Questions and Answers #4 Second Order Linear Equations

Questions and Answers Sheet 4 Second Order Linear Equations Question #1 𝑦 ′′ (𝑥) + 4𝑦 ′ (𝑥) + 3𝑦(𝑥) = 9𝑥 2 − 20 + 30𝑒 2𝑥 , 𝑦(0) = 0, 𝑦 ′ (0) = 2 I'm stuck on finding a particular integral for the non-homogeneous R.H.S.: 9𝑥 2 − 20 + 30𝑒 2𝑥 So 𝑦 = 𝑦𝑐 + 𝑦𝑝 . I found 𝑦𝑐 = 𝐶1 𝑒 −𝑥 + 𝐶2 𝑒 −3𝑥 (where 𝐶1 and 𝐶2 are constants) But what would I try for 𝑦𝑝 ?? If it helps, I know how to do it if it's just a polynomial (i.e. for a 2nd degree polynomial): you set 𝑦𝑝 = 𝐴𝑥 2 + 𝐵𝑥 + 𝐶 and 𝑦𝑝′ = 𝐴𝑥 + 𝐵 and 𝑦𝑝′′ = 𝐴 But in this case we have different things. What would I need to do? I would really appreciate your help. Answer: Hint Let 𝑦𝑝 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 + 𝑘𝑒 2𝑥 substitute this in the equation and proceed through. Question #2 I need some hints for solving 𝑦𝑦 ′′ − (𝑦 ′ )2 = 𝑥𝑦 2 I noticed that the left hand side is close to (𝑦𝑦 ′ )′ : 𝑦𝑦 ′′ − (𝑦 ′ )2 = 𝑥𝑦 2 ⇒ 𝑦𝑦 ′′ + (𝑦 ′ )2 − 2(𝑦 ′ )2 = 𝑥𝑦 2 ⇒ (𝑦𝑦 ′ )′ − 2(𝑦 ′ )2 = 𝑥𝑦 2 But I don't know how to continue expressing the terms as derivatives of some functions. Answer: Take the change of variable 𝑧 = ln 𝑦. So you would end up with the following relations 𝑦 = 𝑒𝑧 𝑦′ = 𝑧′𝑒 𝑧 𝑦 ′′ = 𝑒 𝑧 𝑧 ′′ + (𝑧 ′ )2 𝑒 𝑧 Upon substitution (all 𝑒 𝑧 's will cancel out), leaving you with a 2-nd order differential equation, which is easy to solve: 𝑧 ′′ = 𝑥 Question #3 I'm currently working on some problems concerning the calculus of variations and I have come up with the following differential equation that I now want to solve: 1 + 𝑦 ′ (𝑥)2 − 𝑦 ′′ (𝑥)(𝑦(𝑥) − 𝜆) = 0 I'm only used to solving linear differential equations so I would appreciate some advice on where to start. Answer: Let 𝑢 = 𝑦 − 𝜆, 𝑣 = 𝑢′ = 𝑦 ′ . then we can rewrite the differential equation as 1 + 𝑣2 − 𝑣′ 𝑢 = 0 → 𝑢 𝑑𝑣 𝑑𝑥 = 𝑣′ 𝑢 = 1 + 𝑣2, 𝑑𝑢 𝑑𝑥 = 𝑢′ = 𝑣 which implies 𝑢 𝑑𝑣 𝑑𝑢 = 1+𝑣 2 𝑣 → 2𝑣𝑑𝑣 1+𝑣 2 = 2𝑑𝑢 𝑢 on integration, we get (1 + 𝑣 2 ) = 𝑐𝑢2 → 𝑣 = ±√1 − 𝑐𝑢2 → 𝑑𝑢 √1−𝑐𝑢2 = ±𝑑𝑥 now the integration will depend on the sign of c: if c>0 , you get an inverse sine, sin−1 and if c<0, then 𝑙𝑛 Question #4 We have the following differential equation 𝑢𝑢′′ = 1 + (𝑢′ )2 i found that the general solution of this equation is 𝑢 = 𝑑 cosh ((𝑥 − 𝑏/𝑑) where b and d are constants Please how we found this general solution? Answer: You can separate that equation as 2𝑢′𝑢′′ 1+𝑢2 =2 𝑢′ 𝑢 where both sides are complete differentials which integrate to ln(1 + 𝑢′2 ) = ln(𝑢2 ) + 𝑐 ⇒ 1 + 𝑢′2 = 𝐶𝑢2 Can you continue? Question #5 Show that 𝑦 = 𝑥 2 sin(𝑥) and y=0 are both solutions of 𝑥 2 𝑦 ′′ − 4𝑥𝑦 ′ + (𝑥 2 + 6)𝑦 = 0 and that both satisfy the conditions y(0)=0 and y'(0)=0 . Does this theorem contradict Theorem A? If not, why not? Answer: The existence and uniqueness is usually stated when the coefficient of y'' is 1, so you want to look at 4 6 𝑥 𝑥2 𝑦 ′′ (𝑥) − 𝑦 ′ (𝑥) + (1 + ) 𝑦(𝑥) = 0 There is no contradiction because the coefficients 4 6 − and 1 + 2 𝑥 𝑥 are continuous except at x=0 Question #6 Please suggest a substitution for solving or any method of solving: 𝑦 ′′ (𝑥) cot(𝑦(𝑥)) = 𝑦 ′ (𝑥)2 + 𝑐 Answer: Substituting 𝑢 = sin(𝑦) one gets 𝑢′′ = (sin(𝑦))′′ = (cos(𝑦) 𝑦 ′ )′ = cos(𝑦) 𝑦 ′′ − sin(𝑦) 𝑦 ′2 = 𝑐 sin(𝑦) = 𝑐𝑢 which is, depending on the sign of c, an oscillation equation or an exponential function which is easily solvable for u and thus for y. a1 cos( | c | x) + a2 sin( | c | x)  u = b1 + b2 x  cx cx c1e + c2 e for c  0 for c = 0 for c  0 Question #7 Given the function 𝑦(𝑥)𝐶1 𝑒 𝑥 + 𝐶2 𝑒 𝑥 cos 𝑥 a) Prove that it is a solution to the equation 𝑦 ′′′ − 3𝑦 ′′ + 4𝑦 ′ − 2𝑦 = 0 b) Show that the initial value problem 𝑦(0) = 1; 𝑦 ′ (0) = 0; 𝑦 ′′ (0) = 0 has no solution. Answer: 𝑦(𝑥)𝐶1 𝑒 𝑥 + 𝐶2 𝑒 𝑥 cos 𝑥 𝑦 ′ = 𝐶1 𝑒 𝑥 + 𝐶2 [−𝑒 𝑥 sin 𝑥 + 𝑒 𝑥 cos 𝑥] 𝑦 ′′ = 𝐶1 𝑒 𝑥 + 𝐶2 [−𝑒 𝑥 cos 𝑥 − 𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 sin 𝑥 + 𝑒 𝑥 cos 𝑥] 𝑦 ′′′ = 𝐶1 𝑒 𝑥 + 𝐶2 [𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 cos 𝑥 − 𝑒 𝑥 cos 𝑥 − 𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 sin 𝑥 − 𝑒 𝑥 cos 𝑥 − 𝑒 𝑥 sin 𝑥 + 𝑒 𝑥 cos 𝑥] ∴ 𝑦 ′ = 𝐶1 𝑒 𝑥 + 𝐶2 [𝑒 𝑥 cos 𝑥 − 𝑒 𝑥 sin 𝑥] 𝑦 ′′ 𝐶1 𝑒 𝑥 + 𝐶2 [−2𝑒 𝑥 sin 𝑥] 𝑦 ′′′ = 𝐶1 𝑒 𝑥 + 𝐶2 [−2𝑒 𝑥 cos 𝑥 − 2𝑒 𝑥 sin 𝑥] Substitute in 𝑦 ′′ − 3𝑦 ′′ + 4𝑦 ′ − 2𝑦 ⇒ 𝐶1 𝑒 𝑥 − 2𝐶2 𝑒 𝑥 cos 𝑥 − 2𝐶2 𝑒 𝑥 sin 𝑥 − 3𝐶1 𝑒 𝑥 + 6𝐶2 𝑒 𝑥 sin 𝑥 + 4𝐶1 𝑒 𝑥 + 4𝐶2 𝑒 𝑥 cos 𝑥 − 4𝐶2 𝑒 𝑥 sin 𝑥 − 2𝐶1 𝑒 𝑥 − 2𝐶2 𝑒 𝑥 cos 𝑥 =0 Since all terms cancel out. Hence given y is solution. Question #8 Solve the following differential equation 𝑦 ′′ + 4𝑦 = cos 𝑥 sin 𝑥 Answer: The given equation is a second order linear, non-homogeneous differential equation with constant coefficient. The general solution will be the sum of the complementary solution and particular solution. The complementary solution for the given equation: 𝜆2 + 4 = 0 (𝜆 − 2𝑖)(𝜆 + 2𝑖) = 0 𝜆 = 2𝑖, −2𝑖 fow two complex root 𝛾1 ≠ 𝛾2 where 𝛾1 = 𝛼 + 𝑖 ⋅ 𝛽, 𝛾2 = 𝛼 − 𝑖 ⋅ 𝛽. 𝑦(𝑥) = 𝑒 𝛼⋅𝑥 (𝑐1 ⋅ cos(𝛽 ⋅ 𝑥) + 𝑐2 sin(𝛽 ⋅ 𝑥)) 𝑦𝑐 = 𝑐1 cos(2𝑥) + 𝑐2 sin(2𝑥) The particular solution for the equation is 𝑦𝑝 is of the form 𝑥(𝑎1 ⋅ cos(2𝑥) + 𝑎2 ⋅ sin 2 𝑥) 𝑎1 ⋅ cos(2𝑥) + 𝑎2 ⋅ sin 2 𝑥 is Multiplies by x because sin 2x in the complementary solution compute 𝑑 2 𝑦𝑝 (𝑥) 𝑑 2 ⋅𝑦𝑝 (𝑥) 𝑑𝑥 2 𝑑𝑥 2 = 𝑑2 𝑑𝑥 2 𝑥(𝑎1 ⋅ cos(2𝑥) + 𝑎2 ⋅ sin 2 𝑥) = −4𝑎1 cos 2 𝑥 − 4𝑎1 sin(2𝑥) + 4𝑎2 cos(2𝑥) − 4𝑎2 𝑥 sin(2𝑥) Solve further: Substitute the Particular into the differential equation: 𝑑 2 𝑦𝑝 (𝑥) 𝑑𝑥 2 1 + 4𝑦𝑝 (𝑥) = sin(2𝑥) − 4𝑎1 𝑥 cos 2 𝑥 − 4𝑎1 sin(2𝑥) + 4𝑎2 ⋅ cos(2𝑥) − 4𝑎2 𝑥 sin 2 𝑥 + 2 1 4𝑥(𝑎1 ⋅ cos(2𝑥) + 𝑎2 ⋅ sin 2 𝑥) − sin(2𝑥) ; 2 1 −4𝑎1 sin(2𝑥) + 4𝑎2 cos(2𝑥) = sin(2𝑥) 2 Equate the coefficient of cos 2 𝑥 and sin 2 𝑥; 1 −1 2 8 −4𝑎1 = , 4𝑎2 = 0 ⇒ 𝑎1 = , 𝑎2 = 0 1 𝑦𝑝 (𝑥) = − 𝑥 cos(2𝑥) 8 Hence, the general solution for the given equation is: 𝑦(𝑥) = 𝑦𝑐 (𝑥) + 𝑦𝑝 (𝑥) 1 𝑦(𝑥) = 𝑐1 cos(2𝑥) + 𝑐2 sin(2𝑥) − 𝑥 cos(2𝑥) 8 Question #9 Find the general solution of the give equation 3𝑦 ′′ − 20𝑦 ′ + 12𝑦 = 0 Answer: Given differential equation 3𝑦 ′′ − 20𝑦 ′ + 12𝑦 = 0 A second order linear homogeneous ODE has the form of ay''+by'+cy=0 Write the second order differential equation as a quadratic function: 3r^2-20r+12 Solving the above equation 𝑟= 𝑟= 𝑟= 𝑟= −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 −(−20)±√(−20)2 −4⋅3⋅12 2⋅3 20±√256 6 20±16 6 Taking positive sign 𝑟= 20+16 6 r=6 Taking negative sign 𝑟= 20−16 6 𝑟= 2 3 The expression for the general solution is given by 𝑦 = 𝐶1 𝑒 𝑟𝑡 + 𝐶2 𝑒 𝑟𝑡 Substitute the values in above expression 2 𝑦 = 𝐶1 𝑒 6𝑡 + 𝐶2 𝑒 3𝑡 Therefore, he general solution of the given equation 3𝑦 ′′ − 20𝑦 ′ + 12𝑦 = 0 is 2 𝑦 = 𝐶1 𝑒 6𝑡 + 𝐶2 𝑒 3𝑡 Question #10 Solve the differential equation 𝑦 ′′ − 𝑦 ′ = 𝑒𝑥 𝑒 𝑥 +1 Answer: Given: The differential equation is 𝑦 ′′ − 𝑦 ′ = 𝑒𝑥 𝑒 𝑥 +1 Calculation: A second order linear, non-homogeneous ODE has the form of a(x)y''+b(x)y'+c(x)y=g(x) . The general solution to a(x)y''+b(x)y'+c(x)y=g(x) can be written as 𝑦 = 𝑦𝑝 + 𝑦ℎ 𝑦ℎ = 𝑎(𝑥)𝑦 ′′ + 𝑏(𝑥)𝑦 ′ + 𝑐(𝑥)𝑦 = 0 𝑦ℎ = 𝑦 ′′ − 𝑦 ′ = 0 The characteristic equation is, 𝑟 2 − 𝑟 = 0 ⇒ 𝑟(𝑟 − 1) = 0 r=0,1 𝑦ℎ = 𝑐1 𝑒 𝑥 + 𝑐2 𝑒 0𝑥 = 𝑐1 𝑒 𝑥 + 𝑐2 𝑦𝑝 = 1 𝑟 2 −𝑟 ⇒ 𝑦𝑝 = 1 𝑒𝑥 𝑟(𝑟−1) 𝑒 𝑥 +1 = 1𝑒 𝑥 𝑒 𝑥 +1 = 𝑥𝑒 𝑥 − 𝑒 𝑥 ln(𝑒 𝑥 + 1) − ln(𝑒 𝑥 + 1) Answer 𝑦 = 𝑦𝑝 + 𝑦ℎ 𝑦 = 𝑐1 𝑒 𝑥 + 𝑐2 + 𝑥𝑒 𝑥 − 𝑒 𝑥 − 𝑒 𝑥 ln(𝑒 𝑥 + 1) − ln(𝑒 𝑥 + 1)