Questions and Answers #16 Differential equations

Questions and Answers Sheet 16 Differential equations Question #1 Question about solving a differential equation 𝐷(𝐷 − 3)(𝐷 + 4)[𝑦] = 0, where D is the differential operator, how to get the general solution of y? The solution suggest that it is 𝑦 = 6𝑐1 − 2𝑐2 𝑒𝑥𝑝(−4𝑡) + 3𝑐3 𝑒𝑥𝑝(3𝑡) Answer: Given the arbitrary 6, -2 and 3 in front of the constants, the exercise might want you to solve three successive differential equations -first 𝐷𝑢 = 0, then (𝐷 − 3)𝑣 = 𝑢 and finally (D+4)w=v . This process can be circumnavigated with a few observations. First, the differential equation is third-order and thus has three linearly independent solutions. Second, the three operators 𝐷, 𝐷 − 3 and 𝐷 + 4 all commute. Third, D+a annihilates 0 for any constant a (that is, D+a applied to the zero function returns the zero function). Taken together, this implies that any solution to one of ( D + 0) = 0 (i)  ( D − 3) = 0 (ii) ( D + 4) = 0 (iii)  is also a solution to the original differential equation. To see this for e.g. (i), take a solution to the differential equation 𝐷 𝜙 = 0 and observe 𝐷(𝐷 − 3)(𝐷 + 4)𝜙 = (𝐷 − 3)(𝐷 + 4)𝐷𝜙 = (𝐷 − 3)(𝐷 + 4)0 = (𝐷 − 3)0 = 0, Finally, find the general solutions to (i), (ii) and (iii) and observe they are all linearly independent, and thus form a basis for the solution space we are after. Question #2 How to find 𝑥 𝑑𝑦 𝑑𝑥 4 =𝑦+ ? 𝑥 It's given that x>1, y=1 when x=1 . Answer: 4 You want to solve 𝑥𝑦 ′ = 𝑦 + , which is equivalent to 𝑥 4 ′ 𝑥𝑦 − 𝑦 = 𝑥 𝑥𝑦 ′ − 𝑦 4 = 3 2 𝑥 𝑥 Can you notice derivative of some familiar expression there? Are you able to continue from there? Question #3 Find the values of a and b Given that the curve 𝑦 = 𝑎𝑥 2 + Answer: 𝑦 = 𝑎𝑥 2 + 𝑏 𝑥 𝑏 (2, −2) ⇒ −2 = 𝑎(22 ) + (1) 2 𝑏 𝑥 has a gradient of -5 at the point (2,-2), find the value of a and b. We also have: (derivative) 𝑏 𝑦 ′ = 2𝑎𝑥 − 2 (2) 𝑥 It has a gradient of -5 at that point which means: 𝑏 (2)(2, −5) ⇒ −5 = 2𝑎(2) − 2 2 𝑏 −5 = 4𝑎 − (3) 4 (1) and (3) ⇒ 𝑎 = −1 and b=4 Question #4 Integrating a product, one factor a derivative ∫ 𝑑𝑥(𝑡) 𝑥(𝑡)2 𝑑𝑡 𝑑𝑡 Answer: If integration by parts gave you 0, it's probably because you integrated wrong. The simplest way to do this problem is by substitution: let 𝑑𝑥 u=x(t) . Then 𝑑𝑢 = 𝑥 ′ (𝑡)𝑑𝑡 = 𝑑𝑡. 𝑑𝑡 So we can do a simple substitution: 𝑑𝑥(𝑡) 1 2 ∫ (𝑥(𝑡)) 𝑑𝑡 = ∫ 𝑢2 𝑑𝑢 = 𝑢3 + 𝐶 𝑑𝑡 3 1 3 = (𝑥(𝑡)) + 𝐶. 3 2 As for integration by parts, if we set 𝑢 = (𝑥(𝑡)) , 𝑑𝑣 = 𝑑𝑥(𝑡) 𝑑𝑡 𝑑𝑡 , then we can take v=x(t) , 𝑑𝑢 = 2𝑥(𝑡)𝑥 ′ (𝑡)𝑑𝑡, so we would get 𝑑𝑥(𝑡) 2 3 2 𝑑𝑥(𝑡) ∫ (𝑥(𝑡)) 𝑑𝑥 = (𝑥(𝑡)) − ∫ 2(𝑥(𝑡)) 𝑑𝑡 . 𝑑𝑡 𝑑𝑡 Note that the integral on the right is the same as the integral on the left, but multiplied by -2; if we move it to the left hand side, we obtain  dx(t ) dx(t ) ( x(t )) 2 dt + 2  ( x(t )) 2 |, dt = ( x(t ))3 + C. dt dt Now adding the two integrals and dividing by three we get 𝑑𝑥(𝑡) 2 3 3∫ (𝑥(𝑡)) 𝑑𝑡 = (𝑥(𝑡)) + 𝐶 𝑑𝑡 𝑑𝑥(𝑡) 1 2 3 ∫ (𝑥(𝑡)) 𝑑𝑡 = (𝑥(𝑡)) + 𝑐. 𝑑𝑡 3 Question #5 Solution to a differential equation x'=-y and y'=x Answer: If you differentiate y', you have: y''=-y Which has the solutions: 𝑦 = 𝐶1 cos(𝑡) + 𝐶2 sin(𝑡) Question #6 A-stability of Heun method for ODEs I'm trying to determine the stability region of the Heun method for ODEs by using the equation y'=ky , where k is a complex number. Answer: For the method that you wrote down, you indeed have 𝑦𝑛+1 = (0.25 ⋅ ℎ2 ⋅ 𝑧 2 + ℎ𝑧 + 1)𝑦𝑛 . The expression between parentheses is a function of w=hz , and the stability region consists of the numbers w for which the modulus of the expression between parentheses is at most one: Stability region = {𝑤 ∈ 𝐶: |0.25𝑤 2 + 𝑤 + 1| < 1} So the stability region does not depend on the step size h. In the particular case you are asking about, you can use that 0.25𝑤 2 + 𝑤 + 1 = (0.5𝑤 + 1)2 . You should be able to find the region from this. You are right about how to determine whether the method is A-stable. You should find that the method is not A-stable, because explicit methods are never A-stable (see also the Wikipedia page that you link to). In general, to get a feeling for what the stability region looks like, one may start by restricting to the real axis. If w is real, then 0.25𝑤 2 + 𝑤 + 1 is also real, so the condition |0.25𝑤 2 + 𝑤 + 1| < 1 simplifies to 1 < 0.25𝑤 2 + 𝑤 + 1 < 1 . But what people often do in practice is to plot the region using the computer. A final note: Are you sure you copied the method correctly? The method yn +1 = yn + 0.5  h( f (tn , yn ) + f (tn +1 , yn + h  f (tn , yn )) with no factor 0.5 in front of the last h, is more popular. Question #7 How to solve 𝑑𝑦 𝑑𝑥 = 5𝑥𝑦 + sin 𝑥 ? With y(0)=1 . Use an integrating factor. Answer: Rewrite ODE into form: (5𝑥𝑦 + sin 𝑥)𝑑𝑥 − 1 ⋅ 𝑑𝑦 = 0 Next, let's denote: 𝑀 = 5𝑥𝑦 + sin 𝑥 and N=1 , then: integrating factor u(x) is given by: 𝜕𝑀 𝜕𝑁 − 5 2 𝜕𝑦 𝜕𝑥 𝑢(𝑥) = 𝑒 ∫ 𝑑𝑥 = 𝑒 2𝑥 𝑁 Question #8 Why the ODE 𝑥 ′ (𝑡) = 𝑥 2 (𝑡) doesn't have solutions if 𝑥(0) = 𝑥0 ≠ 0 Answer: 𝑥 ′ (𝑡) = 𝑥 2 (𝑡) 𝑑𝑥 = 𝑥2 𝑑𝑡 𝑑𝑥 = 𝑑𝑡 𝑥2 𝑑𝑥 = ∫ 𝑑𝑡 𝑥2 −1 =𝑡+𝑐 𝑥 −1 𝑥(𝑡) = 𝑡+𝑐 if 𝑥(0) = 𝑥0 then −1 −1 𝑥(0) = = = 𝑥0 (𝑥0 ≠ 0) 0+𝑐 𝑐 −1 𝑐= 𝑥0 −1 𝑥0 𝑥(𝑡) = = −1 𝑡 + ( ) 1 − 𝑥0 𝑡 𝑥0 Let's find the values that x(t) is not defined: 1 − 𝑥0 𝑡 = 0 1 𝑡= 𝑥0 1 𝑖𝑓𝑡 = then x(t) is not defined. x(t) is defined for all other values except 𝑡 = ∫ 𝑥0 1 𝑥0 (𝑥0 ≠ 0). Question #9 A simple question about the solution of homogeneous equation to this differential equation Given that 𝑡, 1 + 𝑡, 𝑡 2 , −𝑡 are the solutions to 𝑦 ′′′ + 𝑎(𝑡)𝑦 ′′ + 𝑏(𝑡)𝑦 ′′ + 𝑐(𝑡)𝑦 = 𝑑(𝑡), what is the solution of homogeneous equation to this differential equation? What i have done is tried the properties of linear differential equation that 𝐿(𝑡) = 𝐿(1 + 𝑡) = 𝐿(𝑡 2 ) = 𝐿(−𝑡) = 𝑑(𝑡) so the homogeneous solution should be independent and i claim that 1, t, 𝑡 2 should be the solution. However, i am not sure hot can i actually conclude that these are the solutions? It seems that it can be quite a number of sets of solution by the linearity. Answer: There are actually an infinite number of solutions to the homogeneous version of the differential equation. You'll notice that y=0 is a solution, and for any two solutions u and v, 𝐿(𝛼𝑢 + 𝛽𝑣) = 𝛼𝐿𝑢 + 𝛽𝐿𝑣 = 0, so linear combinations of solutions are also solutions. Therefore the solutions form a vector space, and it is a fact that the dimension of the space is equal to the order of the differential equation, which is three. To characterize the space it suffices to find three linearly independent solutions so that they can be used as a basis for the space (every other solution will be a combination of the basis solutions). Of course {1, 𝑡, 𝑡 2 } is LI and is a very nice choice of polynomials for a basis. Question #10 Solving Differential Equation (𝑡 + 2)𝑑𝑥 = 2𝑥 2 𝑑𝑡 Answer: First I divide both sides by t+2 to get: 2𝑥 2 𝑑𝑥 = 𝑑𝑡 𝑡+2 Then, divide by 2𝑥 2 to get: 𝑑𝑥 𝑑𝑡 = 2 2𝑥 𝑡+2 This will end up to: 1 𝑑𝑡 ∫ 2𝑥 2 𝑑𝑥 = ∫ 𝑡+2 From now on I am not sure how to continue! I ended up having this equation: 1 3 𝑥 = ln(𝑡 + 2) + 𝑐 5 I need to find x(t) now. Can somone help please? 1 1 1 This is how I got 𝑥 3 : I said because ∫ 2 𝑑𝑥 is ∫ 𝑥 −2 isnt it right? 5 2𝑥 2 Question #11 Solving simple differential equation (𝑡 + 4)𝑑𝑥 = 4(1 + 𝑥 2 )𝑑𝑡 Answer: As you rearranged it in your post, we have 𝑑𝑥 𝑑𝑡 ∫ = 4∫ 2 1+𝑥 𝑡+4 As you discovered, the LHS is arctan. To find the value of the RHS, we may say t+4=u so du=dt . Thus the RHS can be rewritten as 4∫ 𝑑𝑢 𝑢 = 4 log|𝑢| + 𝐶2 Because we want this integral to be in respect to t, we have u=t+4 , so arctan 𝑥 + 𝐶1 = 4 log|𝑡 + 4| + 𝐶2 To confirm that the integral we solved is indeed correct, you can simply differentiate 4 log|𝑡 + 4| + 𝐶2 1 Thus 𝑥 = tan(4 log|𝑡 + 4| + 𝐶2 − 𝐶1 ) and 𝑡 = 𝑒𝑥𝑝 ( (𝑎𝑟𝑐𝑡𝑎𝑛 𝑥 + 𝐶1 − 𝐶2 )) − 4 4 Question #12 Using an integrating factor to solve an ODE 𝑑𝑐 𝑑𝑡 = 1 − 𝑐 − 𝑎𝑐 Answer: 𝑑𝑐 + (1 + 𝑎)𝑐 = 1 𝑑𝑡 multiply by integrating factor 𝑒 (1+𝑎)𝑡 : 𝑑𝑐 𝑒 (1+𝑎)𝑡 + (1 + 𝑎)𝑒 (1+𝑎)𝑡 𝑐 = 𝑒 (1+𝑎)𝑡 𝑑𝑡 𝑑 (1+𝑎)𝑡 (𝑒 𝑐) = 𝑒 (1+𝑎)𝑡 𝑑𝑡 𝑒 (1+𝑎)𝑡 𝑒 (1+𝑎)𝑡 𝑐 = ∫ 𝑒 (1+𝑎)𝑡 𝑑𝑡 = +𝐴 1+𝑎 1 𝑐= + 𝐴𝑒 −(1+𝑎)𝑡 1+𝑎 for some constant A. Presumably there is an initial condition (that R.M. has not told us) to let the lecturer determine the constant A Question #13 𝑛 𝑓(𝑡) = ∑∞ 𝑛=0(−1) 𝑢(𝑡 − 𝑛) show that 𝐿{𝑓(𝑡)} = 1 𝑠(1+𝑒 −𝑠 ) Answer: Here is a simpler approach. Note that for t>0 , we have f(t)+f(t-1)=1. The Laplace transform of a time-shifted function (in this case 𝑓(𝑡 − 1) = 𝑓(𝑡 − 1)𝑢(𝑡 − 1)) is s → 𝑒 −𝑠 𝑓̂, where 𝑓̂ is the Laplace transform of f. Furthermore, the Laplace 1 transform of 1 is just s → . Hence we have 𝑠 1 𝑓̂(𝑠) + 𝑒 −𝑠 𝑓̂(𝑠) = , from which it follows that 𝑓̂ (𝑠) = 𝑠 1 . 𝑠(1+𝑒 −𝑠 ) Question #14 For the sulotion of the OED: 𝑑𝑦 𝑑𝑥 = 1 + 𝑥 + 𝑦 2 + 𝑥𝑦. Answer: 𝑧 Changing 𝑦 = 𝑢 − the equation is cast to the form 2 2 𝑑𝑢 𝑥 5 = 𝑢2 + ( − 1) + 𝑑𝑥 2 2 which does not seem to possess solutions in terms of elementary functions Question #15 2 How to solve 𝑦 ′ + 𝑦 = 𝑒 𝑥 𝑦 3 ? Answer: There is one obvious solution y(x)=0 . 2 𝑦′ + 𝑦 = 𝑒 𝑥 𝑦3 2 𝑦 ′ 𝑒 𝑥 + 𝑦𝑒 𝑥 = 𝑒 2𝑥 𝑦 3 2 (𝑦𝑒 𝑥 )′ = 𝑒 2𝑥 𝑦 3 Substitute 𝑢(𝑥) = 𝑦(𝑥)𝑒 𝑥 2 4𝑥 𝑢′ = 𝑢3 𝑒 3 2 4𝑥 𝑢−3 𝑢′ = 𝑒 3 2 We have divided by 𝑢 3 , so this is ok only for points such that 𝑢(𝑥) ≠ 0 . 1 −2 ′ 1 4𝑥 𝑢3𝑢 = 𝑒3 3 3 4𝑥 ′ ′ (𝑢1/3 ) = ( 3 ) 4 4𝑥 𝑢 1/3 𝑒3 = +𝐶 4 3 4𝑥 𝑒3 𝑢 = (𝐶 + ) 4 4𝑥 𝑦 = (𝐶 + 𝑒3 4 3 ) 𝑒 −𝑥 To get y=0 , we need 𝐶 = − 𝑦=( 𝑒 4𝑥/3 −1 4 3 1 4 ) 𝑒 −𝑥 Question #16 A linear differential equation 𝑥 ′′ + 2𝑐𝑥 ′ + ( Answer: 2 cos ℎ2 𝑡 − 1) 𝑥 = 0 where c is a constant. When c=0 we can guess that 𝑥1 = 1 cos ℎ 𝑡 is a solution. Then by inserting 𝑥 = 𝑥1 ⋅ 𝑢 into equation it simplifies to: 𝑢′′ − 2 tan ℎ 𝑡𝑢′ = 0 From which we conclude that: x(t ) = C1 t + C2 (sin h t + ) cos h t cos h t For more general case when 𝑐 ≠ 0 we use substitution 𝑥 = 𝑤𝑒 −𝑐𝑡 . The equation transforms to: 2 𝑤 ′′ = (1 + 𝑐 2 − ) 𝑤(1) cos ℎ2 𝑡 the trick is to use Darboux theorem. Let's say we have two functions y and z that are solutions to: 𝑦 ′′ = 𝑝(𝑡)𝑦, 𝑧 ′′ = (𝑝(𝑡) − 𝜆)𝑧 Then the function: 𝑦𝑧 ′ 𝑦 = 𝑦′ − 𝑧 Satisfies: 𝑑 𝑧′ 𝑦 ′′ = (𝑝 − 2 ( )) 𝑦(2) 𝑑𝑡 𝑧 Taking 𝑧 = cos ℎ t and 𝑝 = 1 + 𝑐 2 equation (2) turns out to be (1). 2 2 We solve: 𝑦 ′′ = (1 + 𝑐 2 )𝑦 ⇒ 𝑦 = 𝐶1 𝑒 √1+𝑐 𝑡 + 𝐶2 𝑒 −√1+𝑐 𝑡 then we plug in for 𝑦 = 𝑤 and x which yields the final result: x = C1 ( 1 + c 2 − tan h t )e +C2 ( 1 + c 2 + tan h t )e− 1+ c 2 t − ct 1+ c 2 t − ct Question #17 𝜕𝑢 ∫𝜕Ω [𝜕𝑁 𝑑 𝜎 if Δ𝑢 = 0 in Ω ? Is it always possible to solve an equation u''=0 in [0,1] if u'(0) and u'(1) are given?(What is connection to first question?) Is the solution - unique? Answer: For the first one From the divergence theorem for the vector field gradu we get that:   ( grad u ) =  grad u, N    where 𝜈 is the volume element. Using now the ypothesis that 0 = u := ( grad u ) and that u := grad g , N  N Question #18 Solving ODE and finding maxima Consider the initial value problem 𝑦 ′ + 5𝑦 = 5𝑡, 𝑦(0) = 𝑦0 . Find the value of 𝑦0 for which the solution touches, but does not cross, the t-axis. Answer: First, your solution is incorrect. 𝑦 ′ (𝑡) + 5𝑦(𝑡) = 5𝑡 ⇒ 𝑒𝑥𝑝(5𝑡)𝑦 ′ (𝑡) + 5𝑦(𝑡) 𝑒𝑥𝑝(5𝑡) = 5𝑡 𝑒𝑥𝑝(5𝑡) ⇒ 𝑑(𝑦(𝑡)𝑒𝑥𝑝(5𝑡)) 𝑑𝑡 = 5𝑡 𝑒𝑥𝑝(5𝑡) 1 1 5 5 Hence, 𝑦(𝑡)𝑒𝑥𝑝(5𝑡) = (𝑡 − ) 𝑒𝑥𝑝(5𝑡) + 𝑐 ⇒ 𝑦(𝑡) = 𝑡 − + 𝑐 𝑒𝑥𝑝(−5𝑡) The slope is given by 1 y(t ) = 5(t − y (t )) = 5( , −c exp ( −5t )) = 1 − 5c exp ( −5t ) 5 For this to touch but not cross the t axis, we need 𝑦(𝑡0 ) = 0 and 𝑦 ′ (𝑡0 ) = 0, These give us 1 𝑡0 − + 𝑐 𝑒𝑥𝑝(−5𝑡0 ) = 0 5 and 1 − 5𝑐 𝑒𝑥𝑝(−5𝑡0 ) = 0 1 Now obtain 𝑡0 and c by solving these two equations. This gives us 𝑡0 = 0 and 𝑐 = . Hence, the curve is 5 given by 1 𝑒𝑥𝑝(−5𝑡) + . 5 5 Now 𝑦0 = 𝑦(0) = 0 𝑦(𝑡) = 𝑡 − Question #19 Linearizing 𝑦 ′′ + sin(2𝑥 + cos(2𝑦 ′ + 𝑦)) + 1 − sin(𝑦 + 3𝑦 ′ ) = 0 Answer: The easiest linearization I can think of in this case is to use cos(𝜃) ≈ 1, and sin(𝜃) 𝜃 where 𝜃 is small enough. In your case, these assumptions boil down to making the assumption that y, 𝑦 ′ ≈ 0 . With these assumptions, your differential equation now becomes 𝑦 ′′ + sin(2𝑥 + 1) + 1 − (𝑦 + 3𝑦 ′ ) = 0 which in turn becomes 𝑦 ′′ − 3𝑦 ′ − 𝑦 = −(1 + sin(2𝑥 + 1)) Now you can use your differential equation tricks to solve the above equation to get 𝑦(𝑥) = 𝑐1 𝑒𝑥𝑝 (( (( 3+√13 2 ) 𝑥) + 3 − √13 ) 𝑥) + 𝑐2 𝑒𝑥𝑝 2 5 sin(2𝑥+1)−6 cos(2𝑥+1)+61 61 Question #20 How to solve 𝑥 ′ (𝑡) = 𝑥+𝑡 2𝑡−𝑥 ? Answer: Your differential equation is homogenous of order one. You did it right finding 𝑑𝑦 𝑑𝑡 = 𝑓(𝑦)−𝑦 𝑡 . So you can integrate of both sides in last equation till your solution is achieved respect to y(t) and t. Then you can substitute x(t)=y(t)t in your solution and then apply the initial condition.