Questions and Answers #2 Second Order Linear Equations

Questions and Answers Sheet 2 Second Order Linear Equations Question #1 Find the general solution of the given differential equation. 𝑦” − 2𝑦 ′ − 3𝑦 = 3𝑒 2𝑡 Answer: First find solution of homogeneous problems. 𝑟 2 − 2𝑟 − 3 = 0 → 𝑟1,2 = −1,3 → using quadratic formula Homogeneous solution: 𝑦𝑐 = 𝑐1 𝑒 −𝑡 + 𝑐2 𝑒 3𝑡 Let 𝑌 = 𝐴𝑒 2𝑡 because 𝑔(𝑡) = 3𝑒 2𝑡 Plug Y into starting equation to find particular solution: (𝐴𝑒 2𝑡 )′′ − 2(𝐴𝑒 2𝑡 )′ − 3𝐴𝑒 2𝑡 = 3𝑒 2𝑡 4𝐴𝑒 2𝑡 − 4𝐴𝑒 2𝑡 − 3𝐴𝑒 2𝑡 = 3𝑒 2𝑡 −3𝐴𝑒 2𝑡 = 3𝑒 2𝑡 A=-1 𝑌 = −𝑒 2𝑡 Solution: 𝑦 = 𝑦𝑐 = 𝑐1 𝑒 −𝑡 + 𝑐2 𝑒 3𝑡 − 𝑒 2𝑡 Result: 𝑦 = 𝑦𝑐 = 𝑐1 𝑒 −𝑡 + 𝑐2 𝑒 3𝑡 − 𝑒 2𝑡 Question #2 Use the method of variation of parameters to find a particular solution of the given differential equation. 𝑦 ′′ + 𝑦 = csc 2 𝑥 Answer: Consider the differential equation 𝑦 ′′ + 𝑦 = csc 2 𝑥 The objective is to find a particular solution 𝑦𝑝 of the given equation. Homogeneous equation associated with the equation above is y''+y=0 Auxiliary equation associated with homogeneous differential equation is 𝑟2 + 1 = 0 𝑟 2 = −1 𝑟 = ±√−1 𝑟 = ±𝑖 Hence the roots of the auxiliary equation are 𝑟1 = 𝑖, 𝑟2 = −𝑖 Therefore, the complementary function is 𝑦𝑐 (𝑥) = 𝑐1 cos 𝑥 + 𝑐2 sin 𝑥 We have 𝑦1 (𝑥) = cos 𝑥 𝑦2 (𝑥) = sin 𝑥 𝑦1′ (𝑥) = − sin 𝑥 𝑦2′ (𝑥) = cos 𝑥 Hence, 𝑊 = 𝑊(𝑦1 , 𝑦2 )  cos x sin x  W =   − sin x cos x  𝑊 = cos ^2𝑥 + sin ^2𝑥 W=1 Here, 𝑓(𝑥) = csc 2 𝑥 The particular solution is given by 𝑦𝑝 (𝑥) = (− ∫ 𝑦2 (𝑥)𝑓(𝑥)𝑑𝑥 𝑊(𝑥) ) 𝑦1 (𝑥) + (∫ 𝑦1 (𝑥)𝑓(𝑥)𝑑𝑥 𝑊(𝑥) ) 𝑦2 (𝑥) 2 𝑦𝑝 (𝑥) = (− ∫ sin 𝑥 csc 𝑥 𝑑𝑥) cos 𝑥 + (∫ cos 𝑥 csc 2 𝑥 𝑑𝑥) sin 𝑥 𝑦𝑝 (𝑥) = (− ∫ sin 𝑥 sin2 𝑥 𝑑𝑥) cos 𝑥 + (∫ cos 𝑥 sin2 𝑥 𝑑𝑥) sin 𝑥 𝑦𝑝 (𝑥) = (− ∫ csc 𝑥 𝑑𝑥) cos 𝑥 + (∫ cot 𝑥 csc 𝑥 𝑑𝑥) sin 𝑥 𝑦𝑝 (𝑥) = −(cos 𝑥) ln|csc 𝑥 − cot 𝑥| − sin 𝑥 sin 𝑥 𝑦𝑝 (𝑥) = −1 − (cos 𝑥) ln|csc 𝑥 − cot 𝑥| Hence, the particular solution of the differential equation is given by 𝑦𝑝 (𝑥) = −1 − (cos 𝑥) ln|csc 𝑥 − cot 𝑥| Result: 𝑦𝑝 (𝑥) = −1 − (cos 𝑥) ln|csc 𝑥 − cot 𝑥| Question #3 Determine the longest interval in which the given initial value problem is certain to have a unique twice differentiable solution. Do not attempt to find the solution. (𝑥 + 3)𝑦 ′′ + 𝑥𝑦 ′ + (ln|𝑥|)𝑦 = 0, 𝑦(1) = 0, 𝑦 ′ (1) = 1 Answer: We have the differential equation (𝑥 + 3)𝑦 ′′ + 𝑥𝑦 ′ + (ln|𝑥|)𝑦 = 0 which is equivalent to 𝑦 ′′ + 𝑥 𝑥+3 𝑦′ + ln|𝑥| 𝑥+3 𝑦=0 as the form 𝑦 ′′ + 𝑝(𝑥)𝑦 ′ + 𝑞(𝑥)𝑦 = 𝑔(𝑥) which the initial values 𝑦(1) = 0, 𝑦 ′ (1) = 1 Using the notation of Theorem, we find that 𝑥 1, for 𝑝(𝑥) = , it is continuous on (−∞, −3) and (−3, ∞) 2. for 𝑞(𝑥) = 𝑥+3 ln|𝑥| 𝑥+3 , it is continuous on (−∞, −3), (−3,0) and (0, ∞). 3. for g(x)=0 , it is continuous on (−∞, ∞) Since we have 𝑡0 = 1, then the longest interval of the solution is (0, ∞) Question #4 Determine the longest interval in which the given initial value problem is certain to have a unique twicedifferentiable solution. Do not attempt to find the solution. (𝑥 − 2)𝑦 ′′ + 𝑦 ′ + (𝑥 − 2)(tan 𝑥)𝑦 = 0 y(3)=1 y'(3)=6 Answer: The given IVP (𝑥 − 2)𝑦 ′′ + 𝑦 ′ + (𝑥 − 2)(tan 𝑥)𝑦 = 0 1 ⇒ 𝑦 ′′ + 𝑦 ′ + (tan 𝑥)𝑦 = 0 𝑥−2 And the initial condition. 𝑦(3) = 1, 𝑦 ′ (3) = 6 Here 𝑃(𝑥) = 1 𝑥−2 𝑞(𝑥) = tan 𝑥 g(x)=0 constant function is continuous everywhere. 𝜋 𝜋 2 2 Solution: ((2𝑛 + 1) , (2𝑛 + 3) ) Question #5 Solve the second order linear equations: 𝑦 ′′ + 4𝑦 = cos 2 𝑥 Answer: We need to use method of variation of parameters to solve 𝑦 ′′ + 4𝑦 = cos(2𝑥) General form of second order differential equation is 𝑦 ′′ + 𝑃(𝑥)𝑦 ′ + 𝑄(𝑥)𝑦 = 𝑅(𝑥) Here, P(x)=0 Q(x)=4 𝑅(𝑥) = cos(2𝑥) Complementary function: (𝑦𝑐 (𝑥)) For finding 𝑦𝑐 (𝑥) we consider the homogenous equation , 𝑦 ′′ + 4𝑦 = 0 𝑑 This is of the form , (𝐷 2 + 4)𝑦 = 0, 𝐷 = 𝑑𝑥 Now, let D=m, and finding the eigen values, we get, 𝑚2 + 4 = 0 ⇒ 𝑚2 + 4 = 0 ⇒ 𝑚2 = −4 ⇒ 𝑚 = ±√−4 = ±2𝑖 We get the eigen values ±2𝑖 If the eigen values are of the form 𝑎 ± 𝑖𝑏 then the complementary function is of the form, 𝑦𝑐 (𝑥) = 𝑒 𝑎𝑥 (𝐶1 cos(𝑏𝑥) + 𝐶2 sin(𝑏𝑥)) Here, the eigen values are ±2𝑖 , therefore, 𝑦𝑐 = 𝑒 0𝑥 (𝐶1 cos(2𝑥) + 𝐶2 sin(2𝑥)) = 𝐶1 cos(2𝑥) + 𝐶2 sin(2𝑥) Hence, the complementary function is 𝑦𝑐 (𝑥) = 𝐶1 cos(2𝑥) + 𝐶2 sin(2𝑥) Now, we find the particular integral. For that we consider the complementary function 𝑦𝑐 (𝑥) = 𝐶1 𝑢(𝑥) + 𝐶2 𝑣(𝑥) Since, the complementary function is 𝑦𝑐 (𝑥) = 𝐶1 cos(2𝑥) + 𝐶2 sin(2𝑥) therefore, 𝑢(𝑥) = cos(2𝑥) , 𝑣(𝑥) = sin(2𝑥) And get the particular integral as 𝑦𝑝 (𝑥) = 𝐴𝑢(𝑥) + 𝐵𝑣(𝑥) Where 𝐴 = − ∫ 𝑣(𝑥)𝑅(𝑥) 𝑊(𝑥) 𝑑𝑥 and 𝐵 = ∫ 𝑢(𝑥)𝑅(𝑥) 𝑊(𝑥) 𝑑𝑥 𝑅(𝑥) = cos(2𝑥) from (1) W is the wroian is u and v, that is, W ( x) = u ( x) v( x) u( x) v( x) W = cos(2 x) sin(2 x) −2sin(2 x) 2 cos(2 x) = 2 cos 2 (2𝑥) + 2 sin2 (2𝑥) = 2(cos 2(2𝑥) + sin2 (2𝑥)) = 2 (Since, cos 2 𝑥 + sin2 𝑥 = 1 ) Now, we evaluate A and B. 𝐴 = −∫ = −∫ = = = −1 4 −1 𝑣(𝑥)𝑅(𝑥) 𝑑𝑥 𝑊(𝑥) sin(2𝑥) cos(2𝑥) 2 𝑑𝑥 ∫ 2 sin(2𝑥) cos(2𝑥) 𝑑𝑥 ∫ sin(4𝑥) 𝑑𝑥 4 1 cos(4𝑥) 4 4 = cos(4𝑥) 16 Now we find B, 𝐵=∫ 𝑢(𝑥)𝑅(𝑥) 𝑊(𝑥) 𝑑𝑥 =∫ cos(2𝑥) cos(2𝑥) 1 = ∫ cos 2 1 = ∫ 2 2 2 (2𝑥) 1+cos(4𝑥) 2 1 𝑑𝑥 𝑑𝑥 𝑑𝑥 using ( cos 2 𝑥 = 1+cos 2𝑥 2 ) = ∫(1 + cos(4𝑥))𝑑𝑥 4 1 sin(4𝑥) 4 1 4 sin(4𝑥) 4 4 = (𝑥 + = (𝑥 + ) ) Now, we have A and B as, 𝐴= cos(4𝑥) 16 1 sin(4𝑥) 4 4 , 𝐵 = (𝑥 + ) Now, we put these values in particular integral 𝑦𝑝 (𝑥) = 𝐴𝑢(𝑥) + 𝐵𝑣(𝑥) ⇒ 𝑦𝑝 (𝑥) = 𝐴𝑢(𝑥) + 𝐵𝑣(𝑥) = cos(4𝑥) 16 1 cos(2𝑥) + (𝑥 + sin(4𝑥) 𝑥 4 cos(4𝑥) cos(2𝑥) 4 𝑥 1 4 16 = sin(2𝑥) + = sin(2𝑥) + 16 ) sin(2𝑥) 4 sin(4𝑥) sin(2𝑥) + 16 (cos(4𝑥) cos(2𝑥) + sin(4𝑥) sin(2𝑥)) We have the identity , cos(𝑥) cos(𝑦) + sin(𝑥) sin(𝑦) = cos(𝑥 − 𝑦) 𝑥 1 ⇒ 𝑦𝑝 (𝑥) = sin(2𝑥) + (cos(4𝑥) cos(2𝑥) + sin(4𝑥) sin(2𝑥)) 4 𝑥 = sin(2𝑥) + 1 4 𝑥 16 1 4 16 = sin(2𝑥) + 16 cos(4𝑥 − 2𝑥) cos(2𝑥) Now, we have, the solution 𝑦(𝑥) = 𝑦𝑐 (𝑥) + 𝑦𝑝 (𝑥) 𝑥 1 4 16 ⇒ 𝑦(𝑥) = 𝐶1 cos(2𝑥) + 𝐶2 sin(2𝑥) + sin(2𝑥) + = (𝐶1 + 1 16 cos(2𝑥) 𝑥 ) cos(2𝑥) + 𝐶2 sin(2𝑥) + sin(2𝑥) 𝑥 4 = 𝐶3 cos(2𝑥) + 𝐶2 sin(2𝑥) + sin(2𝑥) 4 Question #6 How can the following second-order linear equation be converted into a first-order linear equation? This is our second-order equation: 𝑦 ′′ − 2𝑦 ′ + 2𝑦 = 𝑒 2𝑡 𝑠𝑖𝑛 Answer: Note that if 𝛼 and 𝛽 are the roots of 𝑥 2 + 𝑎𝑥 + 𝑏 = 0, we have 𝛼 = −(𝛼 + 𝛽) and 𝑏 = 𝛼𝛽. The equation 𝑦 ′′ + 𝑎𝑦 ′ + 𝑏𝑦 = 𝑓(𝑡) = 𝑦 ′′ − (𝛼 + 𝛽)𝑦 ′ + 𝛼𝛽𝑦 We now rewrite the left-hand side as 𝑦 ′′ − 𝛼𝑦 ′ − 𝛽𝑦 ′ + 𝛼𝛽𝑦 = (𝑦 ′′ − 𝛼𝑦 ′ ) − 𝛽(𝑦 ′ − 𝛼𝑦) = 𝑓(𝑡) and substitute 𝑧 = 𝑦 ′ − 𝛼𝑦 so that 𝑧 ′ = 𝑦 ′′ − 𝛼𝑦 ′ and finally the transformed equation becomes 𝑧 ′ − 𝛽𝑧 = 𝑓(𝑡) Once z=g(t) is known, y is obtained from 𝑦 ′ − 𝛼𝑦 = 𝑔(𝑡) So one second order equation can be solved by solving two first order equations. Question #7 I'm trying to find particular solution of the second-order linear equation but I can't find 𝑦1 and 𝑦2 according to 𝑦 = 𝑐1 𝑦1 + 𝑐2 𝑦2 𝑥 2 𝑦 ′′ − 2𝑥𝑦 ′ + 2𝑦 = 0, 𝑦(1) = 3, 𝑦 ′ (1) = 1 If r is used, 𝑥 2 𝑟 2 − 2𝑥𝑟 = −2 then 𝑥𝑟(𝑥𝑟 − 2) = −2, I can't go on from there to find 𝑦1 and 𝑦2 Answer: Assume 𝑦 = 𝑥 𝑟 then if 𝑦 = 𝑥 𝑟 then 𝑦 ′ = 𝑟𝑥 𝑟−1 And 𝑦 ′′ = 𝑟(𝑟 − 1)𝑥 𝑟−2 Substitute in 𝑥 2 𝑦 ′′ − 2𝑥𝑦 ′ + 2𝑦 = 0 to get 𝑟(𝑟 − 1)𝑥 𝑟 − 2𝑟𝑥 𝑟 + 2𝑥 𝑟 = 0 divide by y to get 𝑟(𝑟 − 1) − 2𝑟 + 2 = (𝑟 − 1)(𝑟 − 2) = 0 therefore 𝑦 = 𝑎𝑥 + 𝑏𝑥 2 Solve 𝑦(1) = 𝑎 + 𝑏 = 3 𝑦 ′ (1) = 𝑎 + 2𝑏 = 1 to get 𝑦 = 5𝑥 − 2𝑥 2 Question #8 I need to find a second order linear homogeneous equation with constant coefficients that has the given function as a solution Queston a) 𝑥𝑒 −3𝑥 Question b) 𝑒 3𝑥 sin 𝑥 Answer: Question A Let the answer to question a be of the form 𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐𝑦 = 0 Then we want 𝑎(−3)2 − 3𝑏 + 𝑐 = 9𝑎 − 3𝑏 + 𝑐 = 0 (1) And more over, since it is a repeated root we need − 𝑏 2𝑎 = −3 Substituting b=6a into (1) we get 9𝑎 − 18𝑎 + 𝑐 = 0 ⇒= 9𝑎 Thus if we let a be our free variable, all solutions to question a are of the form ay''+6ay'+9ay=0 Question B For question b, you need the characteristic polynomial to have the roots 3 ± 𝑖 . Thus the characteristic polynomial must be of the form 𝑎(𝑥 − [3 + 𝑖])(𝑥 − [3 − 𝑖]) = 𝑎𝑥 2 − 6𝑎𝑥 + 10𝑎 Thus solutions to question b are of the form 𝑎𝑦 ′′ − 6𝑎𝑦 ′ + 10𝑎𝑦 = 0 Note For your solution to question a you got 𝑦 ′′ − 6𝑦 ′ + 9𝑦 = 0 but going back and checking you should get +6y' not -6y'. Otherwise your solutions are of these forms with a=1 but a can be any real number. It is our free variable. Question #9 Is there any known method to solve such second order non-linear differential equation? 1 𝑦𝑛′′ − 𝑛𝑥 =0 √𝑦𝑛 Answer:  u =  Let   v =  Then 𝑑𝑣 𝑑𝑢 = 𝑑𝑣 𝑑𝑥 𝑑𝑢 𝑑𝑥 x3 3 yn2 x dyn yn dx 𝑥 𝑑 2 𝑦𝑛 1 𝑑𝑦𝑛 𝑥 𝑑𝑦𝑛 2 + − ( ) 𝑦𝑛 𝑑𝑥 2 𝑦𝑛 𝑑𝑥 𝑦𝑛2 𝑑𝑥 = 3𝑥 2 3𝑥 3 𝑑𝑦𝑛 3 − 5 𝑑𝑥 𝑦𝑛2 2𝑦𝑛2 = = 𝑥 𝑑2 𝑦𝑛 𝑣 𝑣2 + − 𝑦𝑛 𝑑𝑥2 𝑥 𝑥 3𝑢 3𝑢𝑣 − 𝑥 2𝑥 𝑥2 𝑑2 𝑦𝑛 +𝑣−𝑣 2 2 𝑦𝑛 𝑑𝑥 𝑣 3𝑢(1− ) 2 𝑣 𝑑𝑣 𝑥 2 𝑑 2 𝑦𝑛 3𝑢 (1 − ) = + 𝑣 − 𝑣2 2 𝑑𝑢 𝑦𝑛 𝑑𝑥 2 𝑥 2 𝑑 2 𝑦𝑛 𝑣 𝑑𝑢 = 3𝑢 (1 − ) + 𝑣2 − 𝑣 𝑦𝑛 𝑑𝑥 2 2 𝑑𝑣 𝑑 2 𝑦𝑛 𝑦𝑛 𝑣 𝑑𝑣 = 2 (3𝑢 (1 − ) + 𝑣 2 − 𝑣) 2 𝑑𝑥 𝑥 2 𝑑𝑢 𝑦𝑛 𝑣 𝑑𝑣 𝑛𝑥 (3𝑢 (1 − ) + 𝑣 2 − 𝑣) − =0 2 𝑥 2 𝑑𝑢 √𝑦𝑛 𝑦𝑛 𝑣 𝑑𝑣 𝑛𝑥 (3𝑢 (1 − ) + 𝑣 2 − 𝑣) = 𝑥2 2 𝑑𝑢 √𝑦𝑛 𝑣 𝑑𝑣 3𝑢 (1 − ) + 𝑣2 − 𝑣 = 2 𝑑𝑢 𝑛𝑥 3 3 2 𝑦𝑛 𝑣 𝑑𝑣 3𝑢 (1 − ) + 𝑣 2 − 𝑣 = 𝑛𝑢 2 𝑑𝑢 𝑣 𝑑𝑣 3𝑢 ( − 1) = 𝑣 2 − 𝑣 − 𝑛𝑢 2 Let 𝑤 = 𝑤 𝑑𝑢 2 𝑣 −1 Let 𝑤 = − 1 2 Then 𝑣 = 2𝑤 + 2 𝑑𝑣 𝑑𝑢 =2 6𝑢𝑤 𝑤 𝑑𝑤 𝑑𝑢 𝑑𝑤 𝑑𝑢 𝑑𝑤 = 4𝑤 2 + 6𝑤 + 2 − 𝑛𝑢 𝑑𝑢 2𝑤2 = 𝑤 + 3𝑢 𝑢 + 2−𝑛𝑢 6𝑢 In fact, all Abel equation of the second kind can be transformed into Abel equation of the first kind. 1 Let 𝑤 = Then − 𝑑𝑧 𝑑𝑢 𝑑𝑧 𝑑𝑢 1 𝑑𝑧 𝑧 3 𝑑𝑢 = = 𝑑𝑤 𝑑𝑢 = 2 𝑧 =− 2 3𝑢𝑧 2 1 + 3𝑢𝑧 2 𝑢𝑧 (𝑛𝑢−2)𝑧 3 6𝑢 1 𝑑𝑧 𝑧 2 𝑑𝑢 1 + + − + 2−𝑛𝑢 𝑢𝑧 6𝑢 2−𝑛𝑢 𝑧2 𝑢 6𝑢 − 2𝑧 3𝑢 Question #10 I have the following differential equation: 𝑦 ′′ + 𝑦 = cos(𝑡) cos(2𝑡) Maybe something can be done to cos(𝑡) cos(2𝑡) to make it easier to solve. Any ideas? Answer: I don't see an elegant solution , but it can be done by brute force using the method of variation of parameters. The homogeneous differential equation 𝑦 ′′ (𝑡) + 𝑦(𝑡) = 0 has the two fundamental solutions 𝑦1 (𝑡) = sin(𝑡) and 𝑦2 (𝑡) = cos(𝑡) We must solve y 0 ( )=( 1 y1 cos(t ) cos(2t ) y2 u1 sin t cos t u1 )( ) = ( )( ) y2 u2 − cos t sin t u2 for 𝑢1 and 𝑢2 and find 𝑈1 and 𝑈2 such that 𝑈1′ = 𝑢1 and 𝑈2′ = 𝑢2 A particular solution will then be given by 𝑦0 (𝑡) = 𝑈1 𝑦1 + 𝑈2 𝑦2 and 𝑦 = 𝑦0 + 𝑐1 𝑦1 + 𝑐2 𝑦2 will be the general solution. sin t cos t ) is orthogonal, its inverse is its transpose, so − cos t sin t u sin t − cos t 0 cos2 (t ) cos(2t ) ( 1) = ( )( )=( ) u2 cos t sin t cos(t ) cos(2t ) sin(t ) cos(t ) cos(2t ) Since the matrix A(t ) = ( 1 1 2 4 Note that 𝑣2 (𝑡) = sin(𝑡) cos(𝑡) cos(2𝑡) = sin(2𝑡) cos(2𝑡) = sin(4𝑡), so we can take 𝑈2 (𝑡) = − 1 16 cos(4𝑡) On other hand, we get 1 𝑢1 = cos 2(𝑡) cos(2𝑡) = cos(𝑡) (cos(3𝑡) + cos(𝑡)) 1 1 2 1 1 2 4 2 2 1 1 4 4 = cos(𝑡) cos(3𝑡) + cos(𝑡) cos(𝑡) = (cos(4𝑡) + cos(2𝑡)) + (cos(2𝑡) + 1) = cos(4𝑡) + cos(2𝑡) + 1 4 Integrating this gives 1 1 1 𝑈1 (𝑡) = sin(4𝑡) + sin(2𝑡) + 𝑡 16 4 4 Now we can simplify the particular solution 1 1 4 16 𝑦0 = 𝑈1 𝑦1 + 𝑈2 𝑦2 = ⋯ = 𝑡 sin(𝑡) − cos(3𝑡) + 5 16 cos(𝑡) using some further trig identities 1 You should at least check yourself that 𝑦0′′ + 𝑦0 = (cos(𝑡) + cos(3𝑡)) = cos(𝑡) cos(2𝑡) 2 Therefore the general solution is 1 𝑦(𝑡) = 𝑦0 + 𝑐1 𝑦1 + 𝑐2 𝑦2 = 𝑡 sin(𝑡) − 4 for some real constants 𝑐1 and 𝑐2 . 1 16 cos(3𝑡) + 5 16 cos(𝑡) + 𝑐1 sin(𝑡) + 𝑐2 cos(𝑡)