Questions and Answers #1 First Order Differential Equations

Questions and Answers Sheet 1 First Order Differential Equations Question #1 If 𝑓(𝑥) + 𝑥 2 [𝑓(𝑥)]5 = 34 and f(1)=2 find f'(1) Answer: Assuming the problem is 𝑓(𝑥) + 𝑥 2 [𝑓(𝑥)]5 = 34 and f(1)=2 Differentiate the both sides w.r.t. x using the product rule and the chain rule 𝑓 ′ (𝑥) + 2𝑥[𝑓(𝑥)]5 + 𝑥 2 (5[𝑓(𝑥)]4 𝑓 ′ (𝑥)) = 0 (1) Substituting by x=1 in (1) 𝑓 ′ (1) + 2[𝑓(1)]5 + (5[𝑓(1)]4 𝑓 ′ (1)) =0 (2) Substituting by f(1)=2 in (2) 𝑓 ′ (1) + 64 + 80𝑓 ′ (1) = 0 ⇒ 81𝑓 ′ (1) = −64 −64 ⇒ 𝑓 ′ (1) = 81 Question #2 Find all first-order partial derivatives of the function at any point of the domain. 𝑓(𝑥, 𝑦, 𝑧) = 4𝑥 3 𝑦 2 − 𝑒 𝑧 𝑦 4 + 𝑧3 𝑥2 + 4𝑦 − 𝑥 16 + 2021 Answer: Step 1 To find all first - order partial derivatives of the function :𝑓(𝑥, 𝑦, 𝑧) = 4𝑥 3 𝑦 2 − 𝑒 𝑧 𝑦 4 + 𝑓𝑥 = 12𝑥 2 𝑦 2 − 0 − 𝑓𝑥 = 12𝑥 2 𝑦 2 − 2𝑧 3 𝑥3 2𝑧 3 𝑥3 𝑧3 𝑥2 + 4𝑦 − 𝑥 16 + 2021 + 0 − 16𝑥 15 + 0 − 16𝑥 15 Step 2 𝑓𝑦 = 8𝑥 3 𝑦 − 4𝑒 𝑧 𝑦 3 + 0 + 4(1) − 0 𝑓𝑦 = 8𝑥 3 𝑦 − 4𝑒 𝑧 𝑦 3 + 4 𝑓𝑧 = 0 − 𝑒 𝑧 𝑦 4 + 𝑓𝑧 = −𝑒 𝑧 𝑦 4 + 3𝑧 2 3𝑧 2 𝑥2 +0 𝑥2 Question #3 Look at the Chain Rule formula with a visual explanation of how it works: ′ ′ (ℎ(𝑥)) = (𝑔(𝑓(𝑥))) = 𝑔′ (𝑓(𝑥)) ⋅ 𝑓 ′ (𝑥) g(f(x)) -outside function (f(x)) -inside function g'(f(x)) -derivative of the otside at f(x) f'(x) -derivative of the inside Now, explain what happens in regards to calculating the derivative, if f(x) is also a composite function. Answer: Let f(x) also be a composite function ⇒ 𝑓(𝑥) = 𝑝(𝑞(𝑥)) ∴ ℎ(𝑥) = 𝑔(𝑓(𝑥)) = 𝑔 (𝑝(𝑞(𝑥))) ∴ Using chain rule ℎ′(𝑥) = 𝑔′(𝑝(𝑞(𝑥)) ⋅ 𝑝′(𝑞(𝑥)) ⋅ 𝑞′(𝑥) = 𝑔′ (𝑓(𝑥)) ⋅ 𝑝′ (𝑞(𝑥)) ⋅ 𝑞 ′ (𝑥) Question #4 What is the derivative of ℎ(𝑥) = sin 2 𝑥 cos 2 𝑥 Answer: Step 1 Product Rule of differentiation: If f(x) and g(x) are any two differentiable functions, then 𝑓(𝑔)′ (𝑥) = 𝑓(𝑥) × 𝑔′ (𝑥) + 𝑓 ′ (𝑥) × 𝑔(𝑥) Chain rule of differentiation: 𝑑 𝑑 𝑓(𝑔(𝑥)) = 𝑓 ′ (𝑔(𝑥)) × 𝑔(𝑥) 𝑑𝑥 𝑑 ′ ′ (𝑥) (𝑥) = 𝑓 (𝑔(𝑥)) × 𝑔 × 𝑑𝑥 ′ ′ (𝑥) = 𝑓 (𝑔(𝑥)) × 𝑔 The given function is ℎ(𝑥) = sin 2 𝑥 cos 2 𝑥 𝑑𝑥 Step 2 Differentiate the given function using the product rule as follows. 𝑑 ℎ′ (𝑥) = (sin 2 𝑥 cos 2 𝑥) 𝑑𝑥 𝑑 = sin 2 𝑥 𝑑𝑥 (cos 2 𝑥) + cos 2 𝑥 𝑑 𝑑𝑥 (sin 2 𝑥) (Using product rule) = sin 2 𝑥(− sin 2 𝑥)(2) + cos 2 𝑥(cos 2 𝑥)(2) (Using chain rule) = −2 sin2 2𝑥 + 2 cos 2 2𝑥 = −2(1 − cos 2 2𝑥) + 2 cos 2 2𝑥 = −2 + 4 cos 2 2𝑥 Therefore, the derivative of the given function is ℎ′ (𝑥) = −2 + 4 cos 2 2𝑥 Question #5 a) Solve the differential equation: (𝑥 + 1) 𝑑𝑦 𝑑𝑥 − 3𝑦 = (𝑥 + 1)4 given that y=16 and x=1 , expressing the answer in the form of y=f(x) . b) Hence find the area enclosed by the graphs 𝑦 = 𝑓(𝑥), 𝑦 = (1 − 𝑥)4 and the x-axis . I have found the answer to part a) using the first order linear differential equation method and the answer to part a) is: 𝑦 = (1 + 𝑥)4 . However how would you calculate the area between the two graphs ( 𝑦 = (1 − 𝑥)4 and 𝑦 = (1 + 𝑥)4 ) and the x-axis. Answer: Good job on solving part (a) of the question. That was the tough part. Part (b) is much more easier. As you stated, we get that the function is: 𝑦 = (1 + 𝑥)4 Now, in order to find the area between this function, the function 𝑓(𝑥) = (1 − 𝑥)4 , and the x-axis, we need to the find intersection points: (1 + 𝑥)4 = (1 − 𝑥)4 1 1 ((1 + 𝑥)4 )4 = ((1 − 𝑥)4 )4 1+x=1-x x=0 When x=0, (0 + 1)4 = (0 − 1)4 = 1 .Thus, we get that the two functions intersect at the point (0,1). Next, we look at where each of the two functions intersect the x-axis. We get that: (1 + 𝑥)4 = 0 1+x=0 x=-1 (1 − 𝑥)4 = 0 1-x=0 x=1 Therefore, we get our two endpoints: (-1,0) and (1,0). Thus, we can use construct our integral now: 0 1 𝐴 = ∫ (1 + 𝑥)4 𝑑𝑥 + ∫ (1 − 𝑥)4 𝑑𝑥 −1 0 0 1 (𝑥 + 1)5 (𝑥 − 1)5 1 1 2 [ ] +[ ] = − 0 + 0 − (− ) = 5 5 5 5 5 −1 0 Hope that helped. Comment if you have any questions. Question #6 Are the following are linear equation? 1. 2. 𝑑𝑃 𝑑𝑡 𝑑𝑦 𝑑𝑥 + 2𝑡𝑃 = 𝑃 + 4𝑡 − 2 = 𝑦 2 − 3𝑦 Answer: First order linear differential equation: A differential equation of the form 𝑑𝑥 𝑑𝑦 + 𝑃(𝑥)𝑦 = 𝑄(𝑥) is called first order linear differential equation. 1. Here we have given, 𝑑𝑃 𝑑𝑡 + 2𝑡𝑃 = 𝑃 + 4𝑡 − 2 Which can be written as, 𝑑𝑃 𝑑𝑡 ⇒ + 2𝑡𝑃 − 𝑃 = 4𝑡 − 2 𝑑𝑃 𝑑𝑡 + (2𝑡 − 1)𝑃 = 4𝑡 − 2 Which is of the form 𝑑𝑃 𝑑𝑡 + 𝑅(𝑡)𝑃 = 𝑄(𝑡), where 𝑅(𝑡) = 2𝑡 − 1 and 𝑄(𝑡) = 4𝑡 − 2 Hence, the given equation is a first order linear differential equation. Question #7 Are the following are linear equation? 1. sin(𝑥) 2. 𝑑2𝑦 𝑑𝑥 2 𝑑𝑦 𝑑𝑥 − 3𝑦 = 0 + sin(𝑥) 𝑑𝑦 = cos(𝑥) 𝑑𝑥 Answer: First order linear differential equation: A differential equation of the form 𝑑𝑥 𝑑𝑦 + 𝑃(𝑥)𝑦 = 𝑄(𝑥) is called first order linear differential equation. 1. Here we have given, sin(𝑥) ⇒ 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 − − 3𝑦 = 0 3 sin 𝑥 =0 𝑑𝑦 Which is of the form 𝑑𝑥 + 𝑃(𝑥)𝑦 = 𝑄(𝑥), where 𝑃(𝑥) = − 3 sin 𝑥 = 0 and 𝑄(𝑥) = 0 Hence, the given equation is a first order linear differential equation. Question #8 Classify the equation as separable, linear, both, or neither. 𝑦 𝑑𝑦 𝑑𝑥 = 2𝑥 + 𝑦 Answer: Definition seperable: A first order differential equation is seperable then 𝑑𝑦 𝑑𝑥 = 𝑓(𝑥)𝑔(𝑦) Linear: A first order linear differential equation is of the forem 𝑑𝑦 𝑑𝑥 + 𝑃𝑦 = 𝑄, where P and Q are either constants or functions of y 𝑦 𝑑𝑦 𝑑𝑥 = 2𝑥 + 𝑦 Answer: The first order differential equation is neither seperable nor linear. Explanation: 𝑦 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 2𝑥 + 𝑦 𝑥 =2 +1 𝑦 we cannot factor 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 𝑓(𝑥). 𝑔(𝑦), hence it is not seperable 𝑥 𝑥 𝑦 𝑦 = 2 + 1 . here P=0 and 𝑄 = 2 + 1 here Q is not a constant and also not a function of y. Hence it is not a linear first order differential equation Question #9 Classify the following system of first-order partial differential equations (k=const.): 𝑢 𝑢 𝑢 𝜕𝑤 𝜕𝑥 𝜕𝑣 𝜕𝑥 𝜕𝑢 𝜕𝑥 +𝑣 +𝑣 +𝑣 𝜕𝑤 𝜕𝑦 𝜕𝑣 𝜕𝑦 𝜕𝑢 𝜕𝑦 +𝑤 +𝑘 +𝑘 𝜕𝑢 𝜕𝑥 𝜕𝑤 𝜕𝑦 +𝑤 𝜕𝑣 𝜕𝑦 =0 =0 𝜕𝑤 =0 𝜕𝑥 Answer: Consider the following system of first order partial differential equations with k =constant 𝑢 𝑢 𝑢 𝜕𝑤 𝜕𝑥 𝜕𝑣 𝜕𝑥 𝜕𝑢 𝜕𝑥 +𝑣 +𝑣 +𝑣 𝜕𝑤 𝜕𝑦 𝜕𝑣 𝜕𝑦 𝜕𝑢 𝜕𝑦 +𝑤 +𝑘 +𝑘 𝜕𝑢 𝜕𝑥 𝜕𝑤 𝜕𝑦 +𝑤 𝜕𝑣 𝜕𝑦 =0 =0 𝜕𝑤 =0 𝜕𝑥 To classify the first order partial differential equation given by 𝑢 𝜕𝑤 𝜕𝑥 +𝑣 𝜕𝑤 𝜕𝑦 +𝑤 𝜕𝑢 𝜕𝑥 +𝑤 𝜕𝑣 𝜕𝑦 =0 From the equation it is clear that x, y are independent variables while u, v, w are dependent variables. As dependent variable does not multiply with corresponding derivative hence this is first order partial differential equation. To classify the first order partial differential equation given by 𝑢 𝜕𝑣 𝜕𝑥 +𝑣 𝜕𝑣 𝜕𝑦 +𝑘 𝜕𝑤 𝜕𝑦 =0 From the equation it is clear that x, y are independent variables while u, v, w are dependent variables. As dependent variable does not multiply with corresponding derivative. But here it occurs and hence this is not first order partial differential equation. To classify the first order partial differential equation given by 𝑢 𝜕𝑢 𝜕𝑥 +𝑣 𝜕𝑢 𝜕𝑦 +𝑘 𝜕𝑤 𝜕𝑥 =0 Here also the dependent variable multiplies with corresponding derivative. Hence this is not first order partial differential equation. Question #10 Convert the differential equation 𝑢′′ − 𝑢′ − 2𝑢 = 𝑒 −5𝑡 into a system of first order equations by letting x=u, y=u' x = y = Answer: Given 𝑢′′ − 𝑢′ − 2𝑢 = 𝑒 −5𝑡 Suppose x=u, y=u', x'=u' and y'=u'' ⇒ 𝑥 ′ = 𝑦 𝑎𝑛𝑑 𝑦 ′ = 𝑢′ + 2𝑢 + 𝑒 −5𝑡 + 2𝑢 + 𝑒 −5𝑡 ∵ 𝑢′′ − 𝑢′ − 2𝑢 = 𝑒 −5𝑡 ⇒ 𝑢′′ = 𝑢′ ⇒ 𝑥 ′ = 𝑦 𝑎𝑛𝑑 𝑦 ′ = 𝑦 + 2𝑥 + 𝑒 −5𝑡 ∵ 𝑢′ = 𝑦 𝑎𝑛𝑑 𝑢 = 𝑥 Therefore, the system of first order differential equations is, Answer: x'=y 𝑦 ′ = 𝑦 + 2𝑥 + 𝑒 −5𝑡 Question #11 Determine if the equation is linear or non-linear and indicate the order of the differential equation and would you tell me why please (2𝑥 3 + 𝑦)𝑑𝑥 − 𝑥𝑑𝑦 = 0 𝑦𝑑𝑥 − (𝑥 + 6𝑦 2 )𝑑𝑦 = 0 (𝑥 + 𝑦)𝑑𝑥 + tan(𝑥) 𝑑𝑦 = 0 6𝑥𝑦𝑑𝑥 + (4𝑦 + 9𝑥 2 )𝑑𝑦 = 0 Answer: We have to tell the order of the following differential equations: 1. 2. 3. 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = = 2𝑥 3 +𝑦 𝑥 𝑦 𝑥+6𝑦 2 =− 𝑥+𝑦 tan(𝑥) Part 1 It is a first order differential equation. We can write it as below: 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 2𝑥 3 +𝑦 𝑥 = 2𝑥 2 + 𝑦 𝑥 𝑑𝑦 𝑑𝑥 1 − 𝑦 = 2𝑥 2 𝑥 It has the form of the linear differential equation 𝑑𝑦 𝑑𝑥 + 𝑃𝑦 = 𝑄 where P and Q are functions of x only. Therefore, it is a linear differential equation. Part 2 It is a first order differential equation. We can write it as below: 𝑑𝑦 𝑑𝑥 𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑑𝑦 = = 𝑦 𝑥+6𝑦 2 𝑥+6𝑦 2 𝑦 1 = 𝑥 + 6𝑦 𝑦 1 − 𝑥 = 6𝑦 𝑦 It has the form of the linear differential equation 𝑑𝑥 𝑑𝑦 + 𝑅𝑥 = 𝑆 where R and S are functions of y only. Therefore, it is a linear differential equation. Part 3 It is a first order differential equation. We can write it as below: 𝑑𝑦 𝑑𝑥 𝑦 𝑥 =− =− 𝑑𝑦 𝑑𝑥 + 𝑥+𝑦 tan(𝑥) 𝑥 tan(𝑥) 1 tan(𝑥) − 1 tan(𝑦) 𝑦=− 𝑦 𝑥 tan(𝑥) It has the form of the linear differential equation P and Q are functions of x only. Therefore, it is a linear differential equation. Question #12 Don't understand 𝑥 ′′ = 𝑥 2 − 𝑦 − 𝑥𝑦, 𝑦 ′′ = 𝑦 − 𝑥 as a system of first order differential equations. x'''=tx 𝑑𝑦 𝑑𝑥 + 𝑃𝑦 = 𝑄 where, as a non-autonomous system of first order differential equations. x''''=tx as a autonomous system of first order differential equations. Answer: 1. Let, 𝑥1 = 𝑥, 𝑥2 = 𝑥 ′ then the 𝑥 ′′ = 𝑥 2 − 𝑦 − 𝑥𝑦 equation converted to 𝑥2′ = 𝑥12 − 𝑦 − 𝑥1 𝑦 where 𝑥1′ = 𝑥2 . Let, 𝑦1 = 𝑦, 𝑦2 = 𝑦 ′ then the equation y''=y-x converted to 𝑦2′ = 𝑦1 − 𝑥 where 𝑦2 = 𝑦1′ . 2. x'''=tx is already a non-autonomous system. We need to convert it to first order equation. Let, 𝑥1 = 𝑥, 𝑥2 = 𝑥 ′ , 𝑥3 = 𝑥 ′′ , the above system transformed to 𝑥3′ = 𝑡𝑥1 which is a non-autonomous system of first order differential equation. 3. x''''=tx is not a autonomous system of linear equation. First transfer it to autonomous. Put, x=tv . The equation reduces to 3𝑑 2 𝑣 𝑑𝑡 2 + 𝑥 𝑣 𝑣𝑑 3 𝑑𝑡 3 = 𝑥2 𝑣 𝑥 where 𝑡 = . 𝑣 Now let, 𝑣 = 𝑣1 , 𝑣 ′ = 𝑣2 , 𝑣 ′′ = 𝑣3 then the above equation reduces to 3𝑣3 + 𝑥 𝑣1 , 𝑣3′ = 𝑥2 𝑣1 which is an autonomous system of first order differential equation. Question #13 Don`t have much time, help Solving a first-order differential equation in exercise, find the general solution of the first-order differential equation x>0 by any appropriate method 𝑑𝑦 𝑑𝑥 = 𝑒 2𝑥+𝑦 𝑒 𝑥−𝑦 Answer: 𝑑𝑦 𝑑𝑥 ⇒ ⇒ = 𝑒 2𝑥+𝑦 𝑒 𝑥−𝑦 𝑑𝑦 𝑒 2𝑥 𝑒 𝑦 = 𝑒𝑥 𝑑𝑥 𝑒𝑌 𝑑𝑦 𝑑𝑥 = 𝑒 𝑥 𝑒 2𝑦 ⇒ 𝑒 −2𝑥 𝑑𝑦 = 𝑒 𝑥 𝑑𝑥 Separating the variables ⇒ ∫ 𝑒 −2𝑥 𝑑𝑦 = ∫ 𝑒 𝑥 𝑑𝑥 Integrating both sides ⇒− 𝑒 −2𝑦 2 = 𝑒 𝑥 + 𝑒 𝑐1 ⇒ 𝑒 −2𝑦 = −2𝑒 𝑥 − 2𝑒 𝑐1 ⇒ 2𝑒 𝑥 + 𝑒 −2𝑦 = 𝑐[𝑐 = −2𝑒 𝑐1 ] Question #14 Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. 1 1. The function 𝑦 = 𝑡 + , for t > 0, satisfies the initial value problem 𝑡𝑦 ′ + 𝑦 = 2𝑡, 𝑦(1) = 2. 𝑡 Answer: The initial value problem is 𝑡𝑦 ′ + 𝑦 = 2𝑡, Rearrange the terms, 1 𝑦′ + 𝑦 = 2 𝑡 This is a Linear Differential Equation of the form 𝑑𝑦 𝑑𝑥 + 𝑃𝑡 = 𝑄, whose solution is given as, 𝑦(𝐼. 𝐹. ) = ∫ 𝑄(𝐼. 𝐹. )𝑑𝑥 Here, 𝑃 = 1 𝑡 and, Q=2 The Integrating Factor is calculated as, 1 𝐼. 𝐹. = 𝑒 ∫ 𝑃𝑑𝑥 = 𝑒 ∫𝑥𝑑𝑥 = 𝑒 ln(𝑡) = 𝑡 The solution of the differential equation, 𝑦(𝑡) = ∫ 2(𝑡)𝑑𝑡 𝑦𝑡 = 2 ∫ 𝑡𝑑𝑡 𝑦𝑡 = 2 𝑡2 2 𝑦=𝑡+ +𝐶 𝐶 𝑡 The given initial solution is y(1)=2. Question #15 Explain why or why not determine whether the following statements are true and give an explanation or counterexample. a. The differential equation y'+2y =t is first-order, linear, and separable. b. The differential equation 𝑦 ′ 𝑦 = 2𝑡 2 is first-order, linear, and separable. Answer: a) The given differential is y'+2y=t, Rearrange the terms, y'=t-2y The highest order derivative in the equation is the first derivative, therefore it is a first order differential equation. The derivatives and the variables appear as a simple first power, so the the differential equation is linear. The terms in RHS cannot be obtained as a product of two separate functions, one for each y and t, so it is not separable. Thus, the given differential equation is characterized as, 1. First Order 2. Linear 3. Non Separable Therefore, the given statement is FALSE. Question #16 Find f(x) that satisfies the first-order differential equation: 𝑓 ′ (sin2 𝑥) = cos 2 𝑥 + tan2 𝑥(for 0 < 𝑥 < 1) Thanks! Answer: We can easily observe that 𝑓 ′ (sin2 𝑥) = 1 − sin2 𝑥 + sin2 𝑥 1−sin2 𝑥 We can thus generalise to 𝑓 ′ (𝑦) = 1 − 𝑦 + 𝑦 1−𝑦 which can be solved for f(x) by simply integrating (1). Question #17 find 𝑎𝑛 and f(x) so that ∑∞ 𝑛=0 𝑎𝑛 𝑛! 𝑥 𝑛 satisfy 𝑓 ′ (𝑥) − 𝑓(𝑥) = 𝑥 2 and f(0)=1 𝑎 𝑛 𝑛−1 here I tried to find 𝑓 ′ (𝑥) = ∑∞ from 𝑛=0 (𝑛−1)! 𝑥 𝑑𝑓(𝑥) 𝑑𝑥 − 𝑓(𝑥) = 𝑥 2 using first order differential equation I got 𝑦 = 𝑥 2 + 2𝑥 + 1 → 𝑓(𝑥)? but I don't know what is the relation of the summation and first order differential equation. and how to find f(x) and 𝑎𝑛 can someone give me hint? Answer: Hint: If we assume a Taylor expansion of f(x) at 𝑥0 = 0 by 𝑓(𝑥) = ∑∞ 𝑛=0 𝑎𝑛 𝑛! 𝑥 𝑛 what can we conclude for f(x=0) from the Taylor series for the constant 𝑎0 ? Write the first terms of the series to see what happens. The derivative is given as (note the change of the summation index!) 𝑓 ′ (𝑥) ∞ ∞ 𝑛=1 𝑛=0 𝑎𝑛 𝑎𝑛 + 1 𝑛 =∑ 𝑥 𝑛−1 = ∑ 𝑥 . (𝑛 − 1)! 𝑛! Plug this into the differential equation: an + 1 n  an n  an + 1 an n x −  x = [ − x = 0 x0 + 0 x1 + 1x 2 + 0 x3 + ... n! n! n! n =0 n =0 n ! n =0  f ( x) − f ( x) =  What can you conclude by comparing the coefficients? Use the value of 𝑎0 . Question #18 Find the critical points and sketch the phase portrait of the given autonomous first order differential equation. Classify the each critical point as asmyptotically stable, unstable, or semi-stable. 𝑦 ′ = 𝑦 2 (4 − 𝑦 2 ) Answer: Consider the provided autonomous first order differential equation. 𝑦 ′ = 𝑦 2 (4 − 𝑦 2 ) The provided autonomous first order differential equation can be written as: 𝑑𝑦 𝑑𝑥 = 𝑓(𝑦) = 𝑦 2 (4 − 𝑦 2 ) We can get the critical points by solving f(y) = 0 𝑦 2 (4 − 𝑦 2 ) = 0 𝑦 = 0 𝑜𝑟 𝑦 = ±2 Therefore, the critical points are y = -2, 0 and 2. Now to find the asymptotically stable, unstable or semi-stable as shown below: 𝑦 2 (4 − 𝑦 2 ) > 0 when 4 − 𝑦 2 > 0 Therefore, 𝑦2 < 4 −2 < 𝑦 < 2 Thus, the f(y) is increasing in (-2,2) Now, 𝑦 2 (4 − 𝑦 2 ) < 0 when 4 − 𝑦 2 < 0 Therefore, y^2>4 y<-2 or y>2 (−∞, −2) ∪ (2, ∞) the f(y) is decreasing Now classify them as shown below: at 𝑦 = 0, semi-stable (above increasing), (below increasing): } at 𝑦 = 2, asymptoti stable (above increasing), (below increasing): at 𝑦 = −2, ustable (above increasing), (below increasing): Question #19 Find the general solution of the differential equation 𝑦′ = 𝑥−𝑦+3 𝑥−𝑦 Answer: A differential equation is an equation in x,y and derivatives of y. A first order differential equation is on in which the highest order derivative present in the differential equation is first order. A differential equation in some cases can be simplified to a simpler differential equation with an appropriate substitution. For given differential equation find an appropriate substitution to simplify the differential equation. Given differential equation is 𝑦 ′ = 𝑑𝑢 𝑑𝑥 . This can be rewritten as 1 − 1− 1− − 𝑑𝑢 𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑑𝑢 𝑑𝑥 = = 𝑢+3 𝑢 =1+ 3 𝑢 3 𝑢 −𝑢𝑑𝑢 = 3𝑑𝑥 𝑢2 = 3𝑥 + 𝐶 2 𝑢2 = −6𝑥 − 2𝐶 𝑢 = √−6𝑥 − 2𝐶 𝑥−𝑦+3 𝑥−𝑦 𝑑𝑢 𝑑𝑦 𝑑𝑥 = 𝑑𝑥 . Use the substitution x-y=u. Differentiating this gives 1 − . Substitute this in the differential equation and integrate. 𝑑𝑦 𝑑𝑥 = 𝑥 − 𝑦 = √−6𝑥 − 2𝐶 𝑦 = 𝑥 − √−6𝑥 − 2𝐶 Hence, general solution to the differential equation is 𝑦 = 𝑥 − √−6𝑥 − 2𝐶 Question #20 Find the general solution of these first order differential equations (1– 𝑥)𝑦 ′ = 𝑦 2 Answer: Given that: (1– 𝑥)𝑦 ′ = 𝑦 2 Consider (1 − 𝑥)𝑦 ′ = 𝑦 2 (1 − 𝑥) 𝑑𝑦 𝑦2 = 𝑑𝑦 = 𝑦2 𝑑𝑥 1 1−𝑥 𝑑𝑥 Integrate on both sides ∫ 1 1 𝑑𝑦 = ∫ 𝑑𝑥 𝑦2 1−𝑥 Let 𝑢 = 1 − 𝑥 ⇒ 𝑑𝑢 𝑑𝑥 = −1 -du=dx Hence we have ∫ 1 1 𝑑𝑦 = − ∫ 𝑑𝑢 2 𝑦 𝑢 [∵ ∫ 𝑦 𝑛 𝑑𝑛 = 𝑦 −2+1 −2+1 𝑦 𝑛+1 𝑛+1 + 𝑐] = − ln(1 − 𝑥) + 𝑐 1 − = − ln(1 − 𝑥) + 𝑐 𝑦 ⇒𝑦= 1 ln(1−𝑥)−𝑐 ∴ The general solution is 𝑦= 1 ln(1−𝑥)−𝑐