthe resistance is negligible and the driving EMF iszero. The differential equation governing the chargeon the capacitor in this case isd2qdt2+1LCq = 0.If the capacitor has an initial charge of q0coulombs,and no current is flowing initially, determine the chargeon the capacitor for t>0, and the correspondingcurrent in the circuit. [Hint: Let u = dq/dt anduse the chain rule to show that this implies du/dt =u(du/dq).]19. Repeat the previous problem for the case in which thedriving EMF is E(t) = E0, a constant.1.8Change of VariablesSo far we have introduced techniques for solving separable and first-order linear dif-ferential equations. Clearly, most first-order differential equations are not of these twotypes. In this section, we consider two further types of differential equations that can besolved by using a change of variables to reduce them to one of the types we know howto solve. The key point to grasp, however, is not the specific changes of variables thatwe discuss, but the general idea of changing variables in a differential equation. Furtherexamples are considered in the exercises. We first require a preliminary definition.DEFINITION 1.8.1A function f(x,y) issaidtobehomogeneous of degree zero7iff(tx,ty) = f (x,y)for all positive values of t for which (tx, ty) is in the domain of f .Remark Equivalently, we can say that f is homogeneous of degree zero if it isinvariant under a rescaling of the variables x and y.The simplest nonconstant functions that are homogeneous of degree zero aref(x,y) = y/x, and f (x, y) = x/y.Example 1.8.2 If f(x,y) =x2− y22xy + y2, thenf(tx,ty) =t2(x2− y2)t2(2xy + y2)= f(x, y),so that f is homogeneous of degree zero. In the previous example, if we factor an x2term from the numerator and denominator,then the function f can be written in the formf(x,y) =x2[1 − (y/x)2]x2[2(y/x) + (y/x)2].That is,f(x,y) =1 − (y/x)22(y/x) + (y/x)2.7More generally, f(x, y) issaidtobehomogeneous of degree m if f(tx,ty) = tmf(x, y).“main”2007/2/16page 68iiiiiiii68 CHAPTER 1 First-Order Differential EquationsThus f can be considered to depend on the single variable V = y/x. The followingtheorem establishes that this is a basic property of all functions that are homogeneous ofdegree zero.Theorem 1.8.3 A function f(x,y) is homogeneous of degree zero if and only if it depends on y/x only.ProofSuppose that f is homogeneous of degree zero. We must consider two casesseparately.(a) If x>0, we can take t = 1/x in Definition 1.8.1 to obtainf(x,y) = f(1, y/x),which is a function of V = y/x only.(b) If x<0, then we can take t =−1/x in Definition 1.8.1. In this case we obtainf(x,y) = f(−1, −y/x),which once more depends on y/x only.Conversely, suppose that f (x,y) depends only on y/x. If we replace x by tx and yby ty, then f is unaltered, since y/x = (ty)/(tx), and hence is homogeneous of degreezero.Remark Do not memorize the formulas in the preceding theorem. Just remember thata function f (x,y) that is homogeneous of degree zero depends only on the combinationy/x and hence can be considered as a function of a single variable, say, V , whereV = y/x.We now consider solving differential equations that satisfy the following definition.DEFINITION 1.8.4If f(x,y) is homogeneous of degree zero, then the differential equationdydx= f(x, y)is called a homogeneous first-order differential equation.In general, ifdydx= f(x, y)is a homogeneous first-order differential equation, then we cannot solve it directly. How-ever, our preceding discussion implies that such a differential equation can be written inthe equivalent formdydx= F(y/x), (1.8.1)for an appropriate function F . This suggests that, instead of using the variables x and y,we should use the variables x and V , where V = y/x, or equivalently,y = xV (x). (1.8.2)“main”2007/2/16page 69iiiiiiii1.8 Change of Variables 69Substitution of (1.8.2) into the right-hand side of Equation (1.8.1) has the effect ofreducing it to a function of V only. We must also determine how the derivative termdy/dx transforms. Differentiating (1.8.2) with respect to x using the product rule yieldsthe following relationship between dy/dx and dV/dx:dydx= xdVdx+ V.Substituting into Equation (1.8.1), we therefore obtainxdVdx+ V = F(V),or equivalently,xdVdx= F(V)− V.The variables can now be separated to yield1F(V)− VdV =1xdx,which can be solved directly by integration. We have therefore established the nexttheorem.Theorem 1.8.5The change of variables y = xV (x) reduces a homogeneous first-order differentialequation dy/dx = f(x, y) to the separable equation1F(V)− VdV =1xdx.Remark The separable equation that results in the previous technique can be inte-grated to obtain a relationship between V and x. We then obtain the solution to the givendifferential equation by substituting y/x for V in this relationship.Example 1.8.6 Find the general solution todydx=4x + yx − 4y. (1.8.3)Solution: The function on the right-hand side of Equation (1.8.3) is homogeneous ofdegree zero, so that we have a first-order homogeneous differential equation. Substitutingy = xV into the equation yieldsddx(xV ) =4 + V1 − 4V.That is,xdVdx+ V =4 + V1 − 4V,“main”2007/2/16page 70iiiiiiii70 CHAPTER 1 First-Order Differential Equationsor equivalently,xdVdx=4(1 + V2)1 − 4V.Separating the variables gives1 − 4V4(1 + V2)dV =1xdx.We write this as14(1 + V2)−V1 + V2dV =1xdx,which can be integrated directly to obtain14arctan V −12ln (1 + V2) = ln |x|+c.Substituting V = y/x and multiplying through by 2 yields12arctanyx− lnx2+ y2x2= ln (x2) + c1,which simplifies to12arctanyx− ln (x2+ y2) = c1. (1.8.4)Although this technically gives the answer, the solution is more easily expressed in termsof polar coordinates:x = r cos θ and y = r sin θ ⇐⇒ r = x2+ y2and θ = arctanyx.Substituting into Equation (1.8.4) yields12θ −ln (r2) = c1,or equivalently,ln r =14θ +c2.Exponentiating both sides of this equation givesr = c3eθ/4.For each value of c3, this is the equation of a logarithmic spiral. The particular spiralwith equation r =12eθ/4is shown in Figure 1.8.1.“main”2007/2/16page 71iiiiiiii1.8 Change of Variables 71864224 2xy42Figure 1.8.1: Graph of the logarithmic spiral with polar equation r =12eθ/4,−5π/6 ≤ θ ≤ 22π/6.Example 1.8.7Find the equation of the orthogonal