Questions and Answers #13 Differential equations

Questions and Answers Sheet 13 Differential equations Question #1 Counterintuitive PDE After thinking about it for a while and consulting other students, no one seems to be able to find an example of the following: Given the PDE 𝜕𝑓 𝜕𝑥 = 0 on 𝑈 = (𝑥, 𝑦) ∈ 𝑅2 ; 𝑦 > 0,1 < 𝑥 2 + 𝑦 2 < 4 I am looking for a solution f that does not only depend on y. How can this be?! The exercise is taken form Lee's ''Introduction to smooth manifolds'', p. 517 at the end of the chapter on the Frobenius theorem. (Note: According to the errata, the condition on U is y>0, not x>0 as your copy of the book might state). Answer: The Mean Value Theorem implies that any function f differentiable on U must be constant w.r.t. x for any fixed y. There is an error in the text - it should say y>0 instead of x>0 . Question #2 Help with a generating function and differential equation I have a generating function that I'm trying to create. It has a general form to it: 1 + 𝑐𝑥 + 𝑐(𝑐 − 1)𝑥 2 + 𝑐(𝑐 − 1)(𝑐 − 2)𝑥 3 + ⋯ + 𝑐! 𝑥 𝑐 I would like to see a closed form for this function. I'd like something other than a summation. I'm particularly interested if someone could walk through how to derive the form for me. I was hoping that someone could carry on with what I've done and show me how to solve what's left. I set 𝑎0 = 1 . I then proceed to multiply by (c-n) . My guess is that the math should be something like the following: 𝑎𝑛+1 = 𝑐 ⋅ 𝑎𝑛 − 𝑑𝐴(𝑥) 𝑑𝑥 I'm trying to say that the next term should be c times the previous term minus n times the previous term. The latter is represented by the derivative, I believe. So I guess that this could be correct. Unfortunately, I don't know for sure, and so I'm kind of lost at this point. I would greatly appreciate if someone could take it from here and explain the rest in detail. Assuming that the last part is correct, the function becomes a differential equation: 𝐴(𝑥) − 1 𝑑𝐴(𝑥) = 𝑐 ⋅ 𝐴(𝑥) − 𝑥 𝑑𝑥 Answer: The general form of your sum can be expressed succinctly as ∑(−𝑐)𝑘 (−𝑥)𝑘 where (𝑎)𝑘 is a Pochhammer symbol. We recognize at once that this is a hypergeometric series; specifically, it is a 2𝐹0 : 2𝐹0 (−𝑐, 1: −𝑥) which can be rewritten as a Tricomi confluent hypergeometric function: 1 𝑥 1 𝑈 (1, 𝑐 + 2, ) 𝑥 which can be re-expressed as an incomplete gamma function: 1 1 𝑥 𝑐 𝑒𝑥𝑝 ( ) (𝑐 + 1, ) 𝑥 𝑥 As for the differential equation, it can be obtained from this formula to give: 𝑑2 𝑦 1 𝑑𝑦 𝑥 2 2 − (𝑐 − 2 − ) 𝑥 − 𝑐𝑦 = 0 𝑑𝑥 𝑥 𝑑𝑥 where the solution of interest satisfies the initial conditions 𝑦(0) = 1 and 𝑦 ′ (0) = 𝑐 Question #3 I have heard about a lot of fancy numerical methods for solving ODEs. I know that there are methods that (assuming sufficient smoothness) asymptotically give a low error, like the Runge-Kutta methods. These estimate the solution in a set of points 𝑡0 , 𝑡1 , etc. But what if I want to have a function that is close to the correct solution everywhere, not just in a discrete set of points? I can extend the numerical solution to a piecewise linear function. This will be a continuous function and it will converge to the correct solution if the step-size goes to zero. But the error estimate will be poor in most places unless I use a very small step-size, which rather defeats the purpose of using a high-order method. So how does one go about estimating the solution in practice between the 𝑡𝑖 ? Answer: Remember that whatever method you use for solving 𝑦 ′ = 𝑓(𝑡, 𝑦) , be it Runge-Kutta, Bulirsch-Stoer (extrapolative), Gear/Adams multistep, or fancier methods, one always has a triple of values (𝑡𝑖 , 𝑦𝑖 , 𝑦𝑖′ )(𝑦𝑖′ = 𝑓(𝑡𝑖 , 𝑦𝑖 )) at each evaluation point. Thus, one can always do cubic Hermite interpolation across the points (𝑡𝑖 , 𝑦𝑖 ) and (𝑡𝑖 + 1, 𝑦𝑖 + 1) (note that I am not assuming that the evaluation points are equispaced, as is often the case when doing adaptive solving). If the underlying method is at most third order accurate, cubic Hermite is a good choice. Now, the upshot is that modern implementations of DE solvers always support so-called ''dense output''; briefly, associated with a method with pth order ''accuracy'' (in quotes since ''high order doesn't always imply high accuracy'' ;) ) has with it an associated pth order interpolating function. To use my favorite example, the (4,5) adaptive Runge-Kutta solver based on the Dormand-Prince coefficients has with it an associated fifth order interpolating function. The special properties inherent in the coefficients allow the existence of an associated interpolating function with the same order of ''accuracy''; in general, not all Runge-Kutta coefficients will have an associated ''nice'' interpolating function (but again, one can always do cubic Hermite). Question #4 Solve: 𝑦 ′′ + 2𝑦 ′ − 3𝑦 = 0 Answer: 𝑦′ = 𝑧 𝑦 ′′ = 𝑧 𝑑𝑧 𝑑𝑦 𝑑𝑧 + 2𝑧 − 3𝑦 = 0 𝑑𝑦 𝑧𝑑𝑧 + 2𝑧𝑑𝑦 − 3𝑦𝑑𝑦 = 0 𝑧𝑑𝑧 = (3𝑦 − 2𝑧)𝑑𝑦 𝑧 = 3𝑦 − 2𝑧 𝑧=𝑦 𝑦 = 𝑦 ′ = 𝑦 ′′ 𝑧 Question #5 Find the general solution to the differential equation 𝑑𝑦 = 𝑡 3 + 2𝑡 2 − 8𝑡 𝑑𝑡 Also, part 8B. asks: Show that the constant function y(t)=0 is a solution. I've done a problem like this before, except that it wasn't a constant function. This problem seems like a question that asks: ''show that every member of the family of functions 𝑦 = (ln 𝑥 + 𝐶)/𝑥 is a solution to the differential equation (some diff. equation)'' except it seems a little bit different. Answer: Parts of problems that say ''show that [explicitly given function] is a solution'' can be solved by simply 1 ''plugging in''. E.g., if you're supposed to show that 𝑦(𝑡) = is a solution to the differential equation 𝑡−𝑡 𝑑𝑦 𝑑𝑡 1 1 = 𝑦 2 , then you should compute 𝑑𝑦/𝑑𝑡 = (1−𝑡)2 , compute 𝑦 2 = (1−𝑡)2, and ''plug them in'' to check that the differential equation is satisfied for this choice of y. You're correct that the difference in the second problem is that you no longer have the derivative with respect to t expressed as a function of t. When the equation involves both a function y and its derivative (or derivatives), it can generally be more difficult to solve. In this case, a hint I'll give you is that 𝑑𝑦 𝑑𝑡 = 1/(𝑑𝑡/𝑑𝑦) You can first turn things around and think of t as a function of y. You can find t(y) using similar methods to the first problem (but with a more complicated function), and then find the inverse function to get y(t). Question #6 In Physics, the Simple Harmonic Oscillator is represented by the equation 𝑑 2 𝑥/𝑑𝑡 2 = −𝜔2 𝑥. By using the characteristic polynomial, you get solutions of the form 𝑥(𝑡) = 𝐴𝑒 𝑖𝜔𝑡 + 𝐵𝑒 −𝑖𝜔𝑡 , I get that you Euler's formula 𝑒 𝑖𝜃 = cos 𝜃 + 𝑖𝑠𝑖𝑛 𝜃 , but I can't seem to find my way all the way to the ''traditional form'' of 𝐷 cos 𝜔 𝑡 + 𝐶 sin 𝜔 𝑡. Answer: However, if you assume the function x(t) is real, then they are related as 𝐴 = 𝐵. This can be seen since 𝑥(𝑡) = 𝑥(𝑡) which gives us ¯{𝐴} = 𝐵 (𝑠𝑖𝑛𝑐𝑒 𝑒 𝑖𝜔𝑡 𝑎𝑛𝑑𝑒 −𝑖𝜔𝑡 are orthonormal) Rewriting the complex exponentials as sine and cosine we get 𝑥(𝑡) = 𝐴(cos(𝜔𝑡) + 𝑖 sin(𝜔𝑡)) + 𝐵(cos(𝜔𝑡) − 𝑖 sin(𝜔𝑡)) = (𝐴 + 𝐵) cos(𝜔𝑡) + 𝑖(𝐴 − 𝐵) sin(𝜔𝑡) Now as proved before 𝐴 = 𝐵 and hence (A+B) is real and (A-B) is imaginary and hence we can write (𝐴 + 𝐵) = 𝐶 𝑎𝑛𝑑 𝑖(𝐴 − 𝐵) = 𝐷 where C, D are real number. And the solution becomes 𝑥(𝑡) = 𝐶 cos(𝜔𝑡) + 𝐷 sin(𝜔𝑡) where 𝐶, 𝐷 ∈ 𝑅 Question #7 How to find f(x) if 𝑑𝑓(𝑥) 𝑑𝑥 = 𝑓(𝑥) ? I know 𝑐𝑒 𝑥 is a solution, but how does one find it and how to prove it is the complete solution? Answer: To find it, you can use separation of variables. If you're accustomed to recognizing that 𝑓′ 𝑓 = log(|𝑓|)′ , you can integrate directly. Otherwise, you can use the substitution u=f . Even without knowing what the solutions are, note that if f and g are solutions and g is nonzero, then 𝑓 ′ ( ) = 𝑓 ′𝑔−𝑔′ 𝑓 𝑔 𝑔2 = , so 𝑓 𝑔 is constant, which means f=cg for some constant c. This shows that the multiples of one nonzero solution form the complete solution. In general, a first order homogeneous linear ODE has a one dimensional solution space. Question #8 How would I solve these differential equations? 𝑃0′ (𝑡) = 𝛼𝑃1 (𝑡) − 𝛽𝑃0 (𝑡) 𝑃1′ = 𝛽𝑃0 (𝑡) − 𝛼𝑃1 (𝑡) We also know 𝑃0 (𝑡) + 𝑃1 (𝑡) = 1 Answer: Note that from the equation you have 𝑃0′ (𝑡) = 𝛼𝑃1 (𝑡) − 𝛽𝑃0 (𝑡) = −𝑃1′ (𝑡) which gives us 𝑃0′ (𝑡) + 𝑃1′ (𝑡) = 0 which gives us 𝑃0 (𝑡) + 𝑃1 (𝑡) = 𝑐 . We are given that c=1 . Use this now to eliminate one in terms of the other. For instance, 𝑃1 (𝑡) = 1 − 𝑃0 (𝑡) and hence we get, 𝑃0′ (𝑡) = 𝛼(1 − 𝑃0 (𝑡)) − 𝛽𝑃0 (𝑡) ⇒ 𝑃0′ (𝑡) = 𝛼 − (𝛼 + 𝛽)𝑃0 (𝑡) Let 𝑌0 (𝑡) = 𝑒 (𝛼+𝛽)𝑡 𝑃0 (𝑡) ⇒ 𝑌0′ (𝑡) = 𝑒 (𝛼+𝛽)𝑡 [𝑃0′ (𝑡) + (𝛼 + 𝛽)𝑃0 (𝑡)] = 𝛼𝑒 (𝛼+𝛽)𝑡 𝛼 Hence, 𝑌0 (𝑡) = 𝑒 (𝛼+𝛽)𝑡 + 𝑘𝑖. 𝑒. 𝑃0 (𝑡) = 𝛼 𝛼+𝛽 𝛼+𝛽 −(𝛼+𝛽)𝑡 + 𝑘𝑒 𝑃1 (𝑡) = 1 − 𝑃0 (𝑡) = 𝛽 − 𝑘𝑒 −(𝛼+𝛽)𝑡 𝛼+𝛽 Question #9 𝑦 𝑑𝑦 = 𝑒𝑥 𝑑𝑥 Solve it by ''separating variables'' like this: 𝑦𝑑𝑦 = 𝑒 𝑥 𝑑𝑥 ∫ 𝑦𝑑𝑦 = ∫ 𝑒 𝑥 𝑑𝑥 𝑦 2 /2 = 𝑒 𝑥 + 𝑐 Answer: The basic justification is that integration by substitution works, which in turn is justified by the chain rule and the fundamental theorem of calculus. More specifically, suppose you have: 𝑑𝑦 = 𝑔(𝑥)ℎ(𝑦) 𝑑𝑥 Rewrite as: 1 𝑑𝑦 = 𝑔(𝑥) ℎ(𝑦) 𝑑𝑥 Add the implicit dependency of y on x to obtain 1 𝑑𝑦 = 𝑔(𝑥) ℎ(𝑦(𝑥)) 𝑑𝑥 Now, integrate both sides with respect to x: 1 𝑑𝑦 ∫ 𝑑𝑥 = ∫ 𝑔(𝑥)𝑑𝑥 ℎ(𝑦(𝑥)) 𝑑𝑥 If we do a variable substitution of y for x on the left-hand side (i.e., use the integration by substitution 𝑑𝑦 technique), we replace 𝑑𝑥 with dy. Thus we have 𝑑𝑥 1 ∫ 𝑑𝑦 = ∫ 𝑔(𝑥)𝑑𝑥 , ℎ(𝑦) which is the separation of variables formula. So if you believe integration by substitution, then separation of variables is valid. Question #10 2 Differential equation 𝑓 ′′ (𝑥) + (𝑛−1)(𝑓 ′(𝑥)) sin ℎ(𝑥) =0 Answer: Let u=f' . Then − 𝑢′ 𝑢2 1 ′ = 𝑔 and so ( ) = 𝑔 . Integrate both sides to find u and then integrate once again to 𝑢 find f. Question #11 𝑑𝑦 𝑑𝑡 + 2𝑦 = 1; 𝑦(0) = 5 2 when i solve this i get ln(−4) = 𝑐 now the problem is how to solve ln (-4) ? Answer: Cauchy's problem: 5 (𝑦 ′ (𝑡) + 2𝑦(𝑡) = 1), (𝑦(0) = ) : (1) 2 has unique global solution (because the ODE is linear with constant coefficients); this solution is continuous in R, hence 𝑦 ′ (𝑡) = 1 − 2𝑦(𝑡) is continuous in R and y(t) is of class 𝐶1 ∈ 𝑅 ... Bootstrapping one gets y(t) of class 𝐶 ∞ ∈ 𝑅. 1 Moreover, the ODE in (1) has only one constant solution, namely 𝑦(𝑡) ≔ , and it forces its nonconstant 2 global solutions to be either > 1/2 or < 1/2 everywhere in R: in fact, if a solution 𝜙(𝑡) has 𝜙(𝑡1 ) < 1/2 < 𝜙(𝑡2 ) , then (by continuity) there exists T s.t. 𝜙(𝑇) = 1/2 therefore 𝜙(𝑡) = 𝑦(𝑡) everywhere (by uniqueness), a contradiction. In the present case, since 𝑦(0) = 5/2 > 1/2 , one has 𝑦(𝑡) > 1/2 everywhere in R; hence 𝑦 ′ (𝑡) = 1 − 2𝑦(𝑡) < 0 in R, so that y(t) is strictly decreasing. Now, let t>0 ; one can divide both sides of 𝑦 ′ (𝑡) = 1 − 2𝑦(𝑡) by 1 − 2𝑦(𝑡) ≠ 0 and integrate over [0,t] to get: 𝑡 𝑡 𝑦 ′ (𝜏) ∫ 𝑑𝜏 = ∫ 𝑑𝜏 0 1 − 2𝑦(𝜏) 0 since y(\tau) is of class 𝐶 ∞ and strictly decreasing, one can make the change of variable 𝑛 = 𝑦(𝜏) in the LHside to get: 𝑦(𝑡) 1 ∫ 𝑑 𝜂 = 𝑡, 5/2 1 − 2𝜂 1 𝑦(𝑡) hence: − log|1 − 2𝜂|5/2 = 𝑡; 2 keeping in mind that 1-2y(t) is negative in R, one finds: 4 log = 2𝑡 2𝑦(𝑡) − 1 1 and finally: 𝑦(𝑡) = 2𝑒 −2𝑡 + (2) 2 On the other hand, if t<0 one has to integrate over [t,0] to recover the same elementary expression (2) for the solution. Therefore (2) gives the unique global solution to problem (1). Question #12 How can I solve this non-linear differential equation? 𝑦′ = 1 − 𝑦2 Answer: 𝑦′ = 1 − 𝑦2 Divide by (1 − 𝑦 2 ) 𝑦′ =1 1 − 𝑦2 Integrate both sides: 1 y +1 log | |= t + c 2 y −1 Rearrange 𝑦 = 𝑘𝑒 2𝑡 +1 𝑘𝑒 2𝑡 −1 I'd have thought that solution was right, but we have to figure out a specific solution with y(0) = 0. But this isn't possible with the above equation. Question #13 Second degree linear derivate with exponential function, which general solution? Suppose 𝑦 ′′ − 4𝑦 = 𝑥𝑒 2𝑥 Answer: For any polynomial P, constant k and function u, 𝑃(𝐷)𝑒 𝑘𝑥 𝑢 = 𝑒𝑥𝑝 (𝑘𝑥)𝑃(𝐷 + 𝑘)𝑢 (where D stands for derivative). In your case 𝑃(𝑡) = 𝑡 2 − 4, 𝑘 = 2, and 𝑃(𝑡 + 2) = (𝑡 + 2)2 − 4 = 𝑡 2 + 4 𝑡 so if 𝑦 = 𝑒 2𝑥 𝑢 we have 𝑦 ′′ − 4𝑦 = 𝑒 2𝑥 (𝑢′′ + 4𝑢′ ) . Now you want 𝑦 ′′ − 4𝑦 = 𝑥𝑒 2𝑥 so 𝑢′′ + 4𝑢′ = 𝑥. Writing v=u' , we 𝑥 1 have the first-order linear equation in 𝑣: 𝑣 ′ + 4𝑣 = 𝑥 , which has a solution 𝑣 = − . An antiderivative of this is 𝑢 = 𝑥2 8 − 𝑥 16 , corresponding to the particular solution 𝑦 = 𝑒 2𝑥 ( 𝑥2 8 4 − 𝑥 16 16 ) of your original equation. Question #14 I was toying with equations of the type 𝑓(𝑥 + 𝛼) = 𝑓 ′ (𝑥) where f is a real function. For example if 𝛼 = 𝜋 2 then the solutions include the function 𝑓𝜆,𝜇 (𝑥) = 𝜆 cos(𝑥 + 𝜇). Are there more solutions? On the other hand, if I want to solve the equation for any 𝛼 , I can assume a solution of the form 𝑓(𝑥) = 𝑒 𝜆𝑥 , and find 𝜆 as a complex number that enables me to solve the equation... I was wondering: is the set of the solutions of dimension 2 because the derivative operator creates one dimension and the operator 𝜙: 𝑓(𝑥) → 𝑓(𝑥 + 1) adds another one? Is there some litterature about this kind of equations? Answer: This sort of thing is known as a delay differential equation, and there are lots and lots of books on this subject. They're called ''delay'' because the constant 𝛼 is usually chosen to be negative, so that f'(x) depends on the values of f at earlier times. The equation you have given has an infinite dimensional space of solutions. In particular, if 𝑔: |0, 𝑔| → 𝑅 is any smooth function satisfying 𝑔(𝑛) (𝛼) = 𝑔(𝑛+1) (0) for all 𝑛 ≥ 0 , then there exists a unique extension of g to a smooth solution f defined on the entire real line. For 𝑥 ≥ 0 , this solution is defined by 𝑓(𝑥) = 𝑔(𝑛) (𝑥 − 𝑛𝛼) where 𝑛 = [𝑥/𝛼] while for 𝑥 ≤ 0 it involves iterated integrals of g. Question #15 Suppose we have a mechanical system with 1 degree of freedom, i.e. an ODE (1) 𝑞̈ + 𝑉 ′ (𝑞) = 0, where V: R \rightarrow R is some smooth function (potential energy). We then easily see that any solution of this equation must satisfy 𝑞 2̇ + 𝑉(𝑞) = constant 2 In other words, if we put 𝐸(𝑞, 𝑞̇ ) = 𝑞 2̇ + 𝑉(𝑞) 2 (energy), then the image of every solution of (1) must lie in a level set of E. Answer: You are essentially asking whether the particle can stop at a point inside the level set. To stop, it needs to have both zero velocity and zero acceleration (these conditions are sufficient because the dynamics is of second order). First condition requires 𝑞̇ = 0 implying also 𝑉(𝑞) = 𝐸, while the second holds when 𝑉 ′ (𝑞) = 0. So just choose any potential with a local maximum 𝑉𝑚𝑎𝑥 and choose initial conditions such that 𝐸 = 𝑉𝑚𝑎𝑥 . Physically these correspond to a ball rolling down a hill and then again up a hill. As the ball approaches the top, its kinetic energy decreases as E-V and asymptotically it stops. This interval is proper in the level set because you can get another solution approaching the hill from the other side of the hill. Perhaps the answer is even more obvious in more dimensions (or more degrees of freedom). There the level sets are of dimension 2N-1 and the failure of inclusion is more apparent. Just consider a potential with a local maximum (a boulge at the origin) which you can approach from any direction and asymptotically stop there. Question #16 I want to get a particular solution to the differential equation 𝑦 ′′ + 2𝑦 ′ + 2𝑦 = 2𝑒 𝑥 cos(𝑥) and therefore I would like to 'complexify' the right hand side. This means that I want to write the right hand side as 𝑞(𝑥)𝑒 𝛼𝑥 with q(x) a polynomial. How is this possible? Answer: The point is that (for real x) 2𝑒 𝑥 cos(𝑥) is the real part of 2𝑒 𝑥 𝑒 𝑖𝑥 = 2𝑒 (1+𝑖)𝑥 . Find a particular solution of 𝑦 ′′ + 2𝑦 ′ + 2𝑦 = 2𝑒 (1+𝑖)𝑥 , and its real part is a solution of 𝑦 ′′ + 2𝑦 ′ + 2𝑦 = 2𝑒 𝑥 cos(𝑥) Question #17 Proof for law of complex exponents using only differential equation I just read that an elegant proof exists that the law of exponents also holds for complex numbers (a,b,z all complex): 𝑒 𝑎+𝑏 = 𝑒 𝑎 𝑒 𝑏 , which only uses the definition that 𝑦 = 𝑒 𝑧𝑡 is solution to 𝑑𝑦 𝑑𝑡 = 𝑧𝑦, with initial condition y(0)=1 , so in particular 𝑒 𝑧 = 𝑦(1). I can only find a proofs which use the trig-representation of complex numbers. Answer: If you define 𝑒 𝑧 as the unique solution to the ODE f'(z)=f(z) with initial condition f(0)=1 , then you have by the product rule: (𝑒 𝑧 𝑒 𝑐−𝑧 )′ = 𝑒 𝑧 𝑒 𝑐−𝑧 + 𝑒 𝑧 (−𝑒 𝑐−𝑧 ) = 0 Thus 𝑒 𝑧 𝑒 𝑐−𝑧 is a constant. Using the initial condition 𝑒 0 = 1 we find that 𝑒 𝑧 𝑒 𝑐−𝑧 = 𝑒 𝑐 . Now let z=a and 𝑐 = 𝑎 + 𝑏 and 𝑐 = 𝑎 + 𝑏 and the result follows. Question #18 I'm using the method of undetermined coefficients to find a particular solution of: 𝑦 ′′ + 𝑦 ′ = 𝑥𝑒 −𝑥 Ostensibly, it seems that 𝑦𝑝 should take the form 𝑜𝑓(𝐴𝑥 + 𝐵)𝑒 −𝑥 At least that's the form that I think I've been taught. Problem is that it just doesn't work out for me. I get a value for A, but not for B... Am I choosing an incorrect yp form? Answer: It's often good to start with finding the solution to the homogenous solution 𝑦ℎ′′ + 𝑦ℎ′ = 0 which, as you know, is 𝑦ℎ = 𝑐1 + 𝑐2 𝑒 −𝑥 . Thus you don't need to have a 𝑒 −𝑥 term when you look for the particular solution, since this is already included in the homogenous (if you include it in your particular solution, it will cancel out). Second, if you are prone to small misstakes, you can take terms in your particular solution one by one, since differentiation is linear. Start with 𝐴𝑥𝑒 −𝑥 : 𝑑2 𝑑 𝐴𝑥𝑒 −𝑥 + 𝐴𝑥𝑒 −𝑥 = −𝐴𝑒 −𝑥 2 𝑑𝑥 𝑑𝑥 then 𝐵𝑥 2 𝑒 −𝑥 : 𝑑2 𝑑 𝐵𝑥 2 𝑒 −𝑥 + 𝐵𝑥 2 𝑒 −𝑥 = 2𝐵𝑒 −𝑥 − 2𝐵𝑥𝑒 −𝑥 𝑑𝑥 2 𝑑𝑥 You can see that if you have 𝑦𝑝 = 𝐵𝑥 2 𝑒 −𝑥 + 𝐴𝑥𝑒 𝑥 you will get 𝑦𝑝′′ + 𝑦𝑝′ = (2𝐵 − 𝐴)𝑒 −𝑥 − 2𝐵𝑥𝑒 −𝑥 1 thus 𝐵 = − and 2𝐵 − 𝐴 = −1 − 𝐴 = 0, which gives A=-1 , which gives you your answer. 2 A rule which is convenient when making these kinds of calculations is 𝑛 𝑑𝑛 𝑑 𝑎𝑥 ) 𝑎𝑥 (𝑓(𝑥)𝑒 = 𝑒 ( + 𝑎) 𝑓(𝑥) 𝑑𝑥 𝑛 𝑑𝑥 𝑑 where ( ) 𝑛 𝑑𝑥 is interpreted as 𝑑𝑛 𝑑𝑥 𝑛 −𝑥 For example, using this on 𝐴𝑥𝑒 𝑑2 𝑑𝑥 2 𝐴𝑥𝑒 −𝑥 + 𝑑 𝑑𝑥 𝐴𝑥𝑒 −𝑥 = 𝐴𝑒 −𝑥 ((𝐷 . (writing D for 𝑑 𝑑𝑥 ): 2 − 1) 𝑥 + (𝐷 − 1)𝑥) = 𝐴𝑒^{−𝑥}(𝐷 − 1)((𝐷 − 1)𝑥 + 𝑥)) = 𝐴𝑒^{−𝑥}(𝐷 − 1)(1) = −𝐴𝑒^{−𝑥} You can quickly see that the term 𝐴𝑥𝑒 −𝑥 is not enough. Question #19 Solve 𝑦 ′′ + 𝑠 2 𝑦 = 𝑏 cos 𝑠 𝑥 where s and b are constants. I have tried undetermined coefficients, but it makes such a mess that I keep getting lost, I also tried variation of parameters. Really my issue is staying organized enough to get the solution, my solutions are always just a little off from the book. I have solved the complementary homogenous equation, this gave me a solution with cos 𝑠 𝑥 in it... So that's off limits in the particular solution. I looked for a solution in xsinsx and 𝑥 cos 𝑠 𝑥 , but as I said, it was a mess. Answer: You need to differentiate eg 𝑦 = 𝐴𝑥 cos 𝑠 𝑥 twice using the chain rule. So 𝑦 ′ = 𝐴 cos 𝑠 𝑥 − 𝐴𝑠 𝑥 sin 𝑠 𝑥 And 𝑦 ′′ = −2𝐴𝑠 sin 𝑠 𝑥 − 𝐴𝑠 2 𝑥 cos 𝑠 𝑥 Then 𝑦 ′′ + 𝑠 2 𝑦 = −2𝐴𝑠 sin 𝑠 𝑥 But you need cos 𝑠 𝑥 and I'm sure you can see how to do that now. Because the equation is linear, if the expressions get messy you can deal with the sin and cos terms from 𝑦 = 𝐴𝑥 sin 𝑠 𝑥 + 𝐵𝑥 cos 𝑠 𝑥 separately and get the right linear combination at the end. Sometimes it can be easier to convert everything to exponential functions. Question #20 How to deal with two interdependent integrators? I have two functions, f(t,x) and g(t,u), where 𝑑 𝑑𝑡 𝑢 = 𝑓(𝑡, 𝑥) and 𝑑 𝑑𝑡 𝑥 = 𝑔(𝑡, 𝑢) . I am trying to discretize the integral of this system in order to track x and u. I have succeeded using Euler integration, which is quite simple, since x(t) and u(t) are both known at t: 𝑢(𝑡 + ℎ) = 𝑢(𝑡) + ℎ𝑓(𝑡, 𝑥(𝑡)) 𝑥(𝑡 + ℎ) = 𝑥(𝑡) + ℎ𝑔(𝑡, 𝑢(𝑡)) However, I am now trying to implement mid-point integration to get more accurate results. (Eventually Runge-Kutta but I am stuck here for now.) Answer: I'm expanding my comment. The fact that f does not depend on u and g does not depend on x could be of significance if one tried to solve the two differential equations explicitly. But otherwise we just have a differential equation for a vector-valued function 𝑦(𝑡) ≔ (𝑢(𝑡), 𝑥(𝑡)) , namely 𝑦̇ (𝑡) = (𝑓(𝑦, 𝑢(𝑡)), 𝑔(𝑡, 𝑥(𝑡))) . The general existence and uniqueness theorem as well as the usual numerical methods (e.g. Runge- Kutta) are applicable to this setup. It could be that the functions 𝑡 → 𝑢(𝑡) and 𝑡 → 𝑥(𝑡) have ''different time scales''; e.g., u(⋅) could be a ''slow'' function and x(⋅) could be a rapidly oscillating function. This is the phenomenon of ''stiffness'' which requires special measures in numerical work.