Questions and Answers #5 Second Order Linear Equations

Questions and Answers Sheet 5 Second Order Linear Equations Question #1 Find the particular solution of the following linear second-order inhomogeneous differential equation 𝑑 2 𝑦(𝑡) 𝑑𝑡 2 −2 𝑑𝑦(𝑡) 𝑑𝑡 + 4𝑦(𝑡) = 3𝑡 − 4 Answer: Given second order non homogeneous differential equation is 𝑑 2 𝑦(𝑡) 𝑑𝑡 2 −2 𝑑𝑦(𝑡) 𝑑𝑡 + 4𝑦(𝑡) = 3𝑡 − 4 we need to find particular solution of the given non homogeneous differential equation Solution: Given second order non homogeneous differential equation is 𝑑 2 𝑦(𝑡) 𝑑𝑡 2 −2 𝑑𝑦(𝑡) 𝑑𝑡 + 4𝑦(𝑡) = 3𝑡 − 4 So firstly we find solution of homogeneous part Consider associated homogeneous differential equation is 𝑑 2 𝑦(𝑡) 𝑑𝑡 2 −2 𝑑𝑦(𝑡) 𝑑𝑡 + 4𝑦(𝑡) = 0 Auxiliary equation is 𝑚2 − 2𝑚 + 4 = 0 ⇒𝑚= ⇒𝑚= ⇒𝑚= ⇒𝑚= −(−2) ± √4 − 4(4) 2 2±√4−16 2 2±√12 2 2±√3 2 ⇒ 𝑚 = 1 ± √3 Therefore complementary solution is 𝑦𝑐 = 𝑒 𝑡 (𝑐1 cos(√3𝑡) + 𝑐2 sin(√3𝑡)) Now to find particular solution of the non homogeneous differential equation we use method of undetermined coefficient Let particular solution is of the form 𝑦𝑝 = 𝐴𝑡 + 𝐵 therefore we get 𝑦𝑝 = 𝐴𝑡 + 𝐵 ⇒ ⇒ 𝑑𝑦𝑝 =𝐴 𝑑𝑡 𝑑 2 𝑦𝑝 𝑑𝑡 2 =0 Therefore since it is particular solution it must satisfies the differential equation Thus 𝑑 2 𝑦(𝑡) 𝑑𝑡 2 −2 𝑑𝑦(𝑡) 𝑑𝑡 + 4𝑦(𝑡) = 3𝑡 − 4 ⇒ 0 − 3𝐴 + 4(𝐴𝑡 + 𝐵) = 3𝑡 − 4 ⇒ 4𝐴𝑡 + (−2𝐴 + 4𝐵) = 3𝑡 − 4 comparing both side we get 3  4A = 3  A =   4  −2 A + 4 B = −4  B = −4 + 2 A  B = −8 + 3  B = −5  4 8 8  Therefore particular solution is 𝑦𝑝 = 𝐴𝑡 + 𝐵 ⇒ 𝑦𝑝 = 3𝑡 4 − 5 8 So general solution is 𝑦 = 𝑦𝑐 + 𝑦𝑝 ⇒ 𝑦 = 𝑒 𝑡 (𝑐1 cos(√3𝑡) + 𝑐2 sin(√3𝑡)) + 3𝑡 4 − 5 8 Answer: The particular solution of the linear second order non homogeneous differential equation 4𝑦(𝑡) = 3𝑡 − 4𝑖𝑠𝑦𝑝 = 3𝑡 4 − 5 𝑑 2 𝑦(𝑡) 𝑑𝑡 2 8 Question #2 For what value of k, if any, will 𝑦 = 𝑘𝑒 −2𝑥 + 4 cos(3𝑥) be a solution to the differential equation 𝑦 + 9y = 26e^{-2x Answer: The solutions to the differential equation are those that satisfies the equation. Given: A second order linear differential equation, 𝑦 + 9y = 26e^{-2x 𝑦 = 𝑘𝑒 −2𝑥 + 4 cos(3𝑥) is a solution to the above differential equation. Hence, it will satisfy it 𝑦 ′ = −2𝑘𝑒 −2𝑥 − 12 sin(3𝑥) 𝑦 ′′ = 4𝑘𝑒 −2𝑥 − 36 cos(3𝑥) 𝑦 ′′ + 9𝑦 = 26𝑒 −2𝑥 −2 𝑑𝑦(𝑡) 𝑑𝑡 + 4𝑘𝑒 −2𝑥 − 36 cos(3𝑥) + 9(𝑘𝑒 −2𝑥 + 4 cos(3𝑥)) = 26𝑒 −2𝑥 13𝑘𝑒 −2𝑥 = 26𝑒 −2𝑥 Compare both sides we get 13k=26 k=2 Answer: the value of k is 2 Question #3 Solute the equation: (1 − 𝑥 2 )𝑦 ′′ = 𝑥𝑦 ′ Answer: The given equation is second order linear ordinary differential equation. This solution for the interval, |x|<1, there is another solution for |x|>1. (1 − 𝑥 2 )𝑦 ′′ = 𝑥𝑦 ′ ⇒ (1 − 𝑥 2 ) ⇒ ⇒ 𝑑(𝑦 ′ ) 𝑑𝑛 𝑑(𝑦)′ 𝑦′ 𝑑 𝑑𝑥 (𝑦 ′ ) = 𝑥𝑦 ′ 𝑥𝑦 ′ = 1−𝑥 2 𝑥 = 1−𝑥 2 Integrating ∫ 𝑑𝑥 𝑑𝑦 ′ 𝑦′ =∫ 𝑥 1−𝑥 2 𝑑𝑥 1 ln 𝑦 ′ = − ln(1 − 𝑥 2 ) + ln 𝐶 [ ln 𝑐 constant] 2 𝑐⋅1 ⇒ 𝑦′ = √1−𝑥 2 ⇒ 𝑑𝑦 = 𝑐(𝑑𝑥) √1−𝑥 2 1 ⇒ ∫ 𝑑𝑦 = ∫ 𝑐 √ 𝑑𝑥 1−𝑥 2 ⇒ 1 𝑐⋅𝑦 = sin 𝑥 + 𝑑 [d is constant] sin 𝑥 = 𝑦 𝑐−𝑑 ⇒ 𝑥 = sin ( 𝑦 𝑐−𝑑 ) 1 Solution: 𝑥 = sin(𝐴𝑦 + 𝐵) , 𝐴 = , 𝐵 = −𝑑 𝑐 |n|>1 𝑦′ = 𝑑𝑦 = 𝑐 √𝑛2 −1 𝑐(𝑑𝑥) √𝑛2 −1 ∫ 𝑑𝑦 = 𝑐 ∫ √ 𝑑𝑥 𝑛2 −1 𝑦 = 𝑐 ⋅ ln|𝑛 + √𝑛2 − 1| + 𝑑 c, d constant Question #4 Solve the equation 𝑦 ′′ − 4𝑦 = 2 + 3𝑥 − 𝑥 2 Answer: Given differential equation is 𝑦 ′′ − 4𝑦 = 𝑥 2 + 3𝑥 + 2 This is a second order linear non-homogeneous equation of the form of 𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐𝑦 = 𝑔(𝑥) General solution of this equation is 𝑦 = 𝑦ℎ + 𝑦𝑝 Where 𝑦ℎ is solution of 𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐𝑦 = 0 And 𝑦𝑝 is the solution for any function that satisfies non-homogeneous equation Hence, 𝑦ℎ = 𝑐1 𝑒1𝑚 + 𝑐2 𝑒2𝑚 Where 𝑚1 and 𝑚2 are the roots of Auxiliary equation of y''-4y is 𝑚2 − 4 = 0 m=2, -2 Then yh = c1 e2 + c2 e−2 Let 𝑦 = 𝐴𝑥 2 + 𝐵𝑥 + 𝐶 y=2Ax+B y=2A Similarly 𝐵= −3 4 And (2A-4C)=2 1 ⇒ 2 ⋅ − 4𝐶 = 2 4 1 ⇒ − 4𝐶 = 2 2 𝐶= −3 8 Thus solution is 𝑦 = 𝑐1 𝑒 2𝑥 + 𝑐2 𝑒 −2𝑥 + 𝑥2 4 − 3𝑥 4 − 3 8 Question #5 Find y: 𝑦 ′′ − 9𝑦 ′ + 14𝑦 = 0 Answer: Given, 𝑦" − 9𝑦′ + 14𝑦 = 0 This is a second order linear ordinary differential equation. (𝐷2 − 9𝐷 + 14)𝑦 = 0 Auxiliary equation: 𝑚2 − 9𝑚 + 14 = 0 (𝑚 − 7)(𝑚 − 2) = 0 m=2 or m=7 Thus 𝑦 = 𝑐1 𝑟 2𝑥 + 𝑐2 𝑒 7𝑥 Question #6 Solve the given differential equation by undetermined coefficients. 𝑦 ′′ + 4𝑦 = 4 sin(2𝑥) Answer: 𝑦 ′′ + 4𝑦 = 4 sin(2𝑥) is a second order linear, non-homogenous ODE with constant coefficient. The general solution can be written as 𝑦 = 𝑦𝑐 + 𝑦𝑝 𝑦𝑐 is the solution of the homogeneous ODE y''+4y=0 and 𝑦𝑝 is the particular solution, is any function that satisfied the non-homogeneous equation. Find 𝑦𝑐 by solving y''+4y=0 First find the solution of the equation y''+4y=0 . Solve the characteristic equation 𝑚2 + 4 = 0. 𝑚2 + 4 = 0 𝑚±= 2𝑖 When the roots are complex and distinct then the general solution is of the form: The complex roots 𝛾1 = 𝛼 + 𝑖 ∗ 𝛽, 𝛾2 = 𝛼 − 𝑖 ∗ 𝛽. the solution is of the form 𝑦 = 𝑒 𝛼⋅𝑥 (𝑐1 ⋅ cos(𝛽 ⋅ 𝑥) + 𝑐2 ⋅ sin(𝛽 ⋅ 𝑥)) ⇒ for 𝑚 = ±2𝑖 solution is 𝑦𝑡 = 𝑒 (0)𝑥 (𝑐1 ⋅ cos(2𝑥) + 𝑐2 ⋅ sin(2𝑥)) = 𝑐1 ⋅ cos(2𝑥) + 𝑐2 ⋅ sin(2𝑥) Now, find the particular solution use the formula 1 𝐷2 +𝑎2 sin 𝑎 ⋅ 𝑥 = − 𝑥 2𝑎 cos 𝑎 𝑥 similarly 4⋅ 1 𝐷2 +22 sin 2 𝑥 = 4 ⋅ ( −𝑥 4 cos 2𝑥 ) = −𝑥 cos 2 𝑥 Hence solution is : 𝑦 = 𝑐1 cos(2𝑥) + 𝑐2 sin(2𝑥) − 𝑥 cos 2 𝑥 Question #7 Identify the auxiliary equation of the following differential equation along with its roots and the two fundamental solutions. 𝑦 ′′ + 6𝑦 ′ + 8𝑦 = 0 Answer: Given differential equation is second order linear homogeneous differential equation: y''+6y+8y=0 To find auxiliary equation for any second order differential equation, replace: y'' as 𝑚2 y' as m y as 1 Now, replacing these terms the differential equation turns to: 𝑚2 + 6𝑚 + 8 = 0 which is called as Auxiliary equation for the given differential equation. Roots of auxiliary equation 𝑚2 + 6𝑚 + 8 = 0 Factorizing 𝑚2 + 6𝑚 + 8 gives (m+4) and (m+2) 𝑚2 + 6𝑚 + 8 = 0 ⇒ (𝑚 + 4)(𝑚 + 2) = 0 therefore, roots of the auxiliary equation are m=-4 and m=-2 General solution for y''+6y'+8y=0 When an auxiliary equation has two distinct roots 𝑟1 and 𝑟2 , the general solution is of the form: 𝑦(𝑡) = 𝑐1 𝑒 𝑟1 𝑡 + 𝑐2 𝑒 𝑟2𝑡 , where 𝑐1 and 𝑐2 are arbitrary coefficients. Here the roots of auxiliary equation are m=-4 and m=-2 Hence, its general solution is: 𝑦(𝑡) = 𝑐1 𝑒 −2𝑡 + 𝑐2 𝑒 −4𝑡 Question #8 Find the solution of the following initial value problem: 𝑥 ′′ (𝑡) − 7𝑥 ′ (𝑡) + 10𝑥(𝑡) = 0, 𝑥(0) = 2, 𝑥 ′ (0) = 3 Answer: Since the given differential equation is second order linear homogenous differential equation so we will first find the general solution by finding the roots of auxiliary equation then we will use initial conditions to get the required solution 𝑥 ′′ (𝑡) − 7𝑥 ′ (𝑡) + 10𝑥(𝑡) = 0, 𝑥(0) = 2, 𝑥 ′ (0) = 3 The chracteristic (auxiliary) equation of 𝑥 ′′ (𝑡) − 7𝑥 ′ (𝑡) + 10𝑥(𝑡) = 0 is given by 𝑚2 − 7𝑚 + 10 = 0 ⇒ 𝑚2 − 5𝑚 − 2𝑚 + 10 = 0 ⇒ 𝑚(𝑚 − 5) − 2(𝑚 − 5) = 0 ⇒ (𝑚 − 2)(𝑚 − 5) = 0 ⇒ 𝑚 = 2.5 Hence the roots of the characteristic equation are real and district, so the general solution of 𝑥 ′′ (𝑡) − 7𝑥 ′ (𝑡) + 10𝑥(𝑡) = 0 is given by 𝑥(𝑡) = 𝑐1 𝑒 2𝑡 + 𝑐2 𝑒 5𝑡 ⇒ 𝛼. 𝑐1 and 𝑐2 being arbitrary constants Now differentiating 𝛼 b/s w.r.t. t we get 𝑥 ′ (𝑡) = 2𝑐1 𝑒 2𝑡 + 5𝑐2 𝑒 5𝑡 Now using initial conditions we have x(0)=2 ⇒ 2 = 𝑐1 𝑒 2(0) + 𝑐2 𝑒 5(0) ⇒ 𝑐1 + 𝑐2 = 2 ⇒ 𝑐1 = 2 − 𝑐2 ⇒ 𝐴 Also 𝑥 ′ (0) = 3 ⇒ 2𝑐1 𝑒 2(0) + 5𝑐2 𝑒 5(0) = 3 ⇒ 2𝑐1 + 5𝑐2 = 3 ⇒ 𝐵 Putting value q 𝑐1 from A into B we get 2(2 − 𝑐2 ) + 5𝑐2 = 3 ⇒ 4 − 2𝑐2 + 5𝑐2 = 3 ⇒ 3𝑐2 = −1 ⇒ 𝑐2 = −1 3 Putting value of 𝑐2 in A we get 𝑐1 = 2 − −1 3 1 7 3 3 ⇒ 𝑐1 = 2 + = Hence the solution of the initial value problem is 7 1 3 3 𝑥(𝑡) = 𝑒 2𝑡 − 𝑒 5𝑡 Question #9 Solve the equation: 𝜋 𝜋 4 4 𝑦 ′′ + 2𝑦 ′ + 2𝑦 = 0, 𝑦 ( ) = 2, 𝑦 ′ ( ) = −2 Answer: Given that: The second order differential equation 𝑦 ′′ + 2𝑦 ′ + 2𝑦 = 0 𝜋 𝜋 4 4 The initial condition 𝑦 ( ) = 2 and 𝑦 ′ ( ) = −2 Use: A second order linear, homogeneous ordinary differential equation has the form of 𝑦 ′′ + 𝑝𝑦 ′ + 𝑞𝑦 = 0 The characteristic equation: 𝑟 2 + 𝑝𝑟 + 𝑞 = 0 to the general solution with two complex root 𝑟1 ≠ 𝑟2 where 𝑟1 = 𝛼 + 𝑖 ⋅ 𝛽, 𝑟2 = 𝛼 − 𝑖 ∗ 𝛽. The general solution take the form 𝑒 = 𝑒 𝛼⋅𝑡 (𝑐1 cos(𝛽 ⋅ 𝑡) + 𝑐2 sin(𝛽 ⋅ 𝑡)) The characteristic equation of the given second order, linear, homogeneous ordinary differential equation is 𝑟 ′′ + 2𝑟 ′ + 2 = 0 This is the quadratic polynomial of the form 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 To use the quadratic formula to find the root of the above polynomial. 𝑥= −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 Comparing the given polynomial with 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 = 0 To get a=1, b=2, c=2 Plug the values of a=1, b=2, c=2 in the quadratic formula. To get, 𝑟= = = = = −2±√22 −4(1)(2) 2(1) −2±√4−8 2 2±√−4 2 2±𝑖2 2 2(−1±𝑖) 2 = −1 ± 𝑖 To get 𝑟 = −1 ± 𝑖 The root of the given polynomial are 𝑟1 = −1 + 𝑖, 𝑟2 = −1 − 𝑖 Then, The general solution of the given differential equation: 𝑦 = 𝑒 𝛼⋅𝑡 (𝑐1 cos(𝛽 ⋅ 𝑡) + 𝑐2 sin(𝛽 ⋅ 𝑡)) Plug the values of a=-1 and 𝛽 = 1 To get, 𝑦(𝑡) = 𝑒 −𝑡 (𝑐1 cos 𝑡 + 𝑐2 sin 𝑡) To find y'(t) Differentiate the above equation with respect to t, To get 𝑦 ′ (𝑡) = −𝑒 −𝑡 (𝑐1 cos 𝑡 + 𝑐2 sin 𝑡) + 𝑒 −𝑡 (−𝑐1 sin 𝑡 + 𝑐2 cos 𝑡) 𝜋 𝜋 4 4 Use the initial condition 𝑦 ( ) = 2, and 𝑦 ′ ( ) = −2 Plug 𝑡 = 𝜋 in y(t) 4 𝜋 𝜋 𝜋 𝜋 4 4 To get, 𝑦 ( ) = 𝑒 − 4 (𝑐1 cos ( ) + 𝑐2 sin ( )) 4 𝜋 1 1 √2 √2 2 = 𝑒 − 4 (𝑐1 ( ) + 𝑐2 ( )) 𝜋 − 𝑒 4 𝑐1 2= √2 𝜋 + − 𝑒 4 𝑐2 √2 Plug the 𝑡 = 𝜋 4 in y'(t) To get 𝑝𝑖 𝜋 𝜋 𝜋 4 4 𝜋 𝜋 𝜋 4 4 𝑦 ′ ( ) = −𝑒 − 4 (𝑐1 cos ( ) + 𝑐2 sin ( )) + 𝑒 − 4 (−𝑐1 sin ( ) + 𝑐2 cos ( )) 4 𝜋 𝜋 1 1 √2 √2 1 1 √2 √2 −2 = −𝑒 − 4 (𝑐1 ( ) + 𝑐2 ( )) + 𝑒 − 4 (−𝑐1 ( ) + 𝑐2 ( )) 𝜋 −2 = − −𝑒 4 𝑐1 √2 𝜋 − − 𝑒 4 𝑐2 √2 𝜋 − − 𝑒 4 𝑐1 √2 𝜋 + − 𝑒 4 𝑐2 √2 𝜋 −2 = − −𝑒 4 𝑐1 √2 Multiplying on both side by √2 𝜋 − −𝑒 4 To get, −2√2 = 𝑐1 𝜋 − −𝑒 4 To get 𝑐1 = 2√2 𝜋 − 𝑒 4 Plug the value of 𝑐1 = 1 1 √2 √2 (𝑐1 ( ) + 𝑐2 ( )) 𝜋 2= − 𝑐1 𝑒 4 √2 𝜋 + − 𝑐2 𝑒 4 √2 (2√2) 𝑒 𝜋 −4 𝜋 in the equation 2 = 𝑒 − 4 2= 𝜋 − 𝑒 4 2√2 ( 𝜋) − 𝑒 4 𝜋 √2 𝜋 2= − 2√2𝑒 4 𝑝𝑖 − 𝑒 4 √2 − 𝑐2 𝑒 4 + √2 𝜋 + − 𝑐2 𝑒 4 √2 𝜋 2= 2+ − 𝑐2 𝑒 4 √2 Substract 2 on both side, To get 𝜋 0= − 𝑐2 𝑒 4 √2 𝜋 𝑐2 𝑒 − 4 = 0 𝜋 Divide on both side be 𝑒 − 4 to get 𝑐2 = 0 Plug the value of 𝑐1 = 𝑦(𝑡) = 𝑒 −𝑡 ( 2√2 𝜋 − 𝑒 4 2√2 𝜋 − 𝑒 4 and 𝑐2 = 0 in the general solution of the differential equation cos 𝑡 + 0 sin 𝑡) To get, 𝑦(𝑡) = 2√2𝑒 −𝑡 𝜋 − 𝑒 4 𝜋 Therefore, The general solution of the given differential with the given initial condition 𝑦 ( ) = 2 and 4 𝜋 𝑦 ′ ( ) = −2 is 4 𝑦(𝑡) = 2√2𝑒 −𝑡 𝜋 − 𝑒 4 Question #10 Find the general solution 𝑦 ′′ + 5𝑦 ′ − 6𝑦 = 0 Answer: Given: 𝑦 ′′ + 5𝑦 − 6𝑦 = 0 To find roots of this differential equation, Let m be the root of this equation that is, 𝑚2 + 5𝑚 − 6 = 0 Simplifying for m, 𝑚2 + 5𝑚 − 6 = 0 𝑚2 − 𝑚 + 6𝑚 − 6 = 0 m(m-1)+6(m-1)=0 (m+6)(m-1)=0 m=-6 or m=1 So the roots are, 𝑚1 = −6, 𝑚2 = 1 Since, the general solution of second order linear differential equation is, 𝑦 = 𝐶1 𝑒 𝑚1⋅𝑥 + 𝐶2 𝑒 𝑚2⋅𝑥 Where 𝑚1 and 𝑚2 are real and distinct roots of second order linear differential equation and 𝐶1 , 𝐶2 are constant. So now, The general solution of given differential equation is, 𝑦 = 𝐶1 𝑒 −6𝑥 + 𝐶2 𝑒 (1)𝑥 𝑦 = 𝐶1 𝑒 −6𝑥 + 𝐶2 𝑒 𝑥 Answer: The final answer is given below, 𝑦 = 𝐶1 𝑒 −6𝑥 + 𝐶2 𝑒 𝑥