Math 320 - Fall 2010 HW #7 Selected Solutions Problem 5.1.32 Let ...

More documents

Recommendations

Info

$Lecture notes for Math 522 Spring 2012 (Rudin chapter 7)$

our techniques from earlier in the semester. dW −B(x) = W A(x) dx � � dW B(x) = − W A(x) dx � B(x) ln |W | + C1 = − A(x) dx � B(x) ln |W | = − dx + C A(x) Now, first assume that W > 0. Then we have ln |W | = ln W , and then, taking exp of both sides, and setting K = eC , we obtain � � B(x) W = exp − A(x) dx � e C � � B(x) = K exp − A(x) dx � In this case, note that K = eC is a positive constant. On the other hand, if W < 0, then ln |W | = ln −W , and so taking exp of both sides yields � � B(x) W = −K exp − A(x) dx � Now, since K = e C is positive, −K is negative. Then, we can label the value −K as the new constant K, and obtain the same formula as above W = K exp(− � B(x)/A(x) dx), except now the constant K is negative. Finally, if W = 0, then we don’t need to solve a separable equation (in fact, if you look at the separable equation above, it doesn’t make sense when W = 0 since we’re integrating 1/W ). However, in this case, when W = 0, we can still use our formula W = K exp(− � B(x)/A(x) dx) simply by setting K = 0. (c) [EP10, 5.1 Theorem 3] says something quite peculiar. It claims that the Wronskian W (y1, y2) of two solutions to a homogeneous second order linear equation is either always zero (in which case y1 and y2 are linearly dependent), or, W (y1, y2) is never zero (in which case y1 and y2 are linearly independent). However, this seems to leave a gap - what if the Wronskian is sometimes zero, but not always zero? The formula we proved in [part (b)] explains why this can’t happen. � � B(x) W (x) = K exp − A(x) dx � We know that an exponential function is never zero. Thus, the Wronskian, when it exists, is either always non-zero (if K �= 0) or, always zero (if K = 0). There are no other possibilities, and so [EP10, 5.1 Theorem 3] makes sense. 2
<strong>Problem</strong> 5.1.51 A second-order Euler equation is one of the form where a, b, c are constants. ax 2 y ′′ + bxy ′ + cy = 0, (1) (a) Show that if x > 0, then the substitution v = ln x transforms [Equation 1] into the constantcoefficient linear equation a d2y + (b − a)dy + cy = 0 (2) dv2 dv with independent variable v. (b) If the roots r1 and r2 of the characteristic equation of [Equation 2] are real and distinct, conclude that a general solution of the Euler equation in [Equation 1] is y(x) = c1x r1 + c2x r2 . Observe that [Equation 1] is a second order linear equation, but with non-constant coefficients. The point of this problem is to transform this non-constant equation into an equation with constant coefficients that we can easily solve. One important observation before we get started. The y ′ and y ′′ in [Equation 1] are derivatives with respect to x. We want to transform them into derivatives with respect to v, so we’ll have to apply the chain rule. (a) Note: There’s a slightly shorter way of doing this part; see the paragraph immediately before [part b]. Since v = ln x, we have x = e v . Then, since y is a function of x, we have y = y(x) = y(e v ), and so y is also a function of v. Thus, it makes sense to differentiate y with respect to v, but we must use the chain rule. Recall that if y(x) = y(g(v)), the chain rule can be written dy dv dy � � dx = � dx g(v) dv Where dy dx | g(v) simply denotes the derivative of y with respect to x evaluated at g(v). To obtain an expression d2 y dv 2 , we differentiate the above expression d2y dv2 = d2y dx2 � � dg dx � g(v) dv dv dy � � d + � dx g(v) 2x dv2 This expression follows from [Equation 3]: simply differentiate [Equation 3] with respect to v, using the product rule, and remember to apply the chain rule when differentiating dy � � dx� . g(v) Returning to the problem, suppose that x = e v , and so dx/dv = d(e v )/dv = e v . Then, y(x) = y(e v ), and so applying the chain rule above, we have dy dy � � dx = � dv dx ev dv = dy � � � e dx ev v d2y dv2 = d2y dx2 � � de � ev v dx dy � � d + � dv dv dx ev 2x dv2 = d2y dx2 � � � ev eve v + dy � � de � dx ev v dv = d2y dx2 � � � e ev 2v + dy � � � e dx ev v 3 (3)
Page 1: Math 320 - Fall 2010 HW #7 Selected
Page 5 and 6: Problem (HW 7, Problem 2) Given one

Math 320 - Fall 2010 HW #7 Selected Solutions Problem 5.1.32 Let ...

Create successful ePaper yourself

Delete template?

Save as template?