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

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

Now, we wish to show that a d2y dv2 + (b − a) dy dv + cy = 0, so we simply substitute in the expressions we’ve computed for d 2 y/dv 2 and dy/dv. a d2 � 2 y d y + (b − a)dy + cy = a dv2 dv dx2 � � � e ev 2v + dy � � � e dx ev v � � dy � � + (b − a) � e dx ev v � + cy = a d2y dx2 � � � e ev 2v + b dy � � � dx ev = a d2y dx2 � � � x x 2 + b dy � � � x + cy dx x = ay ′′ x 2 + by ′ x + cy = 0 e v + cy now, recall that x = e v We obtain the third from last line, since d 2 y/dx 2 |x is simply the second derivative of y with respect to x evaluated at x - in other words, our good old y ′′ , and similarly dy/dx|x is simply y ′ . The fact that the last line is 0 is simply because y is a solution to the original [Equation 1]. Alternative: We can get the same derivation as above, going the other way, which has a little clearer notation. Suppose that y(v) is a function such that y(ln x) is a solution to [Equation 1]. We wish to show that ay ′′ (v) + (b − a)y ′ (v) + cy(v) = 0. To start off, note that (y(ln x)) ′ = y ′ (ln x)(1/x) and (y(ln x)) ′′ = y ′′ (ln x)(1/x 2 ) + y ′ (ln x)(−1/x 2 ). Then, we have the following: 0 = ax 2 (y(ln x)) ′′ + bx(y(ln x)) ′ + cy(ln x) = ax 2 (y ′′ (ln x)(1/x 2 ) − y ′ (ln x)(1/x 2 )) + bxy ′ (ln x)(1/x) + cy(ln x) = ay ′′ (ln x) − ay ′ (ln x) + by ′ (ln x) + cy(ln x) = ay ′′ (ln x) + (b − a)y ′ (ln x) + cy(ln x) = ay ′′ (v) + (b − a)y ′ (v) + cy(v) (b) After transforming [Equation 1] to [Equation 2] we obtain a second order linear equation with constant coefficients. Thus, we can solve [Equation 2] simply by finding roots of the characteristic equation ar 2 + (b − a)r + c = 0. Suppose that we’re in a situation where this characteristic equation has distinct real roots r1 and r2. Then we know that a general solution to [Equation 2] is given by y = c1e r1v +c2e r2v (note that the independent variable here is v, since [Equation 2] has independent variable v). But since v = ln x, by substitution we have that y = c1e r1v + c2e r2v = c1e r1 ln x + c2e r2 ln x = c1x r1 + c2x r2 is a solution to [Equation 1], which is what we set out to show. 4
Problem (HW 7, Problem 2) Given one solution y1(x) to the ODE: (i) verify that y1(x) is a solution to the given ODE, and (ii) use the method of Reduction of Order to find the general solution: x 2 y ′′ − 4xy ′ + 6y = 0, x > 0, y1 = x 2 4x 2 y ′′ − 4xy ′ + 3y = 0, x > 0, y1 = x 1/2 x 2 y ′′ + xy ′ + y = 0, x > 0, y1 = cos(ln(x)) For these problems we’ll use the following fact from reduction of order: If y1 is a solution to a second order linear differential equation y ′′ + p(x)y ′ + q(x)y = 0, then we can obtain the general solution yg = vy1, where the function v satisfies the differential equation v ′′ + v ′ (2y ′ 1 + y1p)/y1 = 0 (a) We have y ′ 1 = 2x and y ′′ 1 = 2, so substituting we have x 2 2 − 4x(2x) + 6x 2 = 0, and so indeed y1 is a solution. To find the general solution, we’ll use reduction of order. First, our given equation in standard form is y ′′ − (4/x)y ′ + (6/x 2 )y = 0. Then, we must find the unknown function v that satisfies v ′′ + v ′ (2y ′ 1 + y1p)/y1 = 0. Substituting, we wish to solve the differential equation v ′′ + v ′ � � 2 −4 4x + x x x2 = v ′′ = 0 This equation is easy to solve: Integrating twice yields v = c1x + c2, and so our general solution is given by y = y1v = x 2 (c1x + c2) = c1x 3 + c2x 2 . (b) Proceeding exactly as before, we have y ′ 1 = (1/2)x−1/2 and y ′′ 1 = (−1/4)x−3/2 , and then 4x2 (−1/4)x−3/2− 4x(1/2)x−1/2 +3x1/2 = −x1/2−2x1/2 +3x1/2 = 0, and so y1 is indeed a solution. To find the general solution, we proceed with reduction of order with the standard form equation y ′′ −(1/x)y ′ +(3/(4x2 ))y = 0 � � −1/2 1/2 x + x (−1/x) v ′′ + v ′ x 1/2 = v ′′ = 0 Once again, we simply have to solve v ′′ = 0, and so integrating twice yields v = c1x + c2. Then, our general solution is given by y = vy1 = c1x 3/2 + c2x 1/2 . (c) We first verify that y1 is indeed a solution. We have y ′ 1 = − sin(ln(x))(1/x) and y ′′ 1 = −(cos(ln(x)))(1/x 2 )− sin(ln(x))(−1/x 2 ). Then, substituting, we have x 2 (−(cos(ln(x)))(1/x 2 )− sin(ln(x))(−1/x 2 )) + x(− sin(ln(x))(1/x)) + cos(ln(x)) = − cos(ln x) + sin(ln x) − sin(ln x) + cos(ln x) = 0, and so y1 is a solution to our equation. In standard form our equation is given by y ′′ +y ′ /x+y/x 2 = 0. To find the general solution, we must find the function v that satisfies v ′′ ′ (2(− sin(ln x)(1/x)) + cos(ln x)(1/x)) + v = v cos(ln x) ′′ ′ cos(ln x) − 2 sin(ln x) + v = 0 x cos(ln x) If we set w = v ′ , then this is a first order linear equation, so we can apply our techniques from earlier 5
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Math 320 - Fall 2010 HW #7 Selected Solutions Problem 5.1.32 Let ...

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