Questions and Answers #3 Second Order Linear Equations

Questions and Answers Sheet 3 Second Order Linear Equations Question #1 The question is that find the general solution of differential equation 𝑦 ′′ − 2𝑦 ′ + 𝑦 = 𝑒 𝑥 i know that 𝑦𝑐 = 𝐴𝑥𝑒 𝑥 + 𝐵𝑒 𝑥 then let the 𝑓(𝑥) = 𝑒 𝑥 , so 𝑦 = 𝑝𝑥𝑒 𝑥 as f(x) is in the complementary function. So 𝑦 ′ = 𝑝𝑒 𝑥 (𝑥 + 1)𝑦 ′′ = 𝑝𝑒 𝑥 (𝑥 + 2) then i put it into equation, but the answer is (0)𝑝𝑒 𝑥 = 𝑒 𝑥 , then i can't find the unknown number p. Answer: You're off to a good start. The fact that you ended up with the equation 0 = 𝑒 𝑥 isn't due to any calculation error on your part, but rather is because 𝑦(𝑥) = 𝐴𝑥𝑒 𝑥 + 𝐵𝑒 𝑥 is a solution to the homogeneous system, since then 𝑦 ′ = 𝐴𝑥𝑒 𝑥 + 𝐴𝑒 𝑥 + 𝐵𝑒 𝑥 = 𝑦 + 𝐴𝑒 𝑥 so that 𝑦 ′′ − 2𝑦 ′ + 𝑦 = 𝑦 + 2𝐴𝑒 𝑥 − 2(𝑦 + 𝐴𝑒 𝑥 ) + 𝑦 = 0 Your general idea is a good one, though. Instead, consider 𝑦(𝑥) = (𝐴𝑥 2 + 𝐵𝑥 + 𝐶)𝑒 𝑥 , and try that in your system, solving for what parameters you can, and leaving the rest as free parameters Question #2 Consider, 𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐𝑦 = 0 and 𝑎 ≠ 0. Which of the following statements are always true? 1. A unique solution exists satisfying the initial conditions 𝑦(0) = 𝜋, 𝑦 ′ (0) = √𝜋 2. Every solution is differentiable on the interval (−∞, ∞) 3. If 𝑦1 and 𝑦2 are any two linearly independent solutions, then 𝑦 = 𝐶1 𝑦1 + 𝐶2 𝑦2 is a general solution of the equation. Answer: The general solution to this equation is, 𝑦 = 𝐶1 𝑒 𝜆1𝑥 + 𝐶2 𝑒 𝜆2𝑥 where, 𝜆1 and 𝜆2 are roots of the equation 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 1) if we know y(0) and y'(0) , then we can obtain two linear equations in 𝐶1 and 𝐶2 , giving a unique solution. 2) As this function is exponential, we can say it is differentiable everywhere. Question #3 Solve this differential equation: 𝑦 ′′ − 7𝑦 ′ + 10𝑦 = cos 𝑥 where y is a one-variable function x. What am I supposed to do with the cos x? Answer: You can solve by means of complex numbers, generalizing to As you know that the derivatives of the exponential are the same exponential times a coefficient, the solution must be of the form 𝑐𝑒 𝑖𝑥 . Then (𝑖 2 − 7𝑖 + 10)𝑐𝑒 𝑖𝑥 = 𝑒 𝑖𝑥 and 𝑐= 1 9−7𝑖 = 9+7𝑖 130 Finally, the solution is the real part of 𝑐𝑒 𝑖𝑥 , or 9 cos 𝑥−7 sin 𝑥 130 Question #4 Find second order linear homogeneous ODE with constant coefficients if its fundamental set of solutions is {𝑒 3𝑡 , 𝑡𝑒 3𝑡 } Answer: According to the theory of ODE with constant coefficients, the number of elements of a fundamental set of family of solutions coincides with the order of the ODE. So, your ODE here is second order. Moreover, if we consider the ODE with its standard form as ay''+by'+cy=0 so it associated equation is 𝑎𝑚2 + 𝑏𝑚 + 𝑐 = 0 Here you have m=3 of twice order, so the quadratic equation above is a complete square like (𝑚 − 3)2 = 0 and we know that in these cases we can find another solution, for example, by using the reduction method. This latter method gave us (3t). Hence: (𝑚 − 3)2 → 𝑚2 − 6𝑚 + 9 = 0 → 𝑦 ′′ − 6𝑦 ′ + 9𝑦 = 0 is the equation Question #5 I am trying to solve the following: 𝑦 ′′ + 4𝑦 ′ = tan(𝑡) I have used the method of variation of parameters. Currently I am at a point in the equation where I have this: tan 𝑡 cos 2𝑡 𝑢1 = ∫ 2 Answer: cos 2 𝑡 = 2 cos 2 𝑡 − 1 hence tan 𝑡 cos 2𝑡 1 𝑑𝑡 = ∫ sin 𝑡 cos 𝑡 𝑑𝑡 − tan 𝑡 𝑑𝑡 ∫ 1 2 2 1 = sin2 𝑡 − ln(sec 𝑡) + 𝐶 2 1 2 = (sin2 𝑡 + ln(cos 𝑡)) + 𝐶 2 Question #6 Consider the following linear difference equation 1 1 𝑓𝑘 = 1 + 𝑓𝑘+1 + 𝑓𝑘−1 , 1 ≤ 𝑘 ≤ 𝑛 − 1 2 2 with 𝑓0 = 𝑓𝑛 = 0. How do i find solution? Answer: Apart from trying what Will Jagy said, you can also note that, for n=2,3,4,..., ∑𝑛𝑘=2(𝑛 − 𝑘 + 1) ⋅ (−1) = ∑𝑛𝑘=2(𝑛 − 𝑘 + 1)(𝑓𝑘 − 2𝑓𝑘−1 + 𝑓𝑘−2 ) n −1 = f n +  ((n − k + 1− 2(n − k ) + (n − k − 1)) f k + ((n − 2) − 2(n − 1)) f (1) + (n − 1) f (0) k =2 = 𝑓𝑛 − 𝑛𝑓(1) + (𝑛 − 1)𝑓(0) Thus, for every n=0,1,2,..., 𝑓𝑛 (𝑛) = 𝑛𝑓(1) − (𝑛 − 1)𝑓(0) − ∑𝑛𝑘=2(𝑛 − 𝑘 + 1) = 𝑛𝑓(1) − (𝑛 − 1)𝑓(0) − 𝑛(𝑛−1) 2 Question #7 The differential equation is as follows: (𝑥 + 1)𝑦 ′′ − (2𝑥 + 3)𝑦 ′ + (𝑥 + 2)𝑦 = 0 where y(x) is a function on 𝑅. First, I had to show that 𝑦1 (𝑥) = 𝑒 𝑥 is a solution for the DV. The second question was to find the general solution of the DV. I instantly started on the series of solution solving method, but I think I have to use the 'hint' 𝑦1 (𝑥) = 𝑒 𝑥 is a solution to the DV, but how to do this? Answer: Let 𝑦(𝑥) = 𝑦1 (𝑥)𝑧(𝑥) = 𝑒 𝑥 𝑧(𝑥) where z(x) is your new unknown function. Then 𝑦 ′ (𝑥) = 𝑒 𝑥 𝑧(𝑥) + 𝑒 𝑥 𝑧 ′ (𝑥) and y''(x)=.. , and when you substitute this into your ODE you get something with a z'' term and a z' term, but no z term, so if you let w(x)=z'(x) you get a first-order linear ODE for w , and such equations can always be solved in principle, using an integrating factor. Question #8 I got the sum of A is 0? There is no solution to this? 𝑦 ′′ − 4𝑦 ′ + 4𝑦 = −6𝑒 2𝑡 Answer: Hint: make for the particular solution the ansatz 𝑦𝑝 = 𝐴𝑒 2𝑡 𝑡 2 Question #9 I am having trouble figuring out a method for finding a solution to 2 𝑟 + ( ) (𝑟 ′ )2 − 𝑟 ′′ = 0 𝑟 I have tried substitution of w=r' to obtain 2 𝑑𝑤 𝑟 𝑑𝑟 𝑟 + ( ) 𝑤2 = 𝑤 But then I am not sure how to proceed from there. Any suggestions are welcome. Answer: What if we let 𝑢(𝑡) = 1/𝑟(𝑡)? Then 1 2 1 𝑟 ′ = − 2 𝑢′ 𝑎𝑛𝑑 𝑟 ′′ = 3 (𝑢′ )2 − 2 𝑢′′ 𝑢 Hence 𝑟 + = 𝑢 2 𝑟 (𝑟 ′ )2 ′′ 𝑢 1 1 𝑢 𝑢2 − 𝑟 = + 2𝑢 (− 2 𝑢′ ) − 2 𝑢3 (𝑢′ )2 + 1 𝑢2 𝑢′′ 𝑢+𝑢′′ 𝑢2 Can you take it from here? Question #10 I'm trying to find the general solution to: 𝑦 ′′ + 4𝑦 = 𝑡 2 + 7𝑒 𝑡 The actual problem wants me to find the initial value problem with y(0) = 0 and y'(0) = 2 but I'm confident that I can find the IVP after finding the general solution. What I DO need help with is this: I'm trying to set 𝑌(𝑡) = 𝐴𝑡 2 + 𝐵𝑡 + 𝐶 and solving for A, B, and C for a specific solution but I find two different values for A. (1/4, and 0). I did solve for 𝑒 𝑡 and found the answer to be 7/5. Answer: the characteristic equation is 𝑚2 + 4 = 0𝑚 = ±2𝑖 𝑦𝑐 = 𝐶1 cos 2 𝑡 + 𝐶2 sin 2 𝑡 𝑦𝑝 = 𝐴𝑡 2 + 𝐵𝑡 + 𝐶 + 𝐷𝑒 𝑡 𝑦𝑝′ = 2𝐴𝑡 + 𝐵 + 𝐷𝑒 𝑡 𝑦𝑝′′ = 2𝐴 + 𝐷𝑒 𝑡 substitute in the original equation 2𝐴 + 𝐷𝑒 𝑡 + 4𝐴𝑡 2 + 4𝐵𝑡 + 4𝐶 + 4𝐷𝑒 𝑡 = 𝑡 2 + 7𝑒 𝑡 so A=1/4 B=0 C=-1/8 D=7/5 the general solution is 𝑡2 1 7 𝑦 = 𝐶1 cos 2 𝑡 + 𝐶2 sin 2 𝑡 + − + 𝑒 𝑡 4 8 5