Questions and Answers #11 Differential equations

Questions and Answers Sheet 11 Differential equations Question #1 Solve the following differential equations using the indicated method. SEPARATION OF VARIABLES: 𝑥 2 𝑑𝑥 + 𝑦(𝑥 − 1)𝑑𝑦 = 0 Answer: The given differential equation is: 𝑥 2 𝑑𝑥 + 𝑦(𝑥 − 1)𝑑𝑦 = 0 Now separate the variables: 𝑥2 −𝑦𝑑𝑦 = 𝑑𝑥 𝑥−1 Now integrate both side: 𝑥2 − ∫ 𝑦𝑑𝑦 = ∫ 𝑑𝑥 𝑥−1 𝑦2 𝑥2 − 1 + 1 − =∫ 𝑑𝑥 2 𝑥−1 2 2 𝑦 𝑥 −1 1 − =∫ 𝑑𝑥 + ∫ 𝑑𝑥 2 𝑥−1 𝑥−1 (𝑥 − 1)(𝑥 + 1) 𝑦2 1 − =∫ 𝑑𝑥 + ∫ 𝑑𝑥 2 𝑥−1 𝑥−1 2 𝑦 1 − = ∫(𝑥 + 1)𝑑𝑥 + ∫ 𝑑𝑥 2 𝑥−1 𝑦2 𝑥 2 − = + 𝑥 + ln|𝑥 − 1| + 𝐶 2 2 Hence the general solution of the given differential equation is: 𝑦2 𝑥 2 + + 𝑥 + ln|𝑥 − 1| + 𝐶 = 0 2 2 where C is an arbitrary constant. Question #2 Find a general solution of each of the following differential equations: Answer: 𝑑𝑦 2 + 𝑦 = 𝑥 5𝑦4 𝑑𝑥 𝑥 1 𝑑𝑦 2 1 ⇒ 4 + = 𝑥5 𝑦 𝑑𝑥 𝑥 𝑦 3 1 −3 𝑑𝑦 Let, 3 = 𝑡 ⇒ 4 = ⇒ ⇒ 𝑦 −1 2 𝑦 𝑑𝑥 5 𝑑𝑡 𝑑𝑥 + 𝑡=𝑥 𝑥 𝑑𝑡 6 ⇒ − 𝑡 = −3𝑥 5 𝑑𝑥 𝑥 𝑑𝑥 −6 ln( 1 ) 1 I.F = 𝑒 ∫ 𝑥 𝑑𝑥 = 𝑒 −6 ln 𝑥 = 𝑒 𝑥6 = 6 𝑥 1 1 5 𝑡 ( 6 ) = − ∫ 3𝑥 6 𝑑𝑥 + 𝑐 𝑥 𝑥 1 = −3 ∫ 𝑑𝑥 + 𝑐 = −3 ln(𝑥) + 𝑐 𝑥 ⇒ 𝑡 = 𝑥 6 (−3 ln(𝑥) + 𝑐) 𝑑𝑦 𝑑𝑥 = 𝑥 6 𝑦 4 −2𝑦 𝑥 1 =𝑡 𝑦3 1 ⇒ 3 = 𝑥 6 (−3 ln(𝑥) + 𝑐) 𝑦 1 ⇒ 𝑦(𝑥) = 3 √𝑥 6 (−3 ln(𝑥) + 𝑐) ∴ 3 − √−1 or 𝑦(𝑥) = 3 or 𝑦(𝑥) = 3 √𝑥 6 (−3 ln(𝑥)+𝑐) (−1)2/3 √𝑥 6 (−3 ln(𝑥)+𝑐) Question #3 Find the solution of the following Second Order Differential Equations. 4y''+4y'+y=0 Answer: Calculation: Convert 4y''+4y'+y=0 into characteristics equation. 4𝐷2 + 4𝐷 + 1 = 0 (2𝐷)2 + 2 ⋅ 2𝐷 ⋅ 1 + 12 = 0 (2𝐷 + 1)2 = 0 1 𝐷=− 2 𝑥 𝑥 Thus, the solution of differential equation 4𝑦 ′′ + 4𝑦 + 𝑦 = 0 is 𝑦 = 𝐴𝑒 −2 + 𝐵𝑥𝑒 −2 Question #4 Find the differential. 1 a) 𝑧 = (𝑒 𝑥 2 +𝑦 2 2 − 𝑒 −𝑥 2 −𝑦 2 ) b) 𝑤 = 2𝑧 3 𝑦 sin 𝑥 Answer: Step 1 We have to find the differential 2 2 2 2 1 a) 𝑧 = (𝑒 𝑥 +𝑦 − 𝑒 −𝑥 −𝑦 ) 2 Step 2 1 𝑥 2+𝑦2 1 2 2 2 2 2 2 (2𝑥) − 𝑒 −𝑥 −𝑦 (−2𝑥)]𝑑𝑥 + [𝑒 𝑥 +𝑦 (2𝑦) − 𝑒 −𝑥 −𝑦 (−2𝑦)]𝑑𝑦 [𝑒 2 2 2 2 2 2 2 2 2 2 ⇒ 𝑑𝑧 = 𝑥(𝑒 𝑥 +𝑦 + 𝑒 −𝑥 −𝑦 )𝑑𝑥 + 𝑦(𝑒 𝑥 +𝑦 + 𝑒 −𝑥 −𝑦 )𝑑𝑦 ⇒ 𝑑𝑧 = Question #5 A thermometer is taken from an inside room to the outside, where the air temperature is 5∘ 𝐹. After 1 minute the thermometer reads 55∘ 𝐹, and after 5 minutes it reads 30∘ 𝐹 , what is the initial temperature of the inside room? Answer: Let the temperature of the thermometer at any time t is given by: y(t) The temperature of the surrounding is: A 𝑑𝑦 𝑑𝑡 = −𝑘(𝑦 − 𝐴); where k is the constant of proportionality 𝐴=5 𝑑𝑦 = −𝑘(𝑦 − 5) 𝑑𝑡 Solving this differential equation we get: 𝑦(𝑡) = 5 + 𝐶𝑒 −𝑘𝑡 Now, after 1 minutes thermometer reads: 55∘ 𝐹 𝑦(1) = 55 5 + 𝐶𝑒 −𝑘 = 55 𝐶𝑒 −𝑘 = 50(1) Now after 5 minutes thermometer reads: 30∘ 𝐹 𝑦(5) = 30 5 + 𝐶𝑒 −5𝑘 = 30 𝐶𝑒 −𝑘 = 25(2) Now solving equation (1) and (2): 𝐶 = 59.4611 and 𝑘 = 0.1733 Therefore: 𝑦(𝑡) = 5 + 59.4611𝑒 −0.1733𝑡 Hence the initial temperature is: 𝑦(0) = 64.4611∘ 𝐹 Question #6 Solve for the differential equations and get the general solution. Simplify your answer free from In. 𝑑𝑦 = tan 𝑥 tan 𝑦 𝑑𝑥 Answer: 𝑑𝑦 = tan 𝑥 tan 𝑦 𝑑𝑥 𝑑𝑦 = tan 𝑥 𝑑𝑥 tan 𝑦 𝑑𝑦 = tan 𝑥 𝑑𝑥 sin 𝑦 / cos 𝑦 cos 𝑦 𝑑𝑦 = tan 𝑥 𝑑𝑥 sin 𝑦 cos 𝑦 ∫ 𝑑𝑦 = ∫ tan 𝑥 𝑑𝑥 sin 𝑦 ln(sin 𝑦) = − ln(cos 𝑥) + 𝐶 𝑒 ln(sin 𝑦) = 𝑒 − ln(cos 𝑥)+𝐶 −1 sin 𝑦 = 𝑒 ln(cos 𝑥) /𝑒 𝐶 sin 𝑦 = (cos 𝑥)−1 . 𝑒 𝐶 𝑒𝐶 sin 𝑦 = cos 𝑥 𝑦 = sin−1 ( 𝑒𝐶 cos 𝑥 ) Answer: The general solution is 𝑦 = sin−1 ( Question #7 𝑦 ′′ + 5𝑦 ′ − 6𝑦 = 0 Answer: 𝑒𝐶 cos 𝑥 ) Auxiliary equation of given differential equation is 𝑚2 + 5𝑚 − 6 = 0 Now roots of auxiliary equation 𝑚2 + 5𝑚 − 6 = 0 𝑚2 + 6𝑚 − 𝑚 − 6 = 0 𝑚(𝑚 + 6) − 1(𝑚 + 6) = 0 (𝑚 + 6)(𝑚 − 1) = 0 𝑚 = −6,1 So, general solution of given differential equation is 𝑦 = 𝑐1 𝑒 −6𝑥 + 𝑐2 𝑒 𝑥 Question #8 Solve the following differential equations for a general solution: 𝑦 ′′ + 3𝑦 ′ + 2𝑦 = 𝑒 𝑥 Answer: Given differential equation: 𝑦 ′′ + 3𝑦 ′ + 2𝑦 = 𝑒 𝑥 (1) 𝑦 ′′ + 3𝑦 ′ + 2𝑦 = 0(2) The characteristic equation of (2) is 𝑚2 + 3𝑚 + 2 = 0 −3 ± √9 − 8 𝑚= 2 −3 ± 1 𝑚= 2 −2 𝑚= and − 2 2 𝑚 = −1 and − 2 therefore general solution of (2) 𝑦(𝑥) = 𝑐1 𝑒 −𝑥 + 𝑐2 𝑒 −2𝑥 If we assume a particular solution of non homogeneous equation is of the form. 𝑦𝑝 (𝑥) = 𝑘𝑒 𝑥 then 𝑦𝑝 = 𝑘𝑒 𝑥 , 𝑦𝑝′′ = 𝑘𝑒 𝑥 Putting these values in (1), we have 𝑘𝑒 𝑥 + 3(𝑘𝑒 𝑥 ) + 2(𝑘𝑒 𝑥 ) = 𝑒 𝑥 ⇒ (𝑘 + 2𝑘 + 3𝑘)𝑒 𝑥 = 𝑒 𝑥 ⇒ (𝑘 + 2𝑘 + 3𝑘) = 1 ⇒ 6𝑘 = 1 ⇒ 𝑘 = 1/6 1 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑦𝑝 = 𝑒 𝑥 . 6 Hence: General solution of equation (1) is 1 𝑦(𝑥) = 𝑐1 𝑒 −𝑥 + 𝑐2 𝑒 −2𝑥 + 𝑒 𝑥 where 𝑐1 and 𝑐2 are constant. 6 Question #9 Test whether the differential equations are exact and solve those that are 3𝑥 2 𝑦 2 𝑑𝑥 + (2𝑥𝑦 3 + 4𝑦 3 )𝑑𝑦 = 0 Answer: Given differential equation 3𝑥 2 𝑦 2 𝑑𝑥 + (2𝑥𝑦 3 + 4𝑦 3 )𝑑𝑦 = 0(1) Comparing (1) with Mdx+Ndy=0 we get 𝑀 = 3𝑥 2 𝑦 2 , 𝑁 = 2𝑥 3 𝑥 + 4𝑦 3 Now 𝜕𝑀 𝜕𝑦 = 𝜕 𝜕𝑦 (3𝑥 2 𝑦 2 ) = 3𝑥 2 ⋅ 𝜕 𝜕𝑦 (𝑦 2 ) = 3𝑥 2 × 2𝑦 = 6𝑥 2 𝑦 𝜕𝑁 𝜕 𝜕 𝜕 (2𝑥 3 𝑦 + 4𝑦 3 ) = 2𝑦 (𝑥 3 ) + (4𝑦 3 ) = 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 = 2𝑦 × 3𝑥 2 + 0 = 6𝑥 2 𝑦 Since 𝜕𝑀 𝜕𝑦 = 𝜕𝑁 𝜕𝑥 = 6𝑥 2 𝑦 So the differential equation (1) is exact Since the given differential equation exact so the solution ∫ 𝑀𝑑𝑥(𝑦 = constant) + ∫ 𝑁(the term not contained𝑥)𝑑𝑦 = 𝑐 ⇒ ∫ 3𝑥 2 𝑦 2 𝑑𝑥 + ∫ 4𝑦 3 𝑑𝑦 = 𝐶 𝑦 = constant ⇒ 3𝑦 2 ∫ 𝑥 2 𝑑𝑥 + 4 𝑦4 =𝑐 4 𝑥 3 4𝑦 4 + =𝑐 3 4 2 3 4 ⇒𝑦 𝑥 +𝑦 =𝑐 Hence the solution is 𝑦 2 𝑥 3 + 𝑦 4 = 𝑐 ⇒ 3𝑦 2 × Question #10 Find the solution of the following Differential Equations 𝑦 ′′ − 𝑦 ′ − 6𝑦 = 0, 𝑦(0) = 6, 𝑦 ′ (0) = 13. Answer: Given: 𝑦 ′′ − 𝑦 ′ − 6𝑦 = 0 with y(0)=6, y'(0)=13 we put m for first derivative and 𝑚2 for second derivative then we solve it (𝑚2 − 𝑚 − 6)𝑦 = 0 Solving auxilliary equation 𝑚2 − 𝑚 − 6𝑦 = 0 (𝑚 + 2)(𝑚 − 3) = 0 Solution is 𝑦(𝑥) = 𝐴𝑒 −2𝑥 + 𝐵𝑒 3𝑥 With first initial condition y(0)=6 6 = 𝐴 + 𝐵 … (1) 𝑦 ′ (𝑥) = −2𝐴𝑒 −2𝑥 + 3𝐵𝑒 3𝑥 With second initial condition y'(0)=13 -2A+3B=13 (2) solving (1) and (2) A=1, B=5 𝑦 = 𝑒 −2𝑥 + 5𝑒 3𝑥 Question #11 Find the general solution of the following differential equation. 𝑦 ′′ − 6𝑦 ′ + 13𝑦 = 0 Answer: We have to find the general solution of the following differential equation. 𝑦 ′′ − 6𝑦 ′ + 13𝑦 = 0 Let, 𝑦 = 𝐴𝑒 𝑚𝑥 be the trial solution of (1) Then 𝑦 ′ = 𝑚𝐴𝑒 𝑚𝑥 , 𝑦 ′′ = 𝑚2 𝐴𝑒 𝑚𝑥 The auxiliary equation becomes. 𝑚2 − 6𝑚 + 13 = 0 Using quadratic formula, −(−6) ± √(−6)2 − 4 ⋅ 1 ⋅ 13 2(1) 6 ± √36 − 52 ⇒𝑚= 2 6 ± √−16 ⇒𝑚= 2 6 ± 4𝑖 𝑚= 2 𝑚 = 3 ± 2𝑖 So, the general solution is 𝑦 = 𝑒 3𝑥 𝐶4 cos 2 𝑥 + 𝑐2 sin 2 𝑥, 4,2 are arbitrary constants. 𝑚= Question #12 Find the general solution of the following differential equation. 𝑦 ′′ + 2𝑦 ′ + 5𝑦 = 0 Answer: 𝑦 ′′ + 2𝑦 ′ + 5𝑦 = 0 Let, 𝑦 = 𝐴𝑒 𝑚𝑥 be the trial solution at (2) Hence, the auxiliary equation is 𝑚2 + 2𝑚 + 5 = 0 Using quadratic formula 𝑚= −2±√4−20 2 −2 ± √−16 ⇒𝑚= 2 −2 4𝑖 ⇒𝑚= ± (∴ 𝑖 = √−1) 2 2 𝑚 = −1 ± 2𝑖 Therefore, the general solution is 𝑦 = 𝑒 −𝑥 (𝑐1 cos 2 𝑥 + 𝑐2 sin 2 𝑥) Question #13 Solve the following IVP for the exact equation and find the interval of validity for the solution. 2𝑥𝑦 − 9𝑥 2 + (2𝑦 + 𝑥 2 + 1), 𝑑𝑦/𝑑𝑥 = 0, 𝑦(0) = −2 Answer: For any differential equation that is given to be exact, we must have a function f(x,y) such that: 𝜕𝑓 𝜕𝑓 𝑑[𝑓(𝑥, 𝑦)] = 𝑑𝑥 + 𝑑𝑦 = 𝑀𝑑𝑥 + 𝑁𝑑𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑓 𝜕𝑓 ∴ 𝑀(𝑥, 𝑦) = and 𝑁(𝑥, 𝑦) = 𝜕𝑥 𝜕𝑦 𝑖. 𝑒. 𝑓(𝑥, 𝑦) = ∫ 𝑀(𝑥, 𝑦)𝑑𝑥 + 𝜙1 (𝑦) and 𝑓(𝑥, 𝑦) = ∫ 𝑁(𝑥, 𝑦)𝑑𝑦 + 𝜙2 (𝑥) The given differential equation can be written as: 𝑑𝑦 2𝑥𝑦 − 9𝑥 2 + (2𝑦 + 𝑥 2 + 1) =0 𝑑𝑥 2 2 (2𝑥𝑦 − 9𝑥 )𝑑𝑥 + (2𝑦 + 𝑥 + 1)𝑑𝑦 = 0(𝑖) Comparing to the standard form of 𝑀(𝑥, 𝑦)𝑑𝑥 + 𝑁(𝑥. 𝑦)𝑑𝑦 = 0 we get: 𝑀(𝑥, 𝑦) = 2𝑥𝑦 − 9𝑥 2 ∴ 𝑓(𝑥, 𝑦) = ∫(2𝑥𝑦 − 9𝑥 2 )𝑑𝑥 + 𝜙1 (𝑦) 𝑥2 𝑥3 ) − 9 ( ) + 𝜙1 (𝑦) 2 3 = 𝑥 2 𝑦 − 3𝑥 3 + 𝜙1 (𝑦) … (2) 𝑁(𝑥, 𝑦) = 2𝑦 + 𝑥 2 + 1 = 2𝑦 ( ∴ 𝑓(𝑥, 𝑦) = ∫(2𝑦 + 𝑥 2 + 1)𝑑𝑦 + 𝜙2 (𝑥) 𝑦2 ) + 𝑥 2 (𝑦) + (𝑦) + 𝜙2 (𝑥) 2 = 𝑦 2 + 𝑥 2 𝑦 + 𝑦 + 𝜙2 (𝑥) … (3) Comparing the equations (ii) and (iii), we get: 𝑓(𝑥, 𝑦) = −3𝑥 3 + 𝑥 2 𝑦 + 𝑦 2 + 𝑦 From equation (i), we get: 𝑑[−3𝑥 2 + 𝑥 2 𝑦 + 𝑦 2 + 𝑦] = 0 ∴ −3𝑥 3 + 𝑥 2 𝑦 + 𝑦 2 + 𝑦 = 𝑐 Further using the value y(0)=-2 , we get: −3(0)3 + (0)2 (−2) + (−2)2 + (−2) = 𝑐 −3(0) + (0)(−2) + (4) − 2 = 𝑐 𝑐=2 Finally, the solution will be given by: −3𝑥 3 + 𝑥 2 𝑦 + 𝑦 2 + 𝑦 = 2 The solution function is in the implicit form, and can be converted to an explicit form i.e. y=g(x) , as shown below: −3𝑥 3 + 𝑥 2 𝑦 + 𝑦 2 + 𝑦 = 2 𝑦 2 + 𝑦(1 + 𝑥 2 ) = 2 + 3𝑥 3 2 2 1 1 𝑦 2 + (1 + 𝑥 2 )𝑦 + ( (1 + 𝑥 2 )) = 2 + 3𝑥 2 + ( (1 + 𝑥 2 )) 2 2 1 1 (𝑦 + (1 + 𝑥 2 )) = 2 + 3𝑥 3 + (1 + 𝑥 2 )2 2 4 = 2( 2 1 4(𝑥 + 3𝑥 3 ) + (1 + 𝑥 2 )2 (𝑦 + (1 + 𝑥 2 )) = 2 4 1 4(2 + 3𝑥 3 ) + (1 + 𝑥 2 )2 𝑦 = − (1 + 𝑥 2 ) ± √ 2 4 The function will be valid for any value of 𝑥 ≥ 1 as the term under the root i.e. 2 4(2+3𝑥 3 )+(1+𝑥 2 ) 4 is positive for all values of 𝑥 ≥ −1 . Note that for the x-values to the immediate left of x=-1 , the value of the term is -ve. But beyond a certain x-value the term may again be positive. We do not need to consider those intervals as the x-intercept of the data point corresponding to y(0)=-2 i.e. (0,-2) lies in the x-interval [−1, ∞) in which the term is +ve and the solution is valid. Thus, the initial of validity for the given IVP will be [−1, ∞). Question #14 Find the solution of the following Second Order Differential Equation 𝑦 ′′ − 9𝑦 = 0, 𝑦(0) = 2, 𝑦 ′ (0) = 0 Answer: We have given y''-9y=0 Auxiliary equation of given differential equation 𝑚2 − 9 = 0 (𝑚 − 3)(𝑚 + 3) = 0 𝑚 = 3, −3 Therefore, general solution of differential equation 𝑦 = 𝑐1 𝑒 3𝑥 + 𝑐2 𝑒 −3𝑥 We have y(0)=2 ⇒ 𝑦(0) = 𝑐1 𝑒 3(0) + 𝑐2 𝑒 −3(0) 2 = 𝑐1 + 𝑐2 (𝑖) 𝑦 ′ (𝑥) = 3𝑐1 𝑒 3𝑥 − 3𝑐2 𝑒 −3𝑥 We have y'(0)=0 ⇒ 𝑦 ′ (𝑥) = 3𝑐1 𝑒 3(0) − 3𝑐2 𝑒 −3(0) 0 = 3𝑐1 − 3𝑐2 0 = 𝑐1 − 𝑐2 (2) Now, solve equation (1) and (2) we get 𝑐1 = 1, 𝑐2 = 1 Therefore, particular solution is 𝑦 = 𝑒 3𝑥 + 𝑒 −3𝑥 Question #15 Find general solutions (implicit if necessary, explicit if convenient) of the differential equations in Problem. Primes denote derivatives with respect to x. 𝑑𝑦/𝑑𝑥 = (64𝑥𝑦)1/3 Answer: 𝑑𝑦 1 𝑦3 ∫ 1 1 = (64)3 (𝑥 3 ) 𝑑𝑥 𝑑𝑦 1 𝑦3 2 3𝑦 3 1 = ∫ 4 (𝑥 3 ) 𝑑𝑥 4 3 ⋅ 4𝑥 3 = +𝐶 2 4 2 4 3𝑦 3 = 3𝑥 3 + 𝐶 2 2 4 2 𝑦 3 = 2𝑥 3 + 𝐶 3 3 2 2 𝑦= + 𝐶) 3 Here, C is the constant of integration, 2 Suppose is equal to another constant D 4 (2𝑥 3 3 Rewrite the solution as 4 3 3 2 𝑦 = (2𝑥 + 𝐷) which is the general explicit solution the given differential equation. Question #16 Find the solution of the following Differential Equations by Exact (2𝑦 + 𝑥𝑦)𝑑𝑥 + 2𝑥𝑑𝑦 = 0, 𝑦(3) = √2 Answer: Given: (2𝑥 + 𝑥𝑦)𝑑𝑥 + 2𝑥𝑑𝑦 = 0, 𝑦(3) = √2 this equation is form 𝑀(𝑥, 𝑦)𝑑𝑥 + 𝑁(𝑥, 𝑦)𝑑𝑦 = 0 At this equation is exact if and only it 𝜕𝑀 𝜕𝑁 = 𝜕𝑦 𝜕𝑥 𝜕𝑀 𝜕 (2𝑦 + 𝑥𝑦) = 2 + 𝑥 𝑀 = 2𝑦 + 𝑥𝑦 = 𝜕𝑦 𝜕𝑦 𝜕𝑁 =2 𝜕𝑥 𝜕𝑀 𝜕𝑁 Hence ≠ 𝑁 = 2𝑥 𝜕𝑦 𝜕𝑥 Now we find the integrating factor to made this equation exact. 𝑀𝑦 − 𝑁𝑥 = 2 + 𝑥 − 2 = 𝑥 𝑀𝑦 − 𝑁𝑥 𝑥 1 = = 𝑁 2𝑥 2 1 𝑥 Hence I.F. = 𝑒 ∫2𝑑𝑥 = 𝑒 2 ⇒ 𝑒 𝑥/2 (2𝑦 + 𝑥𝑦)𝑑𝑥 + 2𝑒 𝑥/2 𝑥𝑑𝑦 = 0(2) 𝑃𝑆𝑀 = 2𝑒 2/2 𝑦 + 𝑥𝑒 𝑥/2 𝑦 𝑁 = 2𝑒 𝑥/2 𝑥 𝜕𝑀 = 2𝑒 𝑥/2 + 𝑥𝑒 𝑥/2 𝜕𝑦 𝜕𝑁 1 = 2𝑒 𝑥/2 𝑦 + 2𝑥 ⋅ 𝑒 𝑥/2 ⋅ = 2𝑒 𝑥/2 + 𝑥𝑒 𝑥/2 𝜕𝑥 2 𝜕𝑀 𝜕𝑀 ⇒ = 𝜕𝑦 𝜕𝑥 equation (2) is exact equation: Now solution of equation ∫ 𝑀 𝑑𝑥 + ∫ 𝑀 𝑑𝑦 = 𝑐 ∫(2𝑦𝑒 𝑥/2 + 𝑥𝑦𝑒 𝑥/2 )𝑑𝑥 + ∫ 𝑑𝑦 = 𝑐 2𝑦 ∫ 𝑒 𝑥/2 𝑑𝑥 + 𝑦 ∫ 𝑥𝑒 𝑥/2 𝑑𝑥 = 𝑐 𝑒 𝑥/2 𝑒 𝑥/2 𝑑 + 𝑦 [𝑥 ⋅ − ∫ (𝑥) ⋅ 2𝑒 𝑥/2 𝑑𝑥] = 𝑐 1/2 1/2 𝑑𝑥 𝑥/2 𝑒 4𝑦𝑒 𝑥/2 + 𝑦 [2𝑥𝑒 𝑥/2 − 2 ]=𝑐 1/2 4𝑦𝑒 𝑥/2 + 2𝑥𝑦𝑒 𝑥/2 − 4𝑦𝑒 𝑥/2 = 𝑐 2𝑥𝑦𝑒 𝑥/2 = 𝑐 Here 𝑥 = 3, 𝑦 = √2 2 × 3 × √2𝑒 3/2 = 𝑐 𝑐 = 6√2𝑒 3/2 2𝑥𝑦𝑒 𝑥/2 = 6√2𝑒 3/2 𝑥𝑦𝑒 𝑥/2 = 3√2𝑒 3/2 2𝑦 Question #17 𝑦 ′ = cos 2 𝑥 cos 𝑦 Answer: Given, 𝑦 ′ = cos 2 𝑥 ⋅ cos 𝑦 𝑑𝑦 ⇒ = cos 2 𝑥 ⋅ cos 𝑦 𝑑𝑥 𝑑𝑦 ⇒ = cos 2 𝑥 ⋅ 𝑑𝑥 cos 𝑦 Integrate both side, we get 𝑑𝑦 ⇒∫ = ∫ cos 2 𝑥𝑑𝑥 cos 𝑦 1 + cos 2 𝑥 ⇒ ∫ sec 𝑦 ⋅ 𝑑𝑦 = ∫ ( ) 𝑑𝑥 2 1 sin 2 𝑥 ⇒ ln(sec 𝑥 + tan 𝑦) = (𝑥 + )+𝑐 2 2 𝑥 sin 2 𝑥 + +𝑐 2 4 Hence, solution of above differential equation is: 𝑥 sin 2 𝑥 ln(sec 𝑦 + tan 𝑦) = + +𝑐 2 4 (sec2 𝑦+sec 𝑦 tan 𝑦) sec 𝑦(sec 𝑦+tan 𝑦) 𝑑𝑦 = ∫ 𝑑𝑦 ∫ sec 𝑦 𝑑𝑦 = ∫ (sec ⇒ ln(sec 𝑦 + tan 𝑥) = 𝑦+tan 𝑦) sec 𝑦+tan 𝑦 Let, 𝑣 = sec 𝑦 + tan 𝑦 differentiate both side, we get 𝑑𝑣 = (sec 𝑦 ⋅ tan 𝑦 + sec 2 𝑦)𝑑𝑦 1 ∫ 𝑑𝑣 = ln 𝑣 + 𝑐 𝑣 = ln(sec 𝑦 + tan 𝑦) + 𝑐 ∫ sec 𝑦 𝑑𝑦 = ln(sec 𝑦 + tan 𝑦) + 𝑐 Question #18 𝑑𝑦 + 5𝑦 = 15 𝑑𝑥 Answer: 𝑑𝑦 + 5𝑦 = 15 𝑑𝑥 𝑑𝑦 = 15 − 5𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑦 = 𝑑𝑥 ⇒ = 𝑑𝑥 15 − 5𝑦 5(3 − 𝑦) 𝑑𝑦 ⇒∫ = ∫ 5𝑑𝑥 3−𝑦 ⇒ − log(3 − 𝑦) = 5𝑥 + log 𝑐 ⇒ −5𝑥 = log(3 − 𝑦) + log 𝑐 = log 𝑐 (3 − 𝑦) ⇒ 𝑐(3 − 𝑦) = 𝑒 −5𝑥 𝑒 −5𝑥 𝑒 −5𝑥 3−𝑦 = ⇒𝑦 = 3− 𝑐 𝑐 𝑒 −5𝑥 𝑦 = 3− 𝑐 Question #19 Identify the following linear differential equation. 𝑥 2𝑦′ + 𝑦2 = 0 Answer: Given 𝑥 2 𝑦 ′ + 𝑦 2 = 0 𝑥 2𝑦′ + 𝑦2 = 0 𝑑𝑦 𝑥2 = −𝑦 2 𝑑𝑥 𝑑𝑥 𝑑𝑦 ⇒ 2 =− 2 𝑥 𝑦 𝐼𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑜𝑛 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 𝑑𝑥 𝑑𝑦 ⇒ ∫ 2 = −∫ 2 + 𝐶 𝑥 𝑦 ⇒ ∫ 𝑥 −2 𝑑𝑥 = − ∫ 𝑦 −2 𝑑𝑦 + 𝐶 𝑥 −2+1 𝑦 −2+1 = −( )+𝐶 −2 + 1 −2 + 1 𝑥 −1 𝑦 −1 ⇒ = −( )+𝐶 −1 −1 1 1 ⇒− = +𝐶 𝑥 𝑦 1 1 ⇒ + +𝐶 =0 𝑥 𝑦 ⇒ Question #20 (𝑦 − 𝑥𝑦 2 )𝑑𝑥 + (𝑥 + 𝑥 2 𝑦 2 )𝑑𝑦 = 0; when 𝑦(1) = 1 Answer: Solution: (4 − 𝑥𝑦 2 )𝑑𝑥 + (𝑥 + 𝑥 2 𝑦 2 )𝑑𝑥 = 0 (4𝑑𝑥 + 𝑥𝑑𝑦) − (𝑥𝑦 2 𝑑𝑦 − 𝑥 2 𝑦 2 ) = 0 𝑑𝑥 𝑑(𝑥𝑦) − (𝑥𝑦)2 [ − 𝑑𝑦] = 0 𝑥 𝑑(𝑥𝑦) 𝑑𝑥 ∫ −∫ + ∫ 𝑑𝑦 = 0 (𝑥𝑦)2 𝑥 1 ⇒− − ln 𝑥 + 𝑦 = 𝐶 𝑥⋅𝑦 put 4(1)=1 1 − − ln 1 + 1 = 𝐶 ⇒ −1 − 0 + 1 = 𝐶 1⋅1 𝑐=0 1 Hence, − ⋅ ln 𝑥 + 𝑦 = 0 𝑥⋅𝑦