Questions and Answers #29 Differential equations

Questions and Answers Sheet 29 Differential equations Question #1 Use the superposition approach to solve the non-homogeneous differential equation, 𝑦 ′′ + 6𝑦 ′ + 8𝑦 = 4𝑥 − 𝑒 −2𝑥 Answer: We have to use the superposition approach to solve the non-homogeneous differential equations:𝑦 ′′ + 6𝑦 ′ + 8𝑦 = 4𝑥 − 𝑒 −2𝑥 Take homogeneous eqn for Complemantary function 𝑦 ′′ + 6𝑦 ′ + 8𝑦 = 0 ⇒ (𝐷2 + 6𝐷 + 8)𝑦 = 0 The auxiliary equation is 𝑚2 + 6𝑚 + 8 = 0 ⇒ 𝑚2 + 4𝑚 + 2𝑚 + 8 = 0 ⇒ 𝑚(𝑚 + 4) + 2(𝑚 + 4) = 0 Step 2 ⇒ (𝑚 + 2)(𝑚 + 4) = 0 ⇒ 𝑚 = −2, −4 ∴ 𝑦𝑐 (𝑥) = 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 −4𝑥 Now for particuler solution :Let the particuler Solution 𝑦𝑝(𝑥) = 𝑎1 + 𝑎2 𝑥 + 𝑎3 𝑒 −2𝑥 𝑥 ⇒ 𝑦𝑝′ (𝑥) = 𝑎2 + 𝑎3 𝑒 −2𝑥 − 2𝑎3 𝑒 −2𝑥 𝑥 ⇒ 𝑦3′′ (𝑥) = −2𝑎{3] 𝑒 −2𝑥 + 4𝑎3 𝑒 −2𝑥 𝑥 = −4𝑎3 𝑒 −2𝑥 + 4𝑎3 𝑒 −2𝑥 𝑥 ⇒ −4𝑎3 𝑒 −2𝑥 + 4𝑎3 𝑒 −2𝑥 𝑥 + 6[𝑎2 + 𝑎3 𝑒 −2𝑥 = 2𝑎3 𝑒 −2𝑥 𝑥] + 8[𝑎1 + 𝑎2 𝑥 + 𝑎3 𝑒 −2𝑥 𝑥] = 4𝑥 − 3 + 𝑒 −2𝑥 ⇒ 𝑒 −2𝑥 (−4𝑎3 + 4𝑎3 + 6𝑎3 − 12𝑎3 𝑥 + 8𝑎3 𝑥) + 𝑥(8𝑎2 ) + (6𝑎2 + 8𝑎1 ) = −4𝑥 − 3 + 𝑒 −2𝑥 Equating the coefficient of x, 𝑒 −2𝑥 & constant ⇒ 2𝑎3 = 1 ⇒ 𝑎3 = 1 3 1 2 8𝑎2 = 4 ⇒ 𝑎2 = 6𝑎2 + 8𝑎1 = −3 ⇒ 6𝑎1 = −3 − 6𝑎2 ⇒ 8𝑎1 = −3 × 1 3 ⇒ 6𝑎1 = −6 ⇒ 𝑎1 = −6 8 ⇒ 𝑎1 −3 −4 So, 𝑦𝑝 (𝑥) = −3 1 1 + 𝑥 + 𝑒 −2𝑥 𝑥 4 2 2 then 𝑦(𝑥) = 𝑦𝑐 (𝑥) + 𝑦𝑝 𝑥 1 1 −3 ⇒ 𝑦(𝑥) = 𝑐1 𝑒 −2𝑥 + 𝑐2 𝑒 −4𝑥 + 𝑥 + 𝑒 −2𝑥 𝑥 − 2 2 4 𝑥 𝑥 −3 2 2 4 ⇒ 𝑦(𝑥) = (𝑐1 + ) 𝑒 −2𝑥 + 𝑐2 𝑒 −4𝑥 + − Question #2 Using a change in variables, Show that the boundary value problem: − 𝑑 (𝑝(𝑥)𝑦 ′ ) + 𝑞(𝑥)𝑦 = 𝑓(𝑥), 𝑑𝑥 𝑎 ≤ 𝑥 ≤ 𝑏, 𝑦(𝑎) = 𝛼, 𝑦(𝑏) = 𝛽 Can be transformed to the form: − 𝑑 (𝑝(𝑤)𝑧 ′ ) + 𝑞(𝑤)𝑧 = 𝐹(𝑤), 𝑑𝑤 0 ≤ 𝑤 ≤ 1, 𝑧(0) = 0, 𝑧(1) = 0 Answer: Step 1 Notice that you can define 𝜔(𝑥) as 𝑥 − 𝑎 𝜔(𝑥) − 0 = 𝑏−𝑎 1−0 = 𝜔(𝑥) ⇒ 𝑥(𝜔) = 𝑎 + (𝑏 − 𝑎)𝜔, and then 𝑧(𝜔) = 𝜔 − 𝑦(𝑥(𝜔))−𝛼 𝛽−𝛼 . Finally, perform the respective derivatives using this new variables. The differentiation chain rule will be your best friend in that deal! Question #3 Value of differential equation at non existent points in the original function. 𝑦2 + 𝑦 = 𝑥 3 Answer: Step 1 Generally speaking, the equation 𝑓(𝑦) = 𝑔(𝑥) leads to multiple branches, where if both 𝑓(⋅) and 𝑔(⋅) are differentiable, then a derivation of y w.r.t. x can be defined as 𝑦 ′ 𝑓 ′ (𝑦) = 𝑔′ (𝑥) . The latter derivative makes sense as long as the points lie on the branches of 𝑓(𝑦) = 𝑔(𝑥) and is insignificant elsewhere. 1 Also note that for example in your case, y can be defined as a function of x on each branch (𝑦 = − + 2 √𝑥 3 + 1 4 1 2 4 on one branch and 𝑦 = − − √𝑥 3 + on another), and whether or not this function can be ′ explicitly found, y in 𝑦 = 𝑦′ = 1 3𝑥 2 2ℎ(𝑥)+1 3𝑥 2 2𝑦+1 should be replaced by some h(x), for which . Question #4 Variation of parameters method for differential equations. Change the variable 𝑥 = 𝑒 𝑡 and then find the general solution for the following differential equation 2𝑥 2 𝑦 ′′ − 6𝑥𝑦 ′ + 8𝑦 = 2𝑥 + 2𝑥 2 ln 𝑥 It seems a little suspicious that i can factor out 2 and x first. 𝑥 2 𝑦 ′′ − 3𝑥𝑦 ′ + 4𝑦 = 𝑥(ln(𝑥) + 1) and if we 3 4 𝑥 𝑥2 factor out 𝑥 2 now we end up with 𝑦 ′′ − 𝑦 ′ + 𝑦= 𝑥 ln 𝑥+1 𝑥 (1) Therefore substituting 𝑥 = 𝑒 𝑡 (1) now becomes 𝑦 ′′ − 3 𝑒𝑡 𝑦′ + 4 𝑒 (2𝑡) 𝑦= 𝑒 𝑡 𝑡+1 coefficients in order to use the variation of parameters method am i right? Answer: Answer: After change the variable 𝑥 = 𝑒 𝑡 we get 𝑒𝑡 . We need constant 𝑦 ′′ − 4𝑦 ′ + 4𝑦 = 𝑡𝑒 2𝑡 + 𝑒 𝑡 Then 𝑦 = = 𝑡 3 𝑒 2𝑡 +6𝑒 𝑡 6 + (𝐶1 + 𝐶2 𝑡)𝑒 2𝑡 = 𝑥 2 ln3 𝑥 + 6𝑥 + (𝐶1 + 𝐶2 ln 𝑥)𝑥 2 6 Question #5 Verify 𝑦 = 𝑥 tan 𝑥 is solution to ODE: 𝑥𝑦 ′ = 𝑦 + 𝑥 2 + 𝑦 2 Verify 𝑆𝑦 = 𝑥 tan 𝑥 is a solution to 𝑥𝑦 ′ = 𝑦 + 𝑥 2 + 𝑦 2 Answer: Step 1 𝑦 = 𝑥 tan(𝑥) . Verify that 𝑥𝑦 ′ = 𝑦 + 𝑥 2 + 𝑦 2 . Step 2 Alternate approach: 𝑑 𝑑𝑥 tan(𝑥) = sec 2 (𝑥). tan2 (𝑥) + 1 = sec 2 (𝑥). LHS equals: 𝑥[𝑥 sec 2 (𝑥) + tan(𝑥)] = 𝑥 2 [tan2 (𝑥) + 1] + 𝑦 = 𝑥 2 tan2 (𝑥) + 𝑥 2 + 𝑦 = 𝑦 2 + 𝑥 2 + 𝑦. Question #6 Verify that the following function 𝑦 2 = 𝑒 (2𝑥) + 𝐶 is a solution of the ODE: 𝑦𝑦 ′ = 𝑒 (2𝑥) Answer: Step 1 Let 𝐶 ∈ 𝑅 and 𝐼 ⊂ 𝑅 some interval, such that a function 𝑦: 𝐼 → 𝑅 with 𝑦 2 = 𝑒 2𝑥 + 𝐶 exists and is differentiable. Then by product or chain rule 2𝑦(𝑥)𝑦 ′ (𝑥) = (𝑦 2 )′ (𝑥) = 2𝑒 2𝑥 . Now devide both sides by 2. Question #7 Verify that the function is a solution to the differential equation 𝑢(𝑡) = 𝑒 𝑡 2 ∫𝑡 𝑒 −𝑠2 0 𝑑𝑠 + 𝑒 𝑡 2 is a particular solution to the differential equation: 𝑑𝑢 𝑑𝑡 − 2𝑡𝑢 = 1 My Attempt I will verify this by differentiating the function with respect to t. du dt 2 d t 2 t − s2 (e  e ds + et ) 0 dt t t 2 2 2 d 2 2 = 2tet (  e− s ds) + et (  e− s ds) + 2tet 0 0 dt = I'm having trouble solving the integral because it involves an error function. Could I get some pointers on how to evaluate this? Or is there a different way to verify that the function is a solution? Answer: You do not need to integrate the function, you can use Leibniz Integral rule 𝑑𝑢 Then 𝑑𝑢 𝑑𝑡 = 𝑑𝑡 𝑑 𝑑𝑡 becomes (𝑒 𝑡 2 ∫𝑡 𝑒 −𝑠2 0 𝑡 2 2 𝑑𝑠 + 𝑒 𝑡 ) 2 = 2𝑡𝑒 𝑡 (∫ 𝑒 −𝑠 𝑑𝑠) + 𝑒 𝑡 0 = 2𝑡 (𝑒 𝑡 Hence 2 ∫𝑡 𝑒 −𝑠2 0 𝑑𝑢 𝑑𝑡 2 𝑡 𝑡 𝑑 2 2 2 2 2 2 2 (∫ 𝑒 −𝑠 𝑑𝑠) + 2𝑡𝑒 𝑡 = 2𝑡𝑒 𝑡 (∫ 𝑒 −𝑠 𝑑𝑠) + 𝑒 𝑡 𝑒 −𝑡 + 2𝑡𝑒 𝑡 𝑑𝑡 0 0 2 𝑑𝑠 + 𝑒 𝑡 ) + 1 = 2𝑡𝑢 + 1 − 2𝑡𝑢 = 1 . Question #8 We will study the time-evolution of a finite dimensional quantum system. To this end, let us consider a quantum mechanical system with the Hilbert space 𝐶 𝟚 . We denote by |0⟩ 𝑎𝑛𝑑 |1⟩ the standard basis elements (1,0)𝑇 and (0,1)𝑇 . Let the Hamiltonian of the system in this basis be given by  0 1   0 −i  H = =   1 0   −i 0  and assume that for 𝑡 = 0 the state of the system is just given by 𝜓(𝑡 = 0) = |0⟩ . In the following, we also assume natural units in which ℏ = 1 . We expand the state at time t in the basis |0⟩, |1⟩ so: |Ψ(𝑡) ≫= 𝛼_0(𝑡)|0 ≫ +𝛼_1(𝑡)|1 ≫ Problems: Use Schrödinger's equation in order to derive a differential equations for 𝛼0 and 𝛼1 : (i) Find a solution given the initial conditions. (ii) What is the probability that the system can be measured in|1⟩ at some time t? Answer: Step 1   (t )  |  (t ) =  0   1 (t )  Step 2 Then the Schrödinger equation tells us: 𝑖ℏ i 𝜕 𝜕𝑡 |Ψ(𝑡) ≫= 𝐻|Ψ(𝑡) ≫    0 (t )   0 −i    0 (t )   =   t  1 (t )   −i 0   1 (t )  Although I will say I'm a little confused as to why the Hamiltonian isn't diagonalized in what I'm assuming is the energy basis given this a two level system. Question #9 What is the Solution of the Differential Equation (𝑑𝑃)/𝑑𝑡 = 𝑘𝑃 − 𝐴𝑃2 ? Answer: dP = kP − AP 2 , A  0 dt This is separable: dP  A  = kP 1 − P  dt  k  dP  =  kdt  A  P 1 − P   k  Partial Fraction decomp: 1 =   +  A  P 1− A P P 1 − P  k  k   A    1 − P  +  P = 1  k  A   = 1,  = k A 1   + k dP =  kdt P 1− A P k P  t  A  ln P − ln 1 − k P   =  kt 0    Po       P    Po    ln = kt A    1− k P   A   1 − Po  k   P Po  = e kt A 1− P k A 1 − Po k  A  Po  P = 1 − P  ekt A k   1− P o k    A  Po Po  P 1 +  ekt  = ekt A A  k 1 − Po  1 − Po k k   Po ekt A 1 − Po k P= Po A 1+  e kt k 1− A P o k kPo ekt = ( k − APo ) + APoekt = kPo ekt k − APo (1 − e kt ) Question #10 First using a trigonometric identity, find 𝐿𝑓(𝑡) 𝑓(𝑡) = sin 2 𝑡 cos 2 𝑡 Answer: Linearity Property: suppose 𝑓1 (𝑝) and 𝑓2 (𝑝) are Laplace transformations of 𝐹1 (𝑡) and 𝐹2 (𝑡) respectively. Then: L c1F1 ( t ) + c2 F2 ( t ) = c1L F1 ( t ) + c2 L F2 (t ) = c1 f1 ( p ) + c2 f 2 ( p ) Also, L sin at = a p + a2 2 Given, 𝑓(𝑡) = sin 2 𝑡 cos 2 𝑡 We have to find L  f ( t ) i.e. the Laplace Transformation of giveb f(t) . f ( t ) = sin 2t cos 2t = 1 ( 2sin 2t cos 2t ) = 2 1 = sin 4t 2 = Taking Laplace transformation on both sides, 1  L  f ( t ) = L  sin 4t  = 2  1 = L sin 4t = 2 1 4 = = 2 p 2 + 42 2 = 2 p + 42   The required solution L f ( t ) = 2 p + 42 2 Question #11 (𝑥 − 𝑦)𝑑𝑥 + 𝑥 𝑑𝑦 = 0 Please, solve by using an appropriate substitution. Answer: Given differential equation can be written as: dy y − x y = = −1 dx x x Let, y=upsilonx . Then dy  =  + xd dx dx Substituting y =  x in the given equation we have:  + xd  xd  dx  dx =  −1 = −1 dx x dx   d = −  (we take integration both side) x  d = −   = − ln x + c , (c is an integrating constant) y y = − ln x + c , (we put  = ) x x  y = x ( c − ln x )  The solution of (𝑥 − 𝑦)𝑑𝑥 + 𝑥 𝑑𝑦 = 0 is 𝑦 = 𝑥(𝑐 − ln 𝑥) , where c is an integrating constant Question #12 Find transient terms in this general solution to a differential equation, if there are any  x+2 y = ( x + C)   x−2 Answer: ( 2 + C ) x + 2C x+2 x2 = +  x−2 x−2  x−2 x−2 ( x + C )  Then take the limits of each of these three terms as x →  and see which(if any) approach 0. All three are rational expressions, so it is clear that only 2C is transient term. x−2