Questions and Answers #4 First Order Differential Equations

Questions and Answers Sheet 4 First Order Differential Equations Question #1 Solve the first-order differential equations: (𝑥 2 + 1) 𝑑𝑦 𝑑𝑥 = 𝑥𝑦 Answer: Solution: (𝑥 2 + 1) 1 ⇒ 𝑦 𝑑𝑦 = 𝑥𝑦 𝑑𝑥 𝑑𝑦 = ( 𝑥 𝑥 2 +1 ) 𝑑𝑥 This is a separable linear first order differential which is of the form 𝑁(𝑦)𝑑𝑦 = 𝑀(𝑥)𝑑𝑥 ⇒ 1 𝑦 𝑑𝑦 = 𝑥 𝑥 2 +1 𝑑𝑥 Intagrating both sides 1 𝑥 ∫ 𝑦 𝑑𝑦 = ∫ 𝑥 2+1 𝑑𝑥 Let 𝑥 2 + 1 = 𝑢 ⇒ 2𝑥𝑑𝑥 = 𝑑𝑢 1 ⇒ 𝑥𝑑𝑥 = 𝑑𝑢 2 So, 1 1 ⇒ ln(𝑦) = ∫ 2 𝑑𝑢 = ∫ 𝑢𝑑𝑢 𝑢 2 1 ⇒ ln(𝑦) = ln(𝑢) + 𝑐 2 1 ⇒ ln(𝑦) = ln(𝑥 2 + 1) + 𝑐 2 1 ⇒ ln(𝑦) = ln(𝑥 2 + 1)2 + ln(𝑒 𝑐 ) 1 ⇒ ln(𝑦) = ln (𝑒 𝑐 (𝑥 2 + 1)2 ) 1 ⇒ � = 𝑒 𝑐 (𝑥 2 + 1)2 ⇒ 𝑦(𝑥) = 𝑐 √𝑥 2 + 1 ∴ 𝑦(𝑥) = 𝑐 √𝑥 2 + 1 Question #2 Solve the following first order differential equation 𝑦 ′ = 2𝑥 2 𝑦 2 Here is my solution: 𝑦′ 1 = 2𝑥 2 𝑦2 ∫ 1 ′ 𝑦 𝑑𝑥 = ∫ 2𝑥 2 𝑑𝑥 𝑦2 ∫ 1 𝑑𝑦 = ∫ 2𝑥 2 𝑑𝑥 𝑦2 2 ln|𝑦 2 | = 𝑥 3 + 𝑐 3 2 3 +𝑐 𝑦 2 = 𝑒 3𝑥 2 3 2 3 = 𝑒 3𝑥 . 𝑒 𝑐 = 𝑐𝑒 3𝑥 2 3 𝑦 = √𝑐𝑒 3𝑥 Help!! Answer: 1 ∫ 𝑦2 𝑑𝑦 = ∫ 𝑦 −2 𝑑𝑦 = −𝑦 −1 + 𝐶 You should obtain 1 2 3 𝑦 3 2𝑥 3 − = 𝑥3 + 𝑐 ⇔ 𝑦 = − provided 𝑥 ≠ 0 Question #3 Solve the following first order differential equation 𝑦 ′ = order differential equations 𝑦 ′ = 𝑎𝑥+𝑏 𝑦 𝑛 +𝑐 𝑥+1 𝑦 2 +1 , and give the solution to the general first , where a, b and c are constants. Answer: Given: The first order differential equation is 𝑦 ′ = 𝑥+1 𝑦 2 +1 Calculation: The first order differential equation is 𝑦 ′ = 𝑥+1 𝑦 2 +1 and General first order differential equation 𝑦 ′ = 𝑎𝑥+𝑏 𝑦 𝑛 +𝑐 Multiply by 𝑦 2 + 1 on both sides, 𝑦 ′ (𝑦 2 + 1) = 𝑥 + 1 ∫ 𝑦 ′ (𝑦 2 + 1)𝑑𝑥 = ∫ 𝑥 + 1𝑑𝑥 ∫ 𝑑𝑦 2 (𝑦 + 1)𝑑𝑥 = ∫ 𝑥 + 1𝑑𝑥 𝑑𝑥 ∫ 𝑦 ′ (𝑦 2 + 1)𝑑𝑦 = ∫ 𝑥 + 1𝑑𝑥 𝑦2 3 +𝑦= 𝑥2 2 + 𝑥 + 𝐾 , where K is arbitrary constant General first order differential equation 𝑦 ′ = 𝑎𝑥+𝑏 𝑦 𝑛 +𝑐 Multiply by 𝑦 𝑛 + 𝑐 on both sides, 𝑦 ′ (𝑦 𝑛 + 𝑐) = 𝑎𝑥 + 𝑏 ∫ 𝑦 ′ (𝑦 𝑛 + 𝑐)𝑑𝑥 = ∫ 𝑎𝑥 + 𝑏𝑑𝑥 𝑑𝑦 ∫ 𝑑𝑥 (𝑦 𝑛 + 𝑐)𝑑𝑥 = ∫ 𝑎𝑥 + 𝑏𝑑𝑥 ∫(𝑦 𝑛 + 𝑐)𝑑𝑦 = ∫ 𝑎𝑥 + 𝑏𝑑𝑥 𝑦 𝑛+1 𝑛+1 𝑎𝑥 2 + 𝑐𝑦 = 2 + 𝑏𝑥 + 𝐾, where K is arbitrary constant. Answer: (𝑖) 𝑦3 (𝑖𝑖) 3 +𝑦= 𝑦 𝑛+1 𝑛+1 𝑥2 2 +𝑥+𝐾 + 𝑐𝑦 = 𝑎𝑥 2 2 + 𝑏𝑥 + 𝐾 Question #4 Solve the following first order differential equations: 1. (1 + 𝑥 2 + 𝑦 2 + 𝑥 2 𝑦 2 )𝑑𝑦 = 𝑦 2 𝑑𝑥 2. 𝑒 𝑥 𝑦 𝑑𝑦 𝑑𝑥 = 𝑒 −𝑦 + 𝑒 3𝑥−𝑦 Answer: 1. (1 + 𝑥 2 + 𝑦 2 + 𝑥 2 𝑦 2 )𝑑𝑦 = 𝑦 2 𝑑𝑥 ((1 + 𝑥 2 ) + 𝑦 2 (1 + 𝑥 2 ))𝑑𝑦 = 𝑦 2 𝑑𝑥 (1 + 𝑥 2 )(1 + 𝑦 2 )𝑑𝑦 = 𝑦 2 𝑑𝑥 (1 + 𝑦 2 ) 𝑑𝑥 𝑑𝑦 = (1 + 𝑥 2 ) 𝑦2 ∫ 1 1 𝑑𝑦 + ∫ 𝑑𝑦 = ∫ − 𝑑𝑥 + 𝑐 2 𝑦 1 + 𝑥2 𝑦 −2+1 + 𝑦 = tan−1 𝑥 + 𝑐 −2 + 1 1 − + 𝑦 = tan−1 (𝑥) + 𝑐 𝑦 1 (𝑦 − ) = tan−1 (𝑥) + 𝑐 𝑦 Question #5 Solve the following first order differential equations. 𝑦𝑦 ′ = (𝑦 2 − 1)(𝑥 2 + 1) Answer: Solution: 𝑓(𝑥, 𝑦) = 24𝑥𝑦 − 4𝑥 2 − 20𝑦 2 𝑓𝑥 = 24𝑦 − 16𝑥 3 = 0 ⇒ 3𝑦 = 2𝑥 3 𝑓𝑦 = 24𝑥 − 80𝑦 3 ⇒ 3𝑥 = 10𝑦 3 10 ∴ 3𝑦 = 2 ( 𝑦 3 ) 3 3 ⇒ 3𝑦 = 2 1000 27 𝑦9 Solution: 𝑦𝑦 ′ = (𝑦 2 − 1)(𝑥 2 + 1) ∫( 𝑦 𝑦 2 −1 𝑑𝑦 = ∫(𝑥 2 + 1)𝑑𝑥 Let 𝑦 2 − 1 = 𝑢 ∴ =∫ ∴ 𝑑𝑢 𝑑𝑦 = 2𝑦 ∴ 𝑑𝑢 2 = 𝑦𝑑𝑦 𝑑𝑢 1 1 = ln(𝑢) = ln(𝑦 2 − 1) 2𝑢 2 2 1 𝑥3 ln(𝑦 2 − 1) = +𝑥+𝑐 2 3 ln(𝑦 2 − 1) = 𝑦 2 − 1 = 𝑐𝑒 ∴ 𝑦 = √𝑐𝑒 2𝑥 3 + 2𝑥 + 𝑐 3 2𝑥 3 +2𝑥 3 2𝑥3 +2𝑥 3 ⇒ 𝑦 2 = 𝑐𝑒 2𝑥 3 +2𝑥 3 + 1 𝑜𝑟 𝑦 = −√𝑐𝑒 +1 2𝑥3 +2𝑥 3 +1 Question #6 Solve the following first-order nonlinear ordinary differential equation: 𝑦 ′ (𝑥) + log(𝑦(𝑥)) = −𝑥 − 1 Answer: 𝑦 ′ + log 𝑦 = −𝑥 − 1 𝑑𝑦 = −𝑥 − log 𝑦 − 1 𝑑𝑥 (𝑥 + log 𝑦 + 1) 𝑑𝑥 = −1 𝑑𝑦 Let 𝑢 = 𝑥 + log 𝑦 + 1 , Then 𝑥 = 𝑢 − log 𝑦 − 1 𝑑𝑥 𝑑𝑢 1 = − 𝑑𝑦 𝑑𝑦 𝑦 𝑑𝑢 1 ∴ 𝑢 ( − ) = −1 𝑑𝑦 𝑦 𝑢 𝑑𝑢 𝑢 − = −1 𝑑𝑦 𝑦 𝑢 𝑑𝑢 𝑢 = −1 𝑑𝑦 𝑦 𝑢 𝑑𝑢 𝑢 − 𝑦 = 𝑑𝑦 𝑦 (𝑢 − 𝑦) 𝑑𝑦 𝑑𝑢 = 𝑢𝑦 This belongs to an Abel equation of the second kind. Let 𝑣 = 𝑢 − 𝑦 , Then 𝑦 = 𝑢 − 𝑣 𝑑𝑦 𝑑𝑣 = 1− 𝑑𝑢 𝑑𝑢 ∴ 𝑣 (1 − 𝑣−𝑣 𝑑𝑣 ) = 𝑢(𝑢 − 𝑣) 𝑑𝑢 𝑑𝑣 = 𝑢2 − 𝑢𝑣 𝑑𝑢 𝑣 𝑑𝑣 = 𝑢2 − 𝑢𝑣 𝑑𝑢 𝑣 𝑑𝑣 = (𝑢 + 1)𝑣 − 𝑢2 𝑑𝑢 Let 𝑠 = 𝑢 + 1 , Then 𝑑𝑣 𝑑𝑢 = 𝑑𝑣 𝑑𝑠 𝑑𝑠 𝑑𝑢 = 𝑑𝑣 𝑑𝑠 ∴𝑣 𝑣 𝑑𝑣 𝑑𝑠 𝑑𝑣 𝑑𝑠 = 𝑠𝑣 − (𝑠 − 1)2 = 𝑠𝑣 − 𝑠 2 + 2𝑠 − 1 𝑠2 Let 𝑡 = Then ∴ 𝑠𝑣 𝑣 𝑣 2 𝑑𝑣 𝑑𝑠 𝑑𝑣 𝑑𝑡 , = 𝑑𝑣 𝑑𝑡 𝑑𝑡 𝑑𝑠 =𝑠 𝑑𝑣 𝑑𝑡 = 𝑠𝑣 − 𝑠 2 + 2𝑠 − 1 𝑑𝑣 1 =𝑣−𝑠+2− 𝑑𝑡 𝑠 𝑑𝑣 𝑑𝑡 = 𝑣 ± √2𝑡 + 2 ± 1 √2𝑡 This belongs to an Abel equation of the second kind in the canonical form. Question #7 Suppose that a dose of 𝑥0 (initial) gram of a drug is injected into the bloodstream. Assume that the drug leaves the blood and enters the urine at a rate proportional to the amount of drug present in the blood. In addition, assume that half of the drug dose has entered the urine after 0.75 hour. Find the time at which the amount of drug in the blood stream is 5% of the original drug dose 𝑥0 (initial). I have the first order differential equation. It's finding t that is the problem. Answer: Obviously, the law is exponential decay and you get from 3 1 3 1 4 2 4 2𝑘 𝑥 ( ) = 𝑥0 , 𝑥 (𝑘 ) = 𝑥0 the formula 4 𝑥(𝑡) = 2−3𝑡 𝑥0 where t is measured in hours. Now you have to solve 4 2−3𝑡 = 0.05 ⇔ 𝑡 = − 3 log 0.05 4 log 2 . Question #8 Suppose we have 𝑑𝑦 𝑑𝑥 + 𝑓(𝑥)𝑦 = 𝑟(𝑥) and it has two solutions 𝑦1 (𝑥) and 𝑦2 (𝑥) then how to prove that solution of differential equation 𝑑𝑦 𝑑𝑥 + 𝑓(𝑥)𝑦 = 2𝑟(𝑥) Will be 𝑦1 (𝑥) + 𝑦2 (𝑥) ? I think given differential equations is linear first order equation so its solution will be 𝑦. 𝑒 ∫ 𝑓(𝑥)𝑑𝑥 = ∫ 𝑟 . 𝑒 ∫ 𝑓(𝑥)𝑑𝑥 𝑑𝑥 now do I establish two solution as 𝑦1 and 𝑦2 out of this equation? Answer: I don't know if this is not just placing the solution in and determining that it satisfies..however I will proceed! 𝑦1′′ + 𝑓(𝑥)𝑦1′ = 𝑟(𝑥) 𝑦2′′ + 𝑓(𝑥)𝑦2′ = 𝑟(𝑥). adding the two equations together we obtain 𝑦1′′ + 𝑦2′′ + 𝑓(𝑥)𝑦1′ + 𝑓(𝑥)𝑦2′ = 2𝑟(𝑥) (𝑦1 + 𝑦2 )′′ + 𝑓(𝑥)(𝑦1′ + 𝑦2′ ) = (𝑦1 + 𝑦2 )′′ + 𝑓(𝑥)(𝑦1 + 𝑦2 )′ = 2𝑟(𝑥) . so we find (as not surprising) the solution 𝑦1 + 𝑦2 satisfies the equation. If this is not the kind of process you would like to follow then I can try to improve. Question #9 𝑦 The equation 𝑦 ′ = (ln 𝑦 − ln 𝑥 + 1) is? 𝑥 1. First order, Partial. 2. First order, Non-homogenous. 3. None of above. 4. First order, Homogenous. Answer: y' is the first derivative of y w.r.t x. So, it is a first order differential equation. Question #10 Who know? Solve the following first order differential equations. Use any method: 2𝑥 2 𝑑𝑥 − 3𝑥𝑦 2 𝑑𝑦 = 0; 𝑦(1) = 0 Answer: First order differential equation, 2𝑥 2 𝑑𝑥 − 3𝑥𝑦 2 𝑑𝑦 = 0; 𝑦(1) = 0 Given, 𝑦(1) = 0 2𝑥 2 𝑑𝑥 − 3𝑥𝑦 2 𝑑𝑦 = 0 2𝑥𝑑𝑥 = 3𝑦 2 𝑑𝑦 Integrating both sides, 2 ∫ 𝑥𝑑𝑥 = 3 ∫ 𝑦 2 𝑑𝑦 2 𝑥2 2 =3 𝑦2 3 +𝑐 𝑥 2 = 𝑦3 + 𝑐 Putting 𝑥 = 3 in the above equation, (1)2 = |𝑦(1)|3 + 𝑐 𝑐=1 Thus, the function will be, 1 + 𝑦3 = 𝑥 2 Question #11 Write an equivalent first-order differential equation and initial condition for y. 𝑥 𝑦 = −1 + ∫0 (2𝑡 − 5𝑦(𝑡))𝑑𝑡 What is the equivalent first-order differential equation? y = What is the initial condition? y ( y = ) = Answer: Given, 𝑥 𝑦 = −1 + ∫0 (2𝑡 − 5𝑦(𝑡))𝑑𝑡 Differentiating both side, we get 𝑦 ′ = 0 + 2𝑥 − 5𝑦(𝑥) ⇒ 𝑦 ′ = 2𝑥 − 5𝑦(𝑥) 𝑥 Since, 𝑦(𝑥) = −1 + ∫0 (2𝑡 − 5𝑦(𝑡))𝑑𝑡 0 𝑎 ⇒ 𝑦(0) = −1 + ∫0 (2𝑡 − 5𝑦(𝑡))𝑑𝑡 = −1 + 0 [∵ ∫𝑎 𝑓(𝑥)𝑑𝑥 = 0] ⇒ 𝑦(0) = −1 Given, 𝑥 𝑦 = −1 + ∫0 (2𝑡 − 5𝑦(𝑡))𝑑𝑡 Differentiating both side, we get 𝑦 ′ = 0 + 2𝑥 − 5𝑦(𝑥) ⇒ 𝑦 ′ = 2𝑥 − 5𝑦(𝑥) 𝑥 Since, 𝑦(𝑥) = −1 + ∫0 (2𝑡 − 5𝑦(𝑡))𝑑𝑡 0 𝑎 ⇒ 𝑦(0) = −1 + ∫ (2𝑡 − 5𝑦(𝑡))𝑑𝑡 = −1 + 0 [∵ ∫ 𝑓(𝑥)𝑑𝑥 = 0] 0 𝑎 ⇒ 𝑦(0) = −1 Question #12 Write an equivalent first-order differential equation and initial condition for y. 𝑥 𝑦 = −4 + ∫1 (4𝑡 − 𝑦(𝑡))𝑑𝑡 What is the equivalent first-order differential equation? What is the initial condition? Answer: 𝑥 𝑦 = −4 + ∫1 (4𝑡 − 𝑦(𝑡))𝑑𝑡 Differentiating both side, we get 𝑑𝑦 𝑑𝑥 or, ∴ = 𝑑 𝑑𝑥 𝑑 𝑑𝑥 𝑑𝑦 𝑑𝑥 𝑥 [∫1 (4𝑡 − 𝑦(𝑡))𝑑𝑡 ] 𝑦(𝑥) = 4𝑥 − 𝑦(𝑥) + 𝑦 = 4𝑥 → equivalent at order differential equation Initial condition is at 𝑥 = 1 1 ∴ 𝑦(1) = −4 + ∫ (4𝑡 − 𝑦(𝑡))𝑑𝑡 1 𝑎 𝑦(1) = −4 [∵ ∫𝑎 𝑓(𝑥)𝑑𝑥 = 0] ∴ at { x =1 → initial condition y = −4