Questions and Answers #3 First Order Differential Equations

Questions and Answers Sheet 3 First Order Differential Equations Question #1 Let y be a function of x. Which of the following is a first order linear differential equation? 1. 𝑦 ′ − 𝑥 = 𝑦 2 sin(𝑥) 2. 𝑦 ′ − 𝑥 = 𝑦 sin(𝑥) 3.(𝑦 ′ )2 + 𝑦 = tan(2𝑥) 4. 𝑦 ′ = 𝑦 2 𝑒 𝑥 5. (𝑦 ′ )2 + 𝑦 = 𝑒 𝑥 Answer: The possible differential equations for the first order linear differential equation are: 1. 𝑦 ′ − 𝑥 = 𝑦 2 sin(𝑥) 2. 𝑦 ′ − 𝑥 = 𝑦 sin(𝑥) 3. (𝑦 ′ )2 + 𝑦 = tan(2𝑥) 4. 𝑦 ′ = 𝑦 2 𝑒 𝑥 5. (𝑦 ′ )2 + 𝑦 = 𝑒 𝑥 The general form of the first order linear differential equation is 𝑦 ′ + 𝑞(𝑥)𝑦 = 𝑝(𝑥) … ..(1) The differential equations given in options (a), (c), (d) and (c) are not of the form as the differential equation given in equation (1). So, the first order linear differential equation is given by 𝑦 ′ − 𝑥 = 𝑦 sin(𝑥) Hence, option (b) is correct. Question #2 Make use of vectors to re-write the following second order differential equation into a 1st order differential equation. (note: I do not need to solve it). 2.1𝑥 −2 𝑑2𝑦 𝑑𝑥 2 − 3𝑒 2𝑥 𝑦 4 where 𝑦 = 4.2 and 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = −6𝑦 = −3.1 when 𝑥 = 0.5 . Im not sure were to begin any help is greatly appreciated. Answer: The way to rewrite one second-order equation into two first-order ones is to establish a second function, and tie it in with what's already there. I am sure there are problems with incredibly clever functions to choose that just makes the problem immediately solvable, but the standard choice is 𝑧 = 𝑑𝑦 𝑑𝑥 . That is one equation, and it allows you to rewrite the original equation to get rid of all those second derivatives: 2.1𝑥 −2 2.1𝑥 −2 𝑑2𝑦 𝑑𝑥 2 𝑑𝑧 𝑑𝑥 − 3𝑒 2𝑥 𝑦 4 𝑑𝑦 𝑑𝑥 = −6𝑦 − 3𝑒 2𝑥 𝑦 4 𝑧 = −6𝑦 which gives the two differential equations ( 𝑑𝑧 𝑑𝑥 3 = 2.1 𝑥 2 𝑒 2𝑥 𝑦 4 − 6 2.1 𝑥 2 𝑦) , ( 𝑑𝑦 𝑑𝑥 = 𝑧) : also known as the first order vector differential equation 𝑑 𝑑𝑥 (𝑧, 𝑦) = ( 3 2.1 𝑥 2 𝑒 2𝑥 𝑦 4 − 6 2.1 𝑥 2 𝑦, 𝑧) with initial value (−3.1,4.2) for 𝑥 = 0.5 . Question #3 Maybe you know how to do it? Write an equivalent first-order differential equation and initial condition for y 𝑥 1 𝑦 = ∫ 𝑑𝑡 1 𝑡 What is the equivalent first-order differential equation? y = Answer: 𝑥 𝑦=∫ 1 ⇒ 𝑑𝑦 𝑑𝑥 1 𝑑𝑡 𝑡 = 1 𝑥 For initial condition: 𝑥1 𝑥 𝑦 = ∫1 𝑑𝑥 ⇒ 𝑦 = ln 𝑡 ∫1 = ln 𝑥 − |𝑛| 𝑡 ⇒ 𝑦 = ln 𝑥 [∵ |𝑛| = 0] Here when 𝑥 = 1, 𝑦 = 0 So, initial condition is, 𝑦(1) = 0 So, DE is, y =| 1 |; y (1) = 0 x Question #4 Most systems can be modeled using first order system just like a temperature control system of an incubator. Question 1: How can we know that a certain system can be modeled using first order differential equation? 𝑎1 𝑑𝑦 𝑑𝑡 + 𝑎0 𝑦 = 𝐹(𝑡) (1) A first order differential equation in Equation (1) may be written as: 𝜏 𝑑𝑦 𝑑𝑡 + 𝑦 = 𝑘𝐹(𝑡) (2) Question 2: If a step input if t  0 if t  0 0 F (t ) =  A ( 3) of the system has an output response of a −𝑡 𝑦(𝑡) = 𝑘𝐴 (1 − 𝑒 𝜏 ) (4) then the system can be modeled as first differential equation where 𝑦𝑚𝑎𝑥 = 𝑘𝐴 (5) Is this correct? Given a step input and the graph of the step response of the system. Am i correct if I say that if the output of the system behaves similar to equation (4), then I can model the system using first differential equation? Answer: −1 Your output response 𝑦(𝑡) = 𝑘𝐴 (1 − 𝑒 𝜏 ) for 𝑡 > 0 satisfies the first-order differential equation 𝑦 𝑘 𝜏 𝜏 𝑦 ′ = − + ( ) 𝐹(𝑡). Question #5 Need your help! 1. Identify the substitution method used to solve the given differential equations. Support your answer with the appropriate calculations. 2. Solve for the solution to the given differential equation that satisfies the given condition 𝑦 ′ = 2(3𝑥 + 𝑦)2 − 1; 𝑦(0) = 1 Answer: Given DE is 1. 𝑦 ′ = 2(3𝑥 + 𝑦)2 − 1 𝑦(0) = 1 Let 3𝑥 + 𝑦 = 𝑧 Differential ''x'', we get 3 + 𝑦′ = ⇒ 𝑦′ = 𝑑𝑧 𝑑𝑥 𝑑𝑧 𝑑𝑥 −3 𝑑𝑧 − 3 = 2𝑧 2 − 1 𝑑𝑥 We use separation of variable method to solve this DE on ⇒ on 𝑑𝑧 𝑑𝑥 𝑑𝑧 𝑑𝑥 = 2𝑧 2 + 2 = 2(𝑧 2 + 1) 𝑑𝑧 1+𝑧 2 = 2𝑑𝑥 Integrating we get 𝑑𝑧 ∫ 1+𝑧 2 = 2 ∫ 𝑑𝑥 ⇒ tan−1 (𝑧) = 2𝑥 + 𝑐 ⇒ tan−1 (3𝑥 + 𝑦) = 2𝑥 + 𝑐 (∵ 𝑧 = 3𝑥 + 𝑦) Question #6 (𝑦)2 + 8𝑦 ′ = 4𝑡 is: a. Second order nonlinear differential equation. b. First order linear differential equation. c. Second order linear differential equation. d. First order nonlinear differential equation. Answer: Order of a differential equation is the order of the highest derivative. Linearity: In a differential equation,when the variables and its derivatives are only multiplied by constants,then the differential equation is linear.other wise it is non-linear That is, the variables and their derivatives are always be in simple first power. Given, (𝑦)2 + 8𝑦 ′ = 4𝑡 In this equation,the order of the highest derivative is one. So, it is a first order differential equation. Here the variable y has the power 2. So, the differential equation is nonlinear. Thus, Option (d) is the correct answer. Question #7 𝜕𝑀 𝜕𝑦 ≠ 𝜕𝑁 𝜕𝑥 (𝑥 − 𝑦 3 + 𝑦 2 sin(𝑥))𝑑𝑥 = (3𝑥𝑦 2 + 2𝑦 cos(𝑥))𝑑𝑦 However, neither of the two integrating factors give functions of a variable alone. So what is left to do to solve this? Thank you so much! Answer: Hint: this is most definitely an exact differential equation. The problem is that you have not written it in the form 𝑀(𝑥, 𝑦)𝑑𝑥 + 𝑁(𝑥, 𝑦)𝑑𝑦 = 0 , hence why you are failing to see that. Question #8 dot 𝑥 = 𝑟𝑥 − ℎ𝑥 2 + 𝑞 , where dot 𝑥 = I first tried 𝜏 = 𝑡 𝑡0 and 𝐴 = 𝑥 𝑥0 𝑑𝑥 𝑑𝑡 . I'm assuming r,h,q have different units. . Then I got 𝑑𝐴𝑥0 𝑑𝜏𝑡0 from there. Need any help. Answer: Can use physical dimensions to units: 𝑥 → 𝑚 meter linear 𝑟𝑥 → 𝜔𝑥; 𝑟 → frequency or angular velocity 𝑞 → 𝑣 velocity ℎ= 𝐿 𝑇𝐿2 → 𝜔 𝑚 . Question #9 = 𝑟𝐴𝑥0 − ℎ𝐴2 𝑥02 + 𝑞 . But I am not really sure where to go 𝑚𝑥 ′′ (𝑡) = −𝑘𝑥(𝑡) − 𝑐𝑥 ′ (𝑡) I was wondering if this is the right way of thinking of it.. 1. 𝑐𝑥 ′ (𝑡) = −𝑘𝑥(𝑡) − 𝑚𝑥 ′′ (𝑡) 2.integral of 𝑘𝑥(𝑡) = −𝑐𝑥 ′ (𝑡) − 𝑚𝑥 ′′ (𝑡) 3.derivative of 𝑚𝑥 ′′ (𝑡) = −𝑘𝑥(𝑡) − 𝑐𝑥 ′ (𝑡) If so, how would I solve 2 3 ? Answer: ′ The idea is to use 𝑥 ′′ (𝑡) = (𝑥 ′ (𝑡)) . So instead of having one equation of second order, you introduce a new function, say 𝑦(𝑡) = 𝑥 ′ (𝑡) , and you have two equations of first order: 𝑥 ′ (𝑡) = 𝑦(𝑡) 𝑘 𝑐 𝑚 𝑚 𝑦 ′ (𝑡) = − 𝑥(𝑡) − 𝑦(𝑡) Question #10 ′ 𝑦′ + 𝑒 𝑦 − 𝑥 = 0 that I have simplified like so ′ 𝑒 𝑦 = 𝑥 − 𝑦′ ′ ln 𝑒 𝑦 = ln(𝑥 − 𝑦 ′ ) 𝑦 ′ = ln(𝑥 − 𝑦 ′ ) but I do not know how to solve this further to obtain the general solution. I have done first order linear differential equation strategies so far. How should I get about doing this question with the strategies I have? Answer: Your (revised) simplification doesn't really simplify because it separates the y' s from each other. The productive thing to do would be to rewrite it to 𝑥 = 𝑦′ + 𝑒 𝑦 ′ and work from there. First, let 𝑔(𝑢) = 𝑢 + 𝑒 𝑢 ; then the equation becomes 𝑥 = 𝑔(𝑦 ′ ) or 𝑦 ′ = 𝑔−1 (𝑥) . In this form we get the solution 𝑥 𝑦 = ∫1 𝑔−1 (𝑡)𝑑𝑡 + 𝐶 This integral is not nice, but (noting that 𝑔−1 (1) = 0 , which is why I chose 1 as the lower limit) we can flip it around to 𝑔−1 (𝑥) 𝑦 = 𝑥𝑔−1 (𝑥) − ∫0 𝑔(𝑢)𝑑𝑢 + 𝐶 and g is easy enough to integrate so we get 𝑦 = 𝑥𝑔−1 (𝑥) − 𝑔−1 (𝑥)2 2 − 𝑒𝑔 −1 (𝑥) + 𝐶1 (with 𝐶1 = 𝐶 = 1 ). With a bit of help from Wolfram Alpha we find 𝑔−1 (𝑥) = 𝑥 − 𝑊(𝑒 𝑥 ) where W is the Lambert W function. We can plug this in to get 𝑥2 𝑦= 2 − 𝑊(𝑒 𝑥 )2 2 + 𝑒 𝑥−𝑊(𝑒 𝑥) + 𝐶2 By the definition of W we have 𝑒 −𝑊(𝑒 𝑦= 𝑥 2 −(𝑊(𝑒 𝑥 )−1)2 2 𝑥) = 𝑊(𝑒 𝑥 ) 𝑒𝑥 , so this further simplifies to + 𝐶3 If still not exactly pretty, this is at least a "closed form" modulo the W function. Question #11 Quickly! Need help Solve differential equation, subject to the given initial condition. 𝑥 𝑑𝑦 𝑑𝑥 + (1 + 𝑥)𝑦 = 3; 𝑦(4) = 50 Answer: This is a first order differential equation. This can be solved using the integrating factor method. For a first order differential equation 𝑑𝑦 𝑑𝑥 + 𝑃(𝑥)𝑦 = 𝑄(𝑥) , the integrating factor is 𝑒 ∫ 𝑃(𝑥)𝑑𝑥 . The equation is multiplied by this integrating factor and integrated. First divide the equation by x to write in the form 𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 + 𝑃(𝑥)𝑦 = 𝑄(𝑥) . + (1 + 𝑥)𝑦 = 3 1 3 𝑥 𝑥 + ( + 1) 𝑦 = Calculate the integrating factor for this differential equation. 1 𝑒 ∫(1+𝑥 )𝑑𝑥 = 𝑒 𝑥+ln(𝑥) = 𝑒 𝑥 𝑒 ln(𝑥) = 𝑥𝑒 𝑥 Multiply the differential equation by the integrating factor, simplify and then integrate. ( 𝑑𝑦 𝑑𝑥 1 3 𝑥 𝑥 + ( + 1) 𝑦) 𝑥𝑒 𝑥 = 𝑥𝑒 𝑥 (𝑦𝑥𝑒 𝑥 )′ = 3𝑒 𝑥 𝑦𝑥𝑒 𝑥 = 3𝑒 𝑥 + 𝑐 3 𝑐𝑒 −𝑥 𝑥 𝑥 𝑦= + Substitute the condition 𝑦(4) = 50 to calculate c. 3 𝑐𝑒 −4 4 4 𝑦(4) = + 3 𝑐𝑒 −4 4 4 50 = + 200 = 3 + 𝑐𝑒 −4 197 = 𝑐𝑒 −4 𝑐 = 197𝑒 4 3 197𝑒 4 𝑒 −𝑥 𝑥 𝑥 𝑦= + 3 197𝑒 4−𝑥 𝑥 𝑥 𝑦= + 3 197𝑒 4−𝑥 𝑥 𝑥 Hence, solution to given differential equation with given condition is 𝑦 = + Question #12 Solve and find the soltion to the first order differential equation: 𝑥 𝑑𝑦 𝑑𝑥 + 𝑦 = 𝑒𝑥, 𝑥 > 0 Answer: Given, 𝑥 𝑑𝑦 𝑑𝑥 ⇒ + 𝑦 = 𝑒 𝑥, 𝑥 > 0 𝑑 𝑑 𝑑 (𝑥𝑦) = 𝑒 𝑥 , [𝑥 > 0 ∵ (𝑥𝑦) = 𝑥 + 𝑦 × 1 because by product rule] 𝑑𝑥 𝑑𝑥 𝑑𝑥 Integrating both sides with respect to x ⇒ ∫[ 𝑑 (𝑥𝑦)] 𝑑𝑥 = ∫ 𝑒 𝑥 𝑑𝑥 𝑑𝑥 ⇒ 𝑥𝑦 = 𝑒 𝑥 + 𝑐; where c is constant 1 ⇒ 𝑦 = (𝑒 𝑥 + 𝑐); 𝑥 > 0 𝑥 ⇒𝑦= 𝑒𝑥 𝑥 𝑐 + ; 𝑥 > 0 and c-constant 𝑥 Question #13 Solve by using the Laplace Transform Method? 2 𝑦 ′ + 𝑦 = 6𝑒 𝑡 , 𝑦(0) = 3 Answer: Given first order differential equation 2 𝑦 ′ + 𝑦 = 6𝑒 𝑡 , … (𝑖) 𝑦(0) = 3 claim- to check whether the differential equation can be solved using laplace transform or not solution applying laplace both sided 2 𝐿|𝑦 ′ | + 𝐿|𝑦| = 𝐿|6𝑒 𝑡 | as we know 𝐿|𝑦 ′ | = 𝑠𝑦(𝑠) − 𝑦(0) 𝐿|𝑦| = 𝑦(𝑠) equation (i) becomes, 2 𝑠𝑦(𝑠) − 𝑦(0) + 𝑦(𝑠) = 6𝐿[𝑒 𝑡 ] 1 2 𝐿[𝑒 𝑡 ] = √𝜋𝑒 2 −𝑠2 4 𝜋 1 𝑒𝑟𝑓 ( ) − 𝑖 √𝜋𝑒 2 2 −𝑠2 4 thus, we get that the first order differential equation cannot be solved using laplace transform, because the laplace transform of right hand side is not in the form of standard function. Question #14 Solve for G.S./P.S. for the following differential equations using the method of solution for Bernoulli and Ricatti differential equations. 9𝑥𝑦 ′ = (𝑦 − 1)(𝑦 + 2); 𝑦𝑝 = −2 Answer: 9𝑥𝑦 ′ = (𝑦 − 1)(𝑦 + 2); 𝑦𝑝 = −2 9𝑥 𝑑𝑦 = (𝑦 − 1)(𝑦 + 2) 𝑑𝑥 𝑑𝑦 𝑑𝑥 = (𝑦 − 1)(𝑦 + 2) 9𝑥 𝑑𝑦 𝑑𝑥 ∫ (𝑦−1)(𝑦+2) = ∫ 9𝑥 Let 𝑦 + 2 = 𝑧, 𝑑𝑦 = 𝑑𝑧 ⇒∫ ⇒∫ 𝑑𝑧{𝑧 + !) 𝑑𝑥 ∫ 𝑧 9𝑥 1 𝑧+1 1 1 𝑧 9 𝑑𝑧 − ∫ 𝑑𝑧 = ln(𝑥) ⇒ ln(𝑧 + 1) − ln(𝑧) = 1 ln(𝑥) 9 ⇒ ln(𝑦 + 3) − ln(𝑦 + 2) = 1 ln(𝑥) 9 ⇒ ln 1 1 𝑦+3 𝑦+3 = ln(𝑥)9 ⇒ = 𝑥9 𝑦+2 𝑦+2 Question #15 Solve for G.S./P.S. for the following differential equations using the solutions by a change in variables as suggested by the equations. (𝑡 2 𝑒 𝑡 − 4𝑥 2 )𝑑𝑡 + 8𝑡𝑥𝑑𝑥 = 0 Answer: (𝑡 2 𝑒 𝑡 − 4𝑥 2 )𝑑𝑡 + 8𝑡𝑥𝑑𝑥 = 0 𝑀 = 𝑡 2 𝑒 𝑡 − 4𝑥 2 𝑁 = 8𝑡𝑥 𝑀𝑥 == 8𝑥 𝑁𝑡 = 8𝑥 𝑀𝑥 − 𝑁𝑡 −16𝑥 −2 = = 𝑁 8𝑡𝑥 𝑡 Integrating 1 𝑒 −2 ∫ 𝑡 𝑑𝑡 = 𝑒 −2 ln 𝑡 = 𝑀 = 𝑒𝑡 = 𝑁= 1 𝑡2 4𝑥 2 𝑡2 8𝑥 𝑡 𝑀𝑥 = − 𝑁1 = 8𝑥 𝑡2 −8𝑥 𝑡2 ∴ 𝑀𝑥 = 𝑁𝑡 ∫ (𝑒 𝑡 − ⇒ 𝑒𝑡 + 4𝑥 2 ) 𝑑𝑡 + ∫ 0𝑑𝑥 = 𝐶 𝑡2 4𝑥 2 =𝐶 𝑡 Question #16 Solve in two different methods. (𝑦 − 𝑥 + 3)𝑑𝑥 + (𝑥 + 𝑦 + 3)𝑑𝑦 = 0; 𝑥 = 1; 𝑦 = 4. Answer: (𝑦 − 𝑥 + 3)𝑑𝑥 + (𝑥 + 𝑦 + 3)𝑑𝑦 = 0 𝑀 = 𝑦−𝑥+3 𝑁 =𝑥+𝑦+3 𝑀𝑦 = 1 𝑁𝑥 = 1 ∴ 𝑀𝑦 = 𝑁𝑥 Its solution is ∫(𝑦 − 𝑥 + 3)𝑑𝑥 + ∫(𝑦 + 3)𝑑𝑦 = 𝐶 ⇒ 𝑥𝑦 − 𝑓𝑟𝑎𝑐𝑥 2 2 + 3𝑥 + 𝑦2 + 3𝑦 = 𝐶 2 𝑥 = 1, 𝑦 = 4 1 4 − + 3 + 8 + 12 = 𝐶 2 𝐶= 53 2 𝑥𝑦 − 𝑥2 2 + 3𝑥 + 𝑦2 2 + 3𝑦 = 53 2 Question #17 Solve the equation separable, linear, bernoulli, or homogenous 1. 𝑑𝑦 𝑑𝑥 = 𝑥 4 +4𝑥𝑦 2 2𝑥 3 +𝑥 2 𝑦+𝑦 3 2. (𝑒 −𝑦 cos(𝑥))𝑦 ′ = 𝑥 4 + 6𝑥 2 𝑦 3 3. 𝑦 ′ = 𝑦+𝑦𝑥 3 𝑥+𝑥 2 𝑥2 cos ( 2 ) 𝑦 Answer: Separable equation: A first order differential equation 𝑦’ = 𝑓(𝑥, 𝑦) is called a separable equation if the function f(x,y) can be factored into the product of two functions of x and y 𝑓(𝑥, 𝑦) = 𝑓1 (𝑥)𝑓2 (𝑦) where 𝑓1 (𝑥) and 𝑓2 (𝑦) are continuous functions. Linear differential equation: A first order differential equation is linear when it can be written as: 𝑑𝑦 𝑑𝑥 + 𝑃(𝑥)𝑦 = 𝑄(𝑥) Where P(x) and Q(x) are functions of x. Bernoulli differential equation: A first order Bernoulli differential equation can be written as: 𝑑𝑦 𝑑𝑥 + 𝑃(𝑥)𝑦 = 𝑄(𝑥) Where P(x) and Q(x) are functions of x. Bernoulli differential equation: A first order Bernoulli differential equation can be written as: 𝑑𝑦 𝑑𝑥 + 𝑃(𝑥)𝑦 = 𝑄(𝑥)𝑦 𝑛 Where P(x) and Q(x) are functions of x. Homogeneous differential equation: A first order homogeneous differential equation can be written as: 𝑑𝑦 𝑑𝑥 𝑦 = 𝐹( ) 𝑥 All three equations are written in the table : Equation Separable Linear dy x 4 + 4 xy 2 = 3 dx 2 x + x 2 y + y 3 (e − y cos( x)) y = x 4 + 6 x 2 y 3 y = y + yx y cos( 2 ) 2 x+x x 3 Bernoilli Homogeneous no no no no no no no no no no no no 2 Question #18 Solve the first order differential equation using any acceptable method. 𝑠𝑖𝑛(𝑥)𝑑𝑦/𝑑𝑥 + (𝑐𝑜𝑠(𝑥))𝑦 = 0, 𝑦((7𝑝𝑖)/6) = −2 Answer: We have given First Order Differential Equation, sin(𝑥) 𝑑𝑦/𝑑𝑥 + 𝑦 (cos(𝑥)) = 0 Now, We are using Variable Separable Method for solving the First Order Linear Differential Equation is as follows: First, we need to separate the variable with the respective derivatives: sin 𝑥 𝑑𝑦 = −𝑦 cos 𝑥 𝑑𝑥 𝑑𝑦 𝑦 =− cos 𝑥 sin 𝑥 𝑑𝑥 Now, Integrating both sides of the above equation: ∫ 𝑑𝑦 cos 𝑥 = −∫ 𝑑𝑥 𝑦 sin 𝑥 ln|𝑦| = − ln|sin 𝑥| + ln|𝐶| 𝐶 ln 𝑦 = ln ( ) sin 𝑥 𝑦 sin 𝑥 = 𝐶 Now, We are applying the given Initial Condition is as follow: 7𝜋 𝑦 ( ) = −2 6 −2 × sin ( 7𝜋 )=𝐶 6 −2 × −1 = 𝐶 7𝜋 ∴ sin ( ) = −1 6 𝐶=2 Answer: The resulting curve of the First Order Linear Differential Equation is 𝑦 sin 𝑥 = 2. Question #19 Solve the first order linear differential equation 𝑑𝑦 𝑑𝑥 + 𝑥𝑦 = 𝑥, 𝑦(0) = −6 Answer: A differential equation is an equation is x,y and derivatives of y. A linear differential equation is one in which powers of y or its derivative is equal to 0 or 1. A first order differential equation is one in which only the first derivative of y is present. A first order linear differential equation of the form 𝑑𝑦 𝑑𝑥 + 𝑃(𝑥)𝑦 = 𝑄(𝑥) can be solved using the integrating factor method. The integrating factor for which equation is 𝑒 ∫ 𝑃(𝑥)𝑑𝑥 . The equation is multiplied by the integrating factor, simplified and the integrated. Given differential equation is 𝑑𝑦 𝑑𝑥 + 𝑥𝑦 = (𝑥), 𝑦(0) = −6 . Comparing with integrating factor is 𝑒 ∫ 𝑥𝑑𝑥 which is equal to 𝑒 factor, simplify and then integrate. 𝑥2 𝑒2 ( 𝑑𝑦 𝑑𝑥 𝑥2 + 𝑥𝑦) = 𝑥𝑒 2 𝑥2 𝑥2 𝑑 (𝑦𝑒 2 ) = 𝑥𝑒 2 𝑑𝑥 𝑥2 𝑥2 𝑦𝑒 2 = ∫ 𝑥𝑒 2 𝑑𝑥 𝑥2 2 𝑑𝑦 𝑑𝑥 + 𝑃(𝑥)𝑦 = 𝑄(𝑥) the . Multiply the differential equation by this integrating 𝑥2 𝑥2 𝑦𝑒 2 = ∫ 𝑒 2 𝑑 ( 𝑥2 𝑥2 ) 2 𝑥2 𝑦𝑒 2 = 𝑒 2 + 𝐶 𝑥2 𝑦 = 1 + 𝐶𝑒 − 2 𝑥2 So the general solution is 𝑦 = 1 + 𝐶𝑒 − 2 . Use the initial condition 𝑦(0) = −6 to calculate C. 𝑥2 𝑦 = 1 + 𝐶𝑒 − 2 𝑦(0) = −6 02 −6 = 1 + 𝐶𝑒 − 2 −6 = 1 + 𝐶 𝐶 = −7 𝑦 = 1 − 7𝑒 − 𝑥2 2 Hence, the solution to the given differential equation is 𝑦 = 1 − 7𝑒 −𝑥22 . Question #20 Solve the first-order differential equation State which method you are using! 𝑥 2 𝑦 2 𝑑𝑥 − (𝑥 3 + 1)𝑑𝑦 = 0 Answer: 𝑥 2 𝑦 2 𝑑𝑥 − (𝑥 3 + 1)𝑑𝑦 = 0 (𝑥 3 + 1)𝑑𝑦 = 𝑥 2 𝑦 2 𝑑𝑥 𝑑𝑦 𝑥2 = ( ) 𝑑𝑥 𝑦2 𝑥3 + 1 Variable separable for 𝑥2 1 ∫ 𝑦2 𝑑𝑦 = ∫ 𝑥 3+1 𝑑𝑥 ∫ 1 1 3𝑥 2 𝑑𝑦 = ∫ 3 𝑑𝑥 2 𝑦 3 𝑥 +1 −1 1 = ln(𝑥 3 + 1) + 𝑐 𝑦 3 1 −1 = ln(𝑥 3 + 1) + 𝑐1 𝑦 or 𝑦= 3 −3 ln(𝑥 3 +1) + 𝑐2