Questions and Answers #27 Differential equations

Questions and Answers Sheet 27 Differential equations Question #1 Solve the equation: 𝑥(2𝑦 + 1)𝑑𝑥 = 𝑦(𝑥 2 − 3𝑥 + 2)𝑑𝑦 Answer: Step 1 You want to get all the y stuff to one side, and all the x stuff to the other. These are called "separable equations" for this reason. 𝑥(2𝑦 + 1)𝑑𝑥 = 𝑦(𝑥 2 − 3𝑥 + 2)𝑑𝑦 ⇒ 𝑥2 ⇒∫ 𝑥 𝑦 𝑑𝑥 = 𝑑𝑦 − 3𝑥 + 2 2𝑦 − 1 𝑥 𝑥 2 −3𝑥+2 𝑑𝑥 = ∫ 𝑦 2𝑦−1 𝑑𝑦 Now we integrate. The left side looks like partial fraction decomposition and the right side looks like a u-sub with 𝑢 = 2𝑦 − 1. Question #2 Solve the following differential equation by the form of homogeneous equation. Letting 𝑦 = 𝑣𝑥 The equation: 𝑥 2 𝑑𝑦 𝑑𝑥 + 𝑥𝑦 + 1 = 0 Answer: Step 1 Instead, let 𝑣 = 𝑥𝑦 such that 𝑥2 𝑑𝑦 𝑑𝑥 + 𝑥𝑦 = 𝑥 𝑑𝑣 𝑑𝑥 You can solve 𝑥 and so 𝑥 2 𝑑𝑣 𝑑𝑥 𝑑𝑣 𝑑𝑥 𝑑𝑦 𝑑𝑥 =𝑦+𝑥 𝑑𝑦 𝑑𝑥 + 𝑥𝑦 + 1 = 𝑥 𝑑𝑣 𝑑𝑥 +1=0 = −1 by separating. Question #3 Solve the heat equation using a transform method 𝑘 𝜕2 𝑈 𝜕𝑥 2 = 𝜕𝑈 𝜕𝑡 subject to 𝑈(0, 𝑡) = 1, 𝑡 > 0 𝑈(𝑥, 0) = 𝑒 −𝑥 , 𝑥 > 0 Answer: Step 1 If we were to take the Laplace Transform of this expression, assuming ℒ(𝑈(𝑥, 𝑡))(𝑥, 𝑠) = 𝑉(𝑥, 𝑠) we have 𝑘 𝜕2𝑉 = 𝑠𝑉 − 𝑈(𝑥, 0) = 𝑠𝑉 − 𝑒 −𝑥 𝜕𝑥 2 From here we can solve for V in terms of constants that may depend on s. The homogeneous solution is 𝑉ℎ (𝑥, 𝑠) = 𝐶1 (𝑠)𝑒 𝑥√ 𝑠 𝑘 + 𝐶2 (𝑠)𝑒 −𝑥√ 𝑠 𝑘 The particular solution must take the form of 𝐴(𝑠)𝑒 −𝑥 , so 𝑘𝐴(𝑠)𝑒 −𝑥 = 𝑠𝐴(𝑠)𝑒 −𝑥 − 𝑒 −𝑥 ⇒ 𝐴(𝑠) = 1 𝑠−𝑘 Thus, the general solution is 𝑉(𝑥, 𝑠) = 𝑉ℎ (𝑥, 𝑠) = 𝐶1 (𝑠)𝑒 𝑥√ 𝑠 𝑘 + 𝐶2 (𝑠)𝑒 −𝑥√ 𝑠 𝑘 + 𝑒 −𝑥 𝑠−𝑘 We have to employ an assumption that for any fixed value of t, 𝑈(𝑥, 𝑡) → 0 as x gets large, meaning that the total heat density falls to zero. This translates to the fact that as x gets large, 𝑉(𝑥, 𝑠) should remain bounded (vague issues regarding convergence here), and so any growth in x should be ignored. As such, we can assume that 𝐶1 (𝑠) = 0 Next, using the boundary condition, we transform U to get 𝑉(0, 𝑠) = 1 𝑠 Thus, 𝐶2 (𝑠) + 1 1 𝑘 = ⇒ 𝐶2 (𝑠) = − 𝑠−𝑘 𝑠 𝑠(𝑠 − 𝑘) and so 1 1 𝑠 𝑠−𝑘 𝑉(𝑥, 𝑠) = ( − )𝑒 −𝑥√ 𝑠 𝑘 + 𝑒 −𝑥 𝑠−𝑘 From here, we need to leverage the well-known and famous Laplace Transform of ℒ{ 1 3 2√𝜋𝑡 2 1 𝑒 −4𝑡 } = 𝑒 −√𝑠 𝑠 (this can be shown with a variety of clever calculus tricks) The modification to get our version with 𝑥√ is 𝑘 to include a parameter in the exponent and we can show (through u-substitutions) that ℒ{ 𝑥2 𝑥 𝑒 −4𝑘𝑡 } = 𝑒 3 2√𝑘𝜋𝑡 2 −𝑥√ 𝑠 𝑘 Now we can throw everything together using the convolution theorem. Thus, 𝑡 𝑥 𝑈(𝑥, 𝑡) = ∫ (1 − 𝑒 𝑘(𝑡−𝑢) ) 0 𝑥2 − 4𝑘𝑢 𝑑𝑢 + 𝑒 𝑘𝑡−𝑥 3𝑒 2√𝑘𝜋𝑢2 or a little more cleanly: 𝑈(𝑥, 𝑡) = 𝑡 𝑥 2√𝑘𝜋 ∫ 𝑢 0 𝑥2 1 (1 − 𝑒 𝑘(𝑡−𝑢) )𝑒 −4𝑘𝑢 𝑑𝑢 + 𝑒 𝑘𝑡−𝑥 3/2 From here, the integration is nasty, but doable, albeit in terms of non-elementary functions. Question #4 Solve the ode using fixed-point iteration −[(1 + 𝑢4 )𝑢𝑥 ]𝑥 = sin(𝑥) + sin(5𝑥) , where the domain is [0, 2𝜋]𝑢(0) = 𝑢(2𝜋) = 0 for boundaries. Answer: Step 1 The equation can be written in the form (1 + 𝑢4 )𝑢𝑥𝑥 + 4𝑢3 (𝑢𝑥 )2 = 𝑓(𝑥). Using centred finite differences, we get (1 + 𝑢𝑖4 )(𝑢𝑖+1 − 2𝑢𝑖 + 𝑢𝑖−1 ) + 𝑢𝑖3 (𝑢𝑖+1 − 𝑢𝑖−1 )2 = ℎ2 𝑓𝑖 . Finally, you must solve this nonlinear system using a fixed point method. One possibility could be (𝑘+1) 𝑢𝑖 (𝑘) = (𝑘) 3 (𝑘) 𝑢𝑖+1 +𝑢𝑖−1 2 + (𝑢𝑖 (𝑘) (𝑘) 2 ) (𝑢𝑖+1 −𝑢𝑖−1 ) (𝑘) 4 2(1+(𝑢𝑖 ) ) − ℎ 2 𝑓𝑖 (𝑘) 4 2(1+(𝑢𝑖 ) ) Now, you need to check if it converges and, if it does not converge, rewrite the nonlinear system into another fixed point problem. Question #5 Solve this equation in Sturm-Liouville. (1 − 𝑥 2 )𝑦 ′′ − 𝑥𝑦 ′ + 𝜆𝑦 = 0 My solution: 𝜇(1 − 𝑥 2 )𝑦 ′′ − 𝜇𝑥𝑦 ′ + 𝜇𝜆𝑦 = (𝑝𝑦 ′ )′ + (𝜆𝑟 − 𝑞)𝑦 𝜇(1 − 𝑥 2 )𝑦 ′′ − 𝜇𝑥𝑦 ′ + 𝜇𝜆𝑦 = 𝑝′ 𝑦 ′ + 𝑝𝑦 ′′ + (𝜆𝑟 − 𝑞)𝑦 𝑝′ = 𝜇𝑥, 𝑝 = 𝜇(1 − 𝑥 2 ) ⇒ 𝑝′ 𝑥 1 1 = ⇒ 𝑙𝑛(𝑝) = − 𝑙𝑛|1 − 𝑥 2 | ⇒ 𝑝 = 2 𝑝 1−𝑥 2 √(|1 − 𝑥 2 |) 1 Then 𝜇 = (1−𝑥 2)3 1 1 (1−𝑥 2 )2 1 𝑦 ′′ − (1−𝑥 2 )3 𝑥𝑦 ′ + (1−𝑥 2)3 𝜆𝑦 = 𝑝′ 𝑦 ′ + 𝑝𝑦 ′′ + (𝜆𝑟 − 𝑞)𝑦 Answer: Explanation: 𝑄 1 First, denote 𝑃(𝑥) = 1 − 𝑥 2 , 𝑄(𝑥) = −𝑥 ⇒ 𝑒 ∫𝑃𝑑𝑥 = 𝑒 2 ∫ −2𝑥 𝑑𝑥 1−𝑥2 = 𝑒 log|1−𝑥 2 |1/2 = √1 − 𝑥 2 so your integration factor (this could in fact better be called The S-L factor...) is 1 𝑃 𝑄 𝑒 ∫𝑃𝑑𝑥 = 1 1−𝑥 2 √1 − 𝑥 2 = (1 − 𝑥 2 )−1/2 Multiply now the whole diff. equation by it and get: √1 − 𝑥 2 𝑦 ′′ − 𝑥 √1 − 𝑥2 𝑦′ + 𝜆 1 √1 − 𝑥 2 𝑦=0⟺ ′ ((1 − 𝑥 2 )1/2 𝑦 ′ ) + 𝜆√1 − 𝑥 2 𝑦 = 0 and there you have your diff. eq. in Sturm-Liouville normal form. Question #6 Solve this nonhomegenous ode 𝑦 ′′ + 4𝑦 = cos(2𝑥) Answer: Step 1 Then by variation of parameters we have the particular solution is given by 𝑦𝑝 = cos(2𝑥) 𝑢1 + sin(2𝑥) 𝑢2 where 1 1 𝑢1 = − ∫ cos(2𝑥) sin(2𝑥) d𝑥 = cos(4𝑥) 2 16 and 1 𝑥 𝑢2 = ∫ cos(2𝑥) cos(2𝑥) d𝑥 = + 2 4 sin(4𝑥) 16 . Simplify, we have the general solution 1 𝑦 = 𝑐1 cos(2𝑥) + 𝑐2 sin(2𝑥) + 𝑥 sin(2𝑥). 4 Question #7 Solving 𝑦 + 𝑥𝑦 ′ = 𝑎(1 + 𝑥𝑦) 1 𝑦 ( ) = −𝑎 𝑎 Answer: Step 1 Well, it is not hard to see that we rewrite your DE in the following form: 1) y ′ (𝑥) + (1−a𝑥)y(𝑥) 𝑥 = a 𝑥 Now, let: 2) 𝜇(𝑥) ≔ exp{∫ 1−a𝑥 𝑥 \𝑠𝑝𝑎𝑐𝑒d𝑥} = 𝑥 exp(−a𝑥) When we multiply both sides by 𝜇(𝑥) , use the fact that exp(−a𝑥) (1 − a𝑥) = d d𝑥 (𝑥 exp(−a𝑥)) and apply the reverse product rule we end up with: 3) d d𝑥 (𝑥 exp(−a𝑥) y(𝑥)) = a exp(−a𝑥) Integrate both sides with respect to x: 4) 𝑥 exp(−a𝑥) y(𝑥) = C − exp(−a𝑥) Dividing both sides by 𝜇(𝑥): 5) y(𝑥) = C exp(a𝑥)−1 𝑥 So, we can solve for C: 1 6) y ( ) = 1 C exp(a⋅ a )−1 1 a a = a(C𝑒 − 1) = −a C= 2 𝑒 So, we end up with: 2 7) y(𝑥) = 𝑒 ⋅exp(a𝑥)−1 𝑥 = 2 exp(a𝑥)−𝑒 𝑒𝑥 = 2 exp(a𝑥) 𝑒𝑥 − 1 𝑥 Question #8 Solving 2nd order homogenous linear ODE with a squared coefficient I am attempting to solve a differential equation of the form: 𝑑2 𝑓 𝑑𝑥 2 − 𝛾 2𝑓 = 0 I have set up and solved the characteristic equation as: z2 −  2 z + 0z = 0 z( z −  2 ) = 0 which is satisfied when 𝑧 = 0 and when 𝑧 = 𝛾 2 There are two distinct roots ( 𝑟1 and 𝑟2 ) hence the general solution should be of the form 𝑓(𝑥) = 𝐴𝑒 𝑟1𝑥 + 𝐵𝑒 𝑟2 𝑥 In this case: f ( x) = Ae0 x + Be = A + Be 2 2 x x Apparently, this solution is incorrect as in the answers it is given as: 𝑓(𝑥) = 𝐴𝑒 𝛾𝑥 + 𝐵𝑒 −𝛾𝑥 I would like to know where I went wrong to give me the incorrect solution Answer: Without appealing directly to the characteristic equation, you can also proceed as follows f  −  2 f = 0  ( f  −  f ) + ( f  −  2 f ) = 0  ( f  −  f ) +  ( f  −  f ) = 0  g +  g = 0 where 𝑔 = 𝑓 ′ − 𝛾𝑓. Question #9 Solving 3rd Order ODE by Reduction of Order 𝑥 3 𝑦 ′′′ − 3𝑥 2 𝑦 ′′ + (6 − 𝑥 2 )𝑥𝑦 ′ − (6 − 𝑥 2 )𝑦 = 0 Answer: Step 1 Searching for the other homogeneous solutions by assuming it is the product of some function times another known solution is called variation of parameters. Formally we are plugging in the guess 𝑦 = 𝑣 ⋅ 𝑦1 = 𝑥 ⋅ 𝑣 into the equation 𝐷[𝑦] = 0  y = xv  y = v + xv    y = 2v + xv  y = 3v + xv ⇒ 𝑥 ⋅ 𝐷[𝑣] + (3𝑥 3 𝑣 ′′ − 6𝑥 2 𝑣 ′ + (6 − 𝑥 2 )𝑥𝑣) = 0 ⇒ (𝑥 4 𝑣 ′′′ − 3𝑥 3 𝑣 ′′ + (6 − 𝑥 2 )𝑥 2 𝑣 ′ − (6 − 𝑥 2 )𝑥𝑣) + (3𝑥 3 𝑣 ′′ − 6𝑥 2 𝑣 ′ + (6 − 𝑥 2 )𝑥𝑣) = 𝑥 4 (𝑣 ′′′ − 𝑣 ′ ) = 0 In other words v is the three homogeneous solutions to the above differential equation 𝑣 = 𝐶1 𝑒 𝑥 + 𝐶2 𝑒 −𝑥 + 𝐶3 Your professor was justified in calling this a reduction of order because 𝑣 ′′′ − 𝑣 ′ = 0 is secretly a second order differential equation, with the constant solution already represented by the first homogeneous solution 𝑦1 = 𝑥 . This gives us our final solutions 𝑦 = 𝐶1 𝑥 + 𝐶2 𝑥𝑒 𝑥 + 𝐶3 𝑥𝑒 −𝑥 Question #10 Solving a differential equation connecting slope and derivative 𝑦(𝑥)−𝑦(𝑎) 𝑥−𝑎 = 𝑦 ′ (𝑥) Answer: Step 1 Since you are solving the equation for 𝑦: (𝑎, 𝑥) ⇒ 𝑅 , it is implied that 𝑥 > 𝑎 , hence the differential equation is equivalent to 1 𝑥−𝑎 = 𝑦 ′ (𝑥) 𝑦(𝑥)−𝑦(𝑎) , and this is implies ln[𝑦(𝑎) − 𝑦(𝑥)] = ln(𝑥 − 𝑎) + 𝐴, or ln[𝑦(𝑥) − 𝑦(𝑎)] = ln(𝑥 − 𝑎) + 𝐴. Step 2 These are respectively equivalent to 𝑦(𝑎) − 𝑒 𝐴 (𝑥 − 𝑎) = 𝑦(𝑥) or 𝑦(𝑥) = 𝑒 𝐴 (𝑥 − 𝑎) + 𝑦(𝑎). This does assume, though, that lim 𝑦 (𝑥) = 𝑦(𝑎). 𝑥→𝑎 Question #11 Solving a system of differential equations with the variable x How to find a basis of solutions of the system for 3𝑥 − 1 𝑦′ = ( −𝑥 − 2 𝑥−1 )𝑦 𝑥−2 where one solution is 𝑦 = (𝑦1 − 𝑦1 ) ? Answer: Step 1 Given: 3𝑥 − 1 𝑦′ = ( −𝑥 − 2 𝑥−1 )𝑦 𝑥−2 𝑦1′ = (3𝑥 − 1)𝑦1 + (𝑥 − 1)𝑦2 𝑦2′ = (−𝑥 − 2)𝑦1 + (𝑥 − 2)𝑦2 Step 2 Sum both differential equations: (𝑦1 + 𝑦2 )′ = (2𝑥 − 3)(𝑦1 + 𝑦2 ) This DE is separable. I suppose that 𝑑𝑦 𝑑𝑥 = 𝑦′ . Question #12 Solving a systemof eqns by given initial conditions 𝑥 ′ (𝑡) = 2𝑥 + 8𝑦 𝑦 ′ (𝑡) = −𝑥 − 2𝑦 Which has the I.C. 𝑋(0) = (6, −2) . So I take that it means this: 6 = 2𝑥 + 8𝑦 −2 = −𝑥 − 2𝑦 1 and then it is solved as any other system? We get eigenvector (1, ), but since the eigenvalues of the 2 system are two, ±2𝑖 , we have to find the second generalized eigenvector. 1 2 − 2i 8 6 − 2i ( )( 1 ) = ( ) −1 −2 − 2i −2 − i 2 So the second vector would be (6 − 2𝑖, −2 − 𝑖) . Using the general solutions for the system, we get 1 y (t ) = e 2it (1, ) + e −2it (6 − 2i, −2 − i ) 2 Would this be correct, or did I misinterpret the initial conditions? Answer: Step 1 If 2 8 A= ,  −1 −2 4sin(2t ) sin(2t ) + cos(2t )   . then exp(tA) =  − 1 sin(2t ) cos(2t ) − sin(2t )   2  Step 2 So, if 𝑓(𝑡) = exp(𝑡𝐴) . (6, −2) , you have 𝑓(0) = (6, −2) and 𝑓 ′ (𝑡) = 𝐴. 𝑓(𝑡) . So, take l ( x(t ), y(t )r ) = f (t ) = l (6cos(2t ) − 2sin(2t ), − sin(2t ) − 2cos(2t )r ). Question #13 Solving differential equation 𝑑𝑦 𝑑𝑥 = 1 3𝑒 𝑦 −1 results in strange result? Answer: Explanation: In your work, you should have 3𝑒 𝑦 (1 + 𝑑𝑦 𝑑𝑥 )=1 Hint: let 𝑧 = 𝑒 𝑦 , 𝑑𝑧/𝑑𝑥 = 𝑧 ⋅ 𝑑𝑦/𝑑𝑥 and solve the linear ODE in z, 𝑑𝑧 𝑑𝑥 1 = −𝑧 3 such that 𝑧(0) = 𝑒 0 = 1 , then go back to 𝑦 = log(𝑧) . Question #14 Solving homogenous system with complex eigenvalues 𝑑𝑥 = 2𝑥 + 8𝑦 𝑑𝑡 𝑑𝑦 = −𝑥 − 2𝑦 𝑑𝑡 When I solve the determinant of the matrix, I get 𝜆 = ±2𝑖 . Then , I plug it in the matrix, and get the for the first eigenvalue, 𝜆 = 2𝑖 ∶ 0 = (2 − 2𝑖)𝑥 + 8𝑦 0 = −𝑥 − (2 + 2𝑖)𝑦 This gives 𝑥= −8 2−2𝑖 𝑦 𝑥 = −(2 + 2𝑖)𝑦 Answer: Note that −8 2−2𝑖 −8(2+2𝑖) = (2−2𝑖)(2+2𝑖) = −8(2+2𝑖) 8 = −(2 + 2𝑖), so the two equations are saying the same thing (as they must, if the eigenvalues are correctly computed). So you can simply take, for example, 𝑦 = −1 and 𝑥 = 2 + 2𝑖 to get an eigenvector. Question #15 Solving ordinary second-order differential equation I have this ODE: 𝑥̇ 𝑦̇ 𝑧̈ + (3 + ) 𝑧̇ = 0 𝑥 𝑦 Can it be solved to get z in terms of a and b ? Answer: Step 1 𝑥̇ 𝑦̇ 𝑥 𝑦 𝑧̈ + (3 + ) 𝑧̇ = 0 𝑧̈ 𝑧̇ 𝑥̇ 𝑦̇ 𝑥 𝑦 +3 + = 0 Step 2 ln 𝑧̇ + 3 ln 𝑥 + ln 𝑦 = 𝐶 𝑧̇ 𝑥 3 𝑦 = 𝑐1 You can't go further and integrate this DE. You need to knwo what x(t) and y(t) are. You need further informations. Question #16 Solving partial differential equation in 2d with 3 boundary conditions Consider the following pde: −Δ𝑢 = 0 in (0, 𝜋) × (0,4) with boundary conditions: 𝑢(0, 𝑦) = 𝑢(𝜋, 𝑦) for 𝑦 ∈ (0,4) and 𝑢(𝑥, 0) = cos 𝑥 sin 𝑥 for𝑥 ∈ (0, 𝜋) and 𝑢(𝑥, 4) = cos 𝑥 sin 𝑥 + 4 for 𝑥 ∈ (0, 𝜋) Answer: Step 1 The periodic boundary condition tells you that the solution is 𝜋 -periodic in x-direction, or at least can be interpreted as such if suitably extended outside the rectangle. This means that in that direction you get a Fourier series expansion 𝑢(𝑥, 𝑦) = 𝑎0 (𝑦) 2 + ∑ 𝑎𝑘 (𝑦) cos(2𝑘𝑥) + 𝑏𝑘 (𝑦) sin(2𝑘𝑥). This gives the same result as the separation approach, but starting from a different direction. 1 Comparing the conditions on the upper and lower boundaries gives 𝑏1 (0) = 𝑏1 (4) = , 𝑎0 (4) = 8 all 2 other coefficients are zero at 𝑦 = 0 and 𝑦 = 4 . Further, −Δ𝑢(𝑥, 𝑦) = − 𝑎0′′ (𝑦) 2 + ∑𝑘[4𝑘 2 𝑎𝑘 (𝑦) − 𝑎𝑘′′ (𝑦)] cos(2𝑘𝑥) + [4𝑘 2 𝑏𝑘 (𝑦) − 𝑏𝑘′′ (𝑦)] sin(2𝑘𝑥). Now comparing coefficients and solving the simple DE gives 𝑎0 (𝑦) = 2𝑦, 𝑏1 (𝑦) = cosh(2(𝑦−2)) 2 cosh(4) , all other coefficient functions being zero. Question #17 Solving 2𝑥𝑦 ′ (𝑥 − 𝑦 2 ) + 𝑦 3 = 0 Answer: Step 1 Well, we are trying to solve: 1) n𝑥y ′ (𝑥)(𝑥 − y(𝑥)2 ) + y(𝑥)3 = 0 Solving for 𝑦 ′ (𝑥) , gives: 2) y ′ (𝑥) = − y(𝑥)3 n𝑥(𝑥−y(𝑥)2 ) Now, we can write the DE in terms of x, since dy(𝑥) d𝑥 3) = 1 dx(𝑦) d𝑦 1 dx(𝑦) d𝑦 dy d𝑥 d𝑥 dy =1 . So we get: =− 𝑦3 nx(𝑦)(x(𝑦)−𝑦 2 ) Raise both sides to the power -1, expand, subtract nx(𝑦) 𝑦 from both sides and divide both sides by −x(𝑦)2 4) n 𝑦x(𝑦) − dx(𝑦) d𝑦 x(𝑦)2 Let 𝜇(𝑦) ≔ 5) d𝜇(𝑦) d𝑦 + = 1 x(𝑦) n𝜇(𝑦) 𝑦 n 𝑦3 . This gives = d𝜇(𝑦) d𝑦 =− dx(𝑦) d𝑦 x(𝑦)2 n 𝑦3 Let: n 6) 𝜉(𝑦) ≔ exp{∫ d𝑦} = 𝑦 n 𝑦 Multiply both sides by 𝜉(𝑦) , subsitute n𝑦 n−1 = 7) d d𝑦 d d𝑦 (𝑦 n ) and apply the reserve product rule. This leads to: (𝑦 n 𝜇(𝑦)) = n𝑦 n−3 Integrate both sides with respect to y and divide both sides by 𝜉(𝑦) 8) 𝜇(𝑦) = n 𝑦 2 (n−2) + C 𝑦n Sovling for x(y), gives: 9) 1 x(𝑦) = n 𝑦 2 (n−2) + C 𝑦n So: 10) 1 𝑥 = n y(𝑥)2 (n−2) + C y(𝑥)n Question #18 Solving 𝑦 (2) − 5𝑦 (1) + 4𝑦 = 1 𝑒 𝑥 +1 . Answer: You can also proceed as follows: y − 5 y + 4 y = 1 1 + ex  ( y − 4 y) − ( y − 4 y ) = 1 1 + ex  ( y  − 4 y ) − ( y  − 4 y ) = 1 1 + ex  z − z = 1 1 + ex  (e − x z ) = Question #19 1 e + ex 2x Solving separable ODEs 𝑑𝑥 𝑑𝑡 = 𝑓(𝑥)𝑔(𝑡) with 𝑥(0) = 𝑥0 Answer: Step 1 The point is that if 𝑥 ≡ 𝑥 ∗ then 𝑥 ′ = 0 from direct differentiation and also 𝑓(𝑥)𝑔(𝑡) = 0 , so a constant function equal to a zero of f is a solution to the ODE. For it to be a solution to the IVP then you need 𝑥0 = 𝑥∗ . Step 2 This needs a hypothesis that allows us to ensure that the solution is unique, so that one cannot have a constant solution and a non-constant solution to the same IVP (which would happen if any non-constant solution ever hit a zero of f, since then you could consider the IVP initialized when the non-constant solution hit the zero of f and get two solutions). Here that hypothesis is that f' is continuous. (As an aside, this particular hypothesis is not necessary for uniqueness, but merely f and g being continuous is not sufficient for uniqueness. So some additional hypothesis is required, this is just what they chose to use to write this problem.) Step 3 Basically this is asking you to find a separable equation with non-unique solution to an IVP with 𝑥(0) = 𝑥0 and 𝑓(𝑥0 ) = 0 . Hopefully you have already seen some examples of this otherwise it will be a bit challenging to make one up from scratch. Question #20 Solving system of differential equations : Wolfram Alpha vs theorem I am burning my brain finding the most correct way to solve a system of differential equations. Here is an example :  x = 5 x − 2 y   y = − x + 6 y  x(t )   5 −2  Y (t ) =   . I use a matrix A =   , calculate its eigen values 𝜆1 = 7, 𝜆2 = 4 and eigen y ( t ) − 1 6      −1  2 vectors v1 =   , v2 =   . 1 1 Let's Now 2 options: - I can use a theorem and find the solution is 𝑌(𝑡) = 𝑐1 𝑒 𝜆1𝑡 𝑢1 + 𝑐2 𝑒 𝜆2𝑡 𝑢2 - I can continue using linear algebra, calculate P, D and 𝑃 −1 such that 𝐴 = 𝑃 𝐷 𝑃 −1 where D is a diagonal matrix. 𝑌 ′ = 𝐴𝑌 so 𝑌 = 𝐶𝑒 𝑡𝐴 = 𝐶  e7 t P  0 0  −1 P e 4t  I don't find same results but both are OK (the second is used by Wolfram). Did I misunderstand something? Which one is the best option? Answer: Explanation: If you were to write that last formula correctly as  e7 t Y (t ) = e C = P   0 tA 0  −1  P C , C = Y (0), e 4t  then you get the correspondence with the first formula with 𝑃 −1 𝐶 = (𝑐1 𝑐2 ), and the columns of P being the eigenvectors of A.