Questions and Answers #1 Second Order Linear Equations

Questions and Answers Sheet 1 Second Order Linear Equations Question #1 Find equations of the tangent line and normal line to the given curve at the specified point. 𝑦 = 2𝑥𝑒 𝑥 , (0, 0) Answer: Step 1 𝑓(𝑥) = 2𝑥𝑒 𝑥 Differentiate 𝑓 ′ (𝑥) = 𝑑(2𝑥𝑒 𝑥 ) 𝑑𝑥 𝑓 ′ (𝑥) = 2𝑥 × 2𝑥 × 𝑑(𝑒 𝑥 ) 𝑑𝑥 𝑑(𝑒 𝑥 ) 𝑑𝑥 + 𝑒𝑥 × + 𝑒𝑥 × 𝑑(2𝑥) 𝑑𝑥 𝑑(2𝑥) 𝑑𝑥 Using the product Rule 𝑓 ′ (𝑥) = 2𝑥 × (𝑒 𝑥 ) + 𝑒 𝑥 × (2) 𝑓 ′ (𝑥) = 2𝑥 × 𝑒 𝑥 + 2𝑒 𝑥 𝑓 ′ (𝑥) = 𝑒 𝑥 (2𝑥 + 2) Step 2 Slope of the tangent at (0, 0) = 𝑓 ′ (0) = 𝑒 0 × (0 + 2) = 2 𝑦−0 Equation of the tangent is =2 𝑥−0 y=2x Step 3 Product of slopes of perpendicular lines is -1 1 Therefore slope of the normal line is − Equation of the normal line is 𝑦−0 𝑥−0 2 1 =− 𝑥 2 Question #2 Consider the function 𝑓(𝜇) = ∑𝑖=1(𝑥𝑖 − 𝜇)2 , where 𝑥𝑖 = 𝑖, 𝑖 = 1, 2, ⋯ , 𝑛 What is the first and second derivative os 𝑓(𝜇) ? Answer: 𝑛 𝑛 𝑖=1 𝑖=1 𝑑 𝑓(𝜇) = −2 ∑(𝑥1 − 𝜇) = −2 ∑ 𝑥𝑖 + 2𝑛𝜇 𝑑𝜇 𝑑2 𝑑𝜇2 𝑓(𝜇) = 2𝑛 In the 1st step on line 1, the chain rule is applied. The term within the brackets is differentiated, producing a -1. Then the bracket itself is differentiated, producing the 2 at the front. The 2nd step on line 1 involves no differentiation. Instead, the bracket is split into two terms. The second term has an n because it is simply the summation from i=1 to i=n of a constant. The summation of a constant is equal to n multiplied by the constant. Then for the second line, there are no extra rules. The first term becomes 0 because it's a constant and the second term loses 𝜇. Question #3 If we consider the equation (1 − 𝑥 2 ) 𝑑2𝑦 𝑑𝑥 2 − 2𝑥 𝑑𝑦 𝑑𝑥 + 2𝑦 = 0, − 1 < 𝑥 < 1 how can we find the explicit solution, what should be the method for solution? Answer: As mentioned in comment, this ODE is Legendre differential equation for l=1 𝑑 𝑑𝑥 𝑑𝑦(𝑥) [(1 − 𝑥 2 ) 𝑑𝑥 ] + 𝑙(𝑙 + 1)𝑦(𝑥) = 0 The standard Frobenius method will give us the Legendre polynomials. By inspection, it is sort of obvious y(x)=x is one solution for this ODE. In general, if we already know one solution to a second order homogenous ODE, we don't need Frobenius method to find the other one. Given any second order homogenous ODE in its Sturm-Liouville form 𝑑 𝑑𝑥 𝑑𝑦(𝑥) [𝑝(𝑥) ] + 𝑞(𝑥)𝑦(𝑥) = 0 𝑑𝑥 Let 𝑦1 (𝑥) be a solution we already knew, let 𝑦2 (𝑥) be another solution we seek. We have 𝑑 𝑑𝑥 𝑑𝑦2 [𝑝 (𝑦1 = 𝑦1 [ 𝑑 𝑑𝑥 [𝑝 𝑑𝑦1 − 𝑦2 𝑑𝑥 𝑑𝑦2 𝑑𝑥 𝑑𝑥 )] ] + 𝑞𝑦2 ] − 𝑦2 [ 𝑑 𝑑𝑥 [𝑝 𝑑𝑦1 𝑑𝑥 ] + 𝑞𝑦1 ] = 0 This implies 𝑑𝑦2 (𝑥) 𝑝(𝑥) (𝑦1 (𝑥) 𝑑𝑥 − 𝑦2 (𝑥) 𝑑𝑦1 (𝑥) 𝑑𝑥 ) = 𝑐𝑜𝑛𝑠𝑡. Let say's we fix the constant at the left to 1, we can rewrite above as 𝑑 [ 𝑦2 (𝑥) 𝑑𝑥 𝑦1 (𝑥) 1 ]= = 𝑦1 (𝑥) [∫ 𝑝(𝑥)𝑦12 (𝑥) 𝑥 𝑑𝑡 ⇒ 𝑦2 (𝑥) 𝑝(𝑡)𝑦1 (𝑡)2 ] + 𝑐𝑜𝑛𝑠𝑡 is another solution. Apply this to our ODE, the other solution we seek can be chosen as 𝑥∫ 𝑥 𝑥 1 = 𝑥 [− + log ( 𝑥 1 1 𝑡 2 1−𝑡 𝑑𝑡 = 𝑥 ∫ [ 2 + ( (1−𝑡 2 )𝑡 2 1 1 2 −1 (𝑥) 1+𝑥 1−𝑥 1 + 1 1+𝑡 )] 𝑑𝑡 )] = 𝑥 tanh −1 The one metioned in Claude Leibovici's answer. Question #4 I need to solve the following differential equation 𝑥 2 𝑦 ′′ + (𝑎𝑥 − 𝑏)𝑦 ′ − 𝑎𝑦 = 0 with 𝑎, 𝑏 > 0, 𝑥 ≥ 0 and y(0)=0. The power series method will fail since there is a singularity at x=0, while the form of the equation does not conform with the Frobenius method. Answer: 𝑥 2 𝑦 ′′ + (𝑎𝑥 − 𝑏)𝑦 ′ − 𝑎𝑦 = 0 Start by removing the leading behavior for small x. Near x=0 the leading terms in the equation are −𝑏𝑦 ′ − 𝑎𝑦 ≈ 0 (this is valid as long as in the end, as 𝑥 → 0, 𝑦 ′′ does not grow as fast as 𝑦 ′ /𝑥 2 or 𝑦/𝑥 2 , does not grow 𝑦 as fast ). 𝑥 𝑎𝑏 The solution to the approximate equation is 𝑦 = 𝑒 − 𝑥 which motivates the substitution 𝑎𝑏 𝑦 = 𝑢(𝑥)𝑒 − 𝑥 𝑎𝑏 𝑎 𝑎 𝑦 ′ = 𝑢′ 𝑒 − 𝑥 − 𝑢𝑒 −𝑏𝑥 𝑎 𝑏 𝑦 ′′ = 𝑢′′ 𝑒 −𝑏𝑥 − 2 𝑎𝑏 ′ 𝑢 𝑎 𝑒 −𝑏𝑥 + 𝑎2 𝑏2 𝑎 𝑢𝑒 −𝑏𝑥 and the differential equation becomes 𝑎 𝑎2 𝑏 𝑏2 𝑥 2 𝑢′′ − (2 𝑥 2 − 𝑎𝑥 + 𝑏) 𝑢′ + ( with u(0)=0 𝑥2 − 𝑎2 𝑏 𝑥 + 𝑎 − 𝑎) 𝑢 = 0 𝑘 Now if we let 𝑢(𝑥) = ∑∞ 0 𝑢𝑘 𝑥 ∞ ∞ ∞ ∞ ∞ 0 0 0 0 𝑎 𝑎2 𝑎2 ∑ 𝑘(𝑘 − 1)𝑢𝑘 𝑥 − 2 ∑ 𝑘𝑢𝑘 𝑥 𝑘+1 + 𝑎 ∑ 𝑘𝑢𝑘 𝑥 𝑘 − 𝑏 ∑ 𝑘𝑢𝑘 𝑥 𝑘−1 + 2 ∑ 𝑢𝑘 𝑥 𝑘+2 − ∑ 𝑢𝑘 𝑥 𝑘+1 = 0 𝑏 𝑏 𝑏 𝑘 0 And in each term we can substitute an offset index to always get 𝑥 𝑘 this gives 𝑎 𝑎2 𝑏 𝑏 [𝑘(𝑘 − 1) + 𝑎𝑘)𝑢𝑘 − [2 (𝑘 − 1) + ] 𝑢𝑘−1 − 𝑏(𝑘 + 1)𝑢𝑘+1 + 𝑎2 𝑏2 𝑢𝑘−2 = 0 and this is to be solved with 𝑢−1 = 𝑢0 = 0 there is on free parameter 𝑢1 𝑎𝑢 𝑢2 = 1 𝑏 𝑎 and so forth. So the solutions are this series, which is well behaved near zero, times 𝑒 −𝑏𝑥 All of the coefficients are proportional to 𝑢1 . The overall solution has one free parameter and is then 𝑎 2 22 3 5𝑎3 + 3𝑎2 4 17𝑎4 + 39𝑎3 + 36 5 𝑦 = 𝑢1 𝑒 −𝑎𝑥/𝑏 [𝑥 + 𝑥 2 + 𝑥 + 𝑥 + 𝑥 +⋯] 𝑏 3 𝑏2 12𝑏 2 60𝑏 4 = 𝑢1 [𝑥 + 𝑎2 6𝑏 2 𝑥3 + 𝑎3 +3𝑎2 12𝑏 3 𝑥4 + 3𝑎4 +16𝑎3 +24𝑎2 60𝑏 4 𝑥5 + ⋯ ] As to whether this series is convergent for all positive x, or has a finite radius of convergence, that depends on the values of a and b. Question #5 Let y(t) be a nontrivial solution for the second order differential equation 𝑥̈ + 𝑎(𝑡)𝑥̇ + 𝑏(𝑡)𝑥 = 0 to determine a solution that is linearly independent from y we set 𝑧(𝑡) = 𝑦(𝑡)𝑣(𝑡) Show that this leads to a first order differential equation for 𝑣̇ = 𝑤 Answer: First of all, we say that y and z=yv are linearly independent if: 𝐴𝑦 + 𝐵𝑧 = 0∀ 𝑟𝑖𝑔ℎ𝑡 ⇒ 𝐴 = 𝐵 = 0 Notice that: Notice that: 𝐴𝑦 + 𝐵𝑦 = 𝑦(𝐴 + 𝐵𝑣) If v is constant for all t, then 𝐴 = −𝐵𝑣 ⇒ 𝐴𝑦 + 𝐵𝑦𝑣 = 0 Therefore, we need that v is not constant over time in order to have that y and z are linearly independent. Since y(t) is a solution of 𝑥̈ + 𝑎(𝑡)𝑥̇ + 𝑏(𝑡)𝑥 = 0 , then: 𝑦̈ + 𝑎(𝑡)𝑦̇ + 𝑏(𝑡)𝑦 = 0 Let's consider z(t)=y(t)v(t). If z(t) is a solution of 𝑥̈ + 𝑎(𝑡)𝑥̇ + 𝑏(𝑡)𝑥 = 0 then: 𝑧̈ + 𝑎(𝑡)𝑧̇ + 𝑏(𝑡)𝑧 = 0 ⇒ (𝑦̈ 𝑣) + 𝑎(𝑡)(𝑦̇ 𝑣) + 𝑏(𝑡)𝑦𝑣 = 0 ⇒ 𝑦̈ 𝑣 + 2𝑦𝑣̇̇ + 𝑦𝑣̈ + 𝑎(𝑡)(𝑦̇ 𝑣 + 𝑦𝑣̇ ) + 𝑏(𝑡)𝑦𝑣 = 0 ⇒ 𝑣(𝑦̈ + 𝑎(𝑡)𝑦̇ + 𝑏(𝑡)𝑦) + 2𝑦𝑣̇̇ + 𝑎(𝑡)𝑦𝑣̇ = 0 𝑣 ⋅ 0 + 2𝑦𝑣̇̇ + 𝑦𝑣̈ + 𝑎(𝑡)𝑦𝑣̇ = 0 𝑦𝑣̈ + (2𝑦̇ + 𝑎(𝑡)𝑦)𝑣̇ = 0 Now let's introduce 𝑤 = 𝑣̇ . The previous equation can be rewritten as: (𝑣̇ = 𝑤), (𝑦𝑤̇ = −(2𝑦̇ + 𝑎(𝑡)𝑦)𝑤): This means that the non constant function v satisfies the previous system of ODEs. Question #6 I have this second-order ode equation: 𝑦 ′′ − 4𝑦 ′ + 13𝑦 = 0 I've identified it as x missing case as 𝑦 ′′ = 𝑓(𝑦 ′ , 𝑦) = 4𝑦 ′ − 13𝑦, so I'm substituting witth: 𝑦 ′ = 𝑃, 𝑦 ′′ = 𝑃 𝑑𝑦 2 𝑑2𝑥 = 𝑓(𝑃, 𝑦) = 4𝑃 − 13𝑦 At this point I have 𝑃 𝑑𝑝 𝑑𝑦 = 4𝑃 − 13𝑦, which seems a non-linear first-order ODE. This is currently beyond the scope of my course, so I'm unsure if I should continue and search online for solving techniques, or did I already do something wrong? Answer: We have auxiliary equation: 𝑚2 − 4𝑚 + 13 = 0 → 𝑚 = 2 ± 3𝑖 Thus the general solution is: 𝑦 = 𝑒 2𝑥 (𝐴 cos(3𝑥) + 𝐵 sin(3𝑥)) Question #7 Find the linear approximation of the function 𝑓(𝑥) = √1 − 𝑥 at a = 0 and use it to approximate the numbers √0.9 and √0.99. (Round your answers to four decimal places.) L(x) = √0.9 ≈ √0.99 ≈ Answer: 𝑓(𝑥) = √1 − 𝑥 f(0) = 1 pt (0,1) 1 1 𝑓 ′ (𝑥) = ⋅ (1 − 𝑥)−2 ⋅ −1 2 1 1 = (− ) ⋅ (𝑥 − 1)−2 2 1 𝑇(𝑥) = 1 − ⋅ (𝑥) 2 To estimate 0.9 we need to put in 0.1 and 0.01 1 𝑇(0.1) = 1 − ⋅ 0.1 = 1 − 0.05 = 0.95