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Mathematics: RC6901B46 1 Mathematics : MTH - 102 TERM PAPER Give a short summary of the method of solution of exact differential equation. Using equation of your own choice. Show that differential equations can be solved by more than one method. Compare the amount of work involved in each case. Submitted By :- saurabh srivastava Acknowledgment The completion of any term paper requires coordination ,dedication and a combined effect of various sources. First of I Thank God for making me capable of completing this term paper. Thereafter I Mathematics: RC6901B46 2 thank Mr. Gurpindar our Math lecturer for spapring his valuable precious time in guiding me in this term paper. CONTENTS Differential Equations Order & Degree Mathematics: RC6901B46 3 Exact Differential Equation Method For Solving Exact Differential equation Equations Reducible To Exact Equations Other Methods Of Solving Differential equations Linear Differential equation Homogeneous differential equation Reducible To Homogeneous Form Reference Differential Equations :A differential equation is an equation which involves differential coefficients or differentials. Ex :- 𝑒 𝑥 𝑑𝑥 + 𝑒 𝑦 𝑑𝑦 = 0 d2x/dx2 + n2x = 0 Differntial equation in which all the differential coefficients have reference to a single independent variable is known as an ordinary differential equation. Order & Degree :- Mathematics: RC6901B46 4 The order of a differential equatio is the order of thee highest derivative appearing . The degree of a differential equation is the degree of the highest derivative occurring in it, after the equation has been expressed in a form free of radicalsand fractions. Ex :- 𝑒 𝑥 𝑑𝑥 + 𝑒 𝑦 𝑑𝑦 = 0 Order :- 1 Degree :- 1 d2x/dx2 + n2x = 0 Order :- 2 Degree :- 1 Exact Differential Equation :An exact differential equation is formed directly differentiating its primitive(solution) without any other process. Mdx + Ndy = 0 Is said to be an exact differential equation if it satisfies the following condition 𝝏𝑴 𝝏𝒚 Where 𝝏𝑴 𝝏𝒚 constant and = 𝝏𝑵 𝝏𝒙 denotes the differential coefficient of M with respect to y keeping x 𝝏𝑵 𝝏𝒙 , the differential coefficient of N with respect to x, keeping y constant. Method For Solving Exact Differential Equation :Step 1 : Integrate M w.r.t. x keeping y constant. Mathematics: RC6901B46 Step 2 : Integrate y, only those terms of N which do not contain x. Step 3 : Result of 1 + Result of 2 = constant. ∫ 𝑀𝑑𝑥 + ∫(𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑁 𝑛𝑜𝑡 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝑥)𝑑𝑦 = 𝐶 (y constant) Ex :- Solve (5𝑥 4 + 3𝑥 2 𝑦 2 − 2𝑥𝑦 3 )𝑑𝑥 + (2𝑥 3 𝑦 − 3𝑥 2 𝑦 2 − 5𝑦 4 )𝑑𝑦 = 0 Solution :- Here M = 5𝑥 4 + 3𝑥 2 𝑦 2 − 2𝑥𝑦 3 N = 2𝑥 3 𝑦 − 3𝑥 2 𝑦 2 − 5𝑦 4 𝜕𝑀 𝜕𝑦 Since Now 𝜕𝑁 = 6𝑥 2 𝑦 − 6𝑥𝑦 2 ∂M ∂y = ∂N ∂x 𝜕𝑥 = 6𝑥 2 𝑦 − 6𝑥𝑦 2 , the given equation is exact. ∫ 𝑀𝑑𝑥 + ∫(𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑁 𝑛𝑜𝑡 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑖𝑛𝑔 𝑥)𝑑𝑦 = 𝐶 (y constant) ∫(5𝑥 4 + 3𝑥 2 𝑦 2 − 2𝑥𝑦 3 )𝑑𝑥 + ∫ −5 𝑦 4 𝑑𝑦 = 𝐶 𝑥 5 + 𝑥 3𝑦2 − 𝑥 2𝑦3 − 𝑦5 = 𝐶 Ans. Equations reducible to The Exact Equations :Sometimes a differential equation which is not exact may become so, on multiplication by a suitable function known as the integrating factor. RULE 1 :If ( 𝜕𝑀 𝜕𝑦 − 𝜕𝑁 𝜕𝑥 )/N is a function of x alone, say f(x), then I F = 𝒆∫ 𝒇(𝒙)𝒅𝒙 RULE 2 :- 5 Mathematics: RC6901B46 If ( 𝜕𝑁 𝜕𝑥 − 𝜕𝑀 𝜕𝑦 6 )/𝑀𝑖𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑦 𝑎𝑙𝑜𝑛𝑒, 𝑠𝑎𝑦 𝑓(𝑦), 𝑡ℎ𝑒𝑛 I F = 𝒆∫ 𝒇(𝒚)𝒅𝒚 RULE 3 :If M is of the form M = 𝑦𝑓 1(𝑥𝑦) and N is of the form N = 𝑥𝑓 2(𝑥𝑦) then, IF = 𝟏 𝑴𝒙−𝑵𝒚 RULE 4 :Homogeneous equation . If 𝑀𝑑𝑥 + 𝑁𝑑𝑦 = 0 be a homogeneous equation in x and y, then IF = 𝟏 𝑴𝒙+𝑵𝒚 Other Methods Of Solving Differential Equations :Linear Differential Equation:A differential equation is said to be linear if the dependent variable and its differential coefficients occur only in the first degree and not multiplied together. Thus the standard form of a linear equation of the first order, commonly known as Leibnitz’s linear equations, is 𝑑𝑦 𝑑𝑥 + 𝑃𝑦 = 𝑄 Where P, Q are the functions of x. To solve the equation, multiply both sides by IF = 𝒆∫ 𝑷𝒅𝒙 𝐝𝐲 𝐝𝐱 i.e 𝐞∫ 𝐏𝐝𝐱 + 𝐲𝐞∫ 𝐏𝐝𝐱 = 𝐐𝐞∫ 𝐏𝐝𝐱 𝐝 𝐝𝐱 (𝐲𝐞∫ 𝐏𝐝𝐱 ) = 𝐐𝐞∫ 𝐏𝐝𝐱 Mathematics: RC6901B46 Integrating both sides , we get y𝒆∫ 𝑷𝒅𝒙 = ∫ 𝑸𝒆∫ 𝑷𝒅𝒙 𝒅𝒙 + 𝑪 as the required solution. Ex :- Solve (𝑥 + 1) 𝑑𝑦 𝑑𝑥 − 𝑦 = 𝑒 3𝑥 (𝑥 + 1)2 Sol. Dividing throughout by (𝑥 + 1), given equation becomes 𝑑𝑦 𝑑𝑥 − 𝑦 𝑥+1 Here P = − = 𝑒 3𝑥 (𝑥 + 1) which is Leibnitz’s equation. 1 𝑥+1 and ∫ 𝑃𝑑𝑥 = − log(𝑥 + 1) = log(𝑥 + 1)-1 Therefore, IF = 𝐞∫ 𝐏𝐝𝐱 1 = 𝑒 𝑙𝑜𝑔𝑥+1 = 1 𝑥+1 Thus the solution of above equation is 𝑦(𝐼𝐹) = ∫[𝑒 3𝑥 (𝑥 + 1)](𝐼𝐹) 𝑑𝑥 + 𝑐 Or 𝑦 𝑥+1 = ∫ 𝑒 3𝑥 𝑑𝑥 1 = 𝑒 3𝑥 + 𝑐 3 Or 1 𝑦 = ( 𝑒 3𝑥 + 𝑐) (𝑥 + 1) 3 Homogeneous Equations :- Ans. 7 Mathematics: RC6901B46 These are of the form 𝑑𝑦 𝑑𝑥 = 8 𝑓(𝑥,𝑦) ∅(𝑥,𝑦) Where 𝑓(𝑥, 𝑦)𝑎𝑛𝑑 ∅(𝑥, 𝑦)are homogeneous functions of the same degree in x and y. To solve homogeneous equations (i) Put y=vx Then 𝑑𝑦 𝑑𝑥 =𝑣+𝑥 𝑑𝑣 𝑑𝑥 (ii) Separate the variables v and x, and integrate. Equations Reducible To Homogeneous Form:The equation of the form 𝑑𝑦 𝑑𝑥 = 𝑎𝑥+𝑏𝑦+𝑐 𝑎′ 𝑥+𝑏′ 𝑦+𝑐′ Can be reduced to the homogeneous form as follows:Case 1 :𝑎 𝑎′ = 𝑏 𝑏′ Putting 𝑥 = 𝑋 + ℎ , 𝑦 = 𝑌+k (h,k being constants) So that 𝑑𝑥 = 𝑑𝑋 𝑑𝑦 = 𝑑𝑌 Now the given equation becomes 𝑑𝑌 𝑑𝑋 = 𝑎𝑋+𝑏𝑌+(𝑎ℎ+𝑏𝑘+𝑐) 𝑎′ 𝑋+𝑏′ 𝑌+(𝑎′ ℎ+𝑏′ 𝑘+𝑐 ′ ) ----------------------(1) Choose h,k so that above equation may become homogeneous Put 𝑎ℎ + 𝑏𝑘 + 𝑐 = 0 Mathematics: RC6901B46 𝑎′ ℎ + 𝑏 ′ 𝑘 + 𝑐 ′ = 0 ℎ So that Or = 𝑏𝑐 ′ −𝑏′𝑐 𝑘 𝑐𝑎′ −𝑐′𝑎 = 1 𝑎𝑏′ −𝑏𝑎′ 𝑏𝑐 ′ −𝑏′𝑐 ℎ= 𝑎𝑏′ −𝑏𝑎′ 𝑐𝑎′ −𝑐′𝑎 𝑘= 𝑎𝑏′ −𝑏𝑎′ Thus when 𝑎𝑏 ′ − 𝑏𝑎′ ≠ 0 (1) becomes 𝑑𝑋 = 𝑑𝑌 𝑎𝑋+𝑏𝑌 𝑎′ 𝑋+𝑏′𝑌 Which is homogeneous in X,Y and can be solved by putting 𝑌 = 𝑣𝑋 Case 2 :When 𝑎 𝑎′ ≠ 𝑏 𝑏′ i.e. 𝑎𝑏 ′ − 𝑏𝑎′ = 0, the above method fails as h and k become infinite or indeterminate. Now, 𝑎 𝑎′ = 𝑏 𝑏′ = 1 𝑚 (say) 𝑎′ = 𝑎𝑚 𝑏 ′ = 𝑏𝑚 Then given equation becomes 𝑑𝑦 𝑑𝑥 (𝑎𝑥=𝑏𝑦)+𝑐 = (𝑎𝑥+𝑏𝑦)𝑚+𝑐′ Put 𝑎𝑥 + 𝑏𝑦 = 𝑡, so that 𝑎+𝑏 𝑑𝑦 𝑑𝑥 = 𝑑𝑡 𝑑𝑥 -----------------------------------(2) 9 Mathematics: RC6901B46 𝑑𝑦 Or 1 10 𝑑𝑡 = ( − 𝑎) 𝑏 𝑑𝑥 𝑑𝑥 Therefore (2) becomes 1 𝑑𝑡 𝑡+𝑐 ( − 𝑎) = 𝑚𝑡=𝑐′ 𝑏 𝑑𝑥 𝑑𝑡 Or =𝑎+ 𝑑𝑥 = 𝑏𝑡+𝑏𝑐 𝑚𝑡+𝑐 ′ (𝑎𝑚+𝑏)𝑡+𝑎𝑐 ′ +𝑏𝑐 𝑚𝑡+𝑐′ So that the variables are separable. In this solution , putting 𝑡 = 𝑎𝑥 + 𝑏𝑦, we get the required solution for the given differential equation. Ex :- Solve (3𝑦 + 2𝑥 + 4)𝑑𝑥 − (4𝑥 + 6𝑦 + 5)𝑑𝑦 = 0 Sol. Given equation is 𝑑𝑦 𝑑𝑥 = Putting 2𝑥 + 3𝑦 = 𝑡 So that 2+3 𝑑𝑦 = 𝑑𝑥 𝑑𝑡 𝑑𝑥 Therefore (1) becomes 1 𝑑𝑡 𝑡+4 ( − 2) = 2𝑡+5 3 𝑑𝑥 Or 𝑑𝑡 𝑑𝑥 =2+ = Or 2𝑡+5 7𝑡+22 3𝑡+12 2𝑡+5 7𝑡+22 2𝑡+5 𝑑𝑡 = 𝑑𝑥 Integrating both sides, (2𝑥+3𝑦)+4 2(2𝑥+3𝑦)+5 ------------------------(1) Mathematics: RC6901B46 11 2𝑡+5 ∫ 7𝑡+22 𝑑𝑡 = ∫ 𝑑𝑥 + 𝑐 2 9 1 Or ∫ ( − . ) 𝑑𝑡 = 𝑥 + 𝑐 7 7 7𝑡+22 Or 2 7 𝑡− 9 49 log(7𝑡 + 22) = 𝑥 + 𝑐 Putting 𝑡 = 2𝑥 + 3𝑦, we have 14(2𝑥 + 3𝑦) − 9𝑙𝑜𝑓(14𝑥 + 21𝑦 + 22) = 49𝑥 + 49𝑐 Or 21𝑥 − 42𝑦 + 9 log(14𝑥 + 21𝑦 + 22) = 𝑐′ which is the required solution. Work Comparison:There is no much difference in the work done in all the methods. Just the main task is to find out the solution. In Exact differential equation type mainly the its type should be satisfied. While in others also their type should be fulfilled first then only the appropriate method can be applied. In Linear equations main concentration is on Integrating Factor while in Homogeneous Equation substituting of the variables with certain appropriate term is done so that we can solve it very easily. For equations reducible to Homogeneous form its two cases should be accomplished first then only the method of homogeneous equation is applied. Reference: http://en.wikipedia.org/exactdifferentialequation http://www.mathforu.com http://www.mathsworld.com Mathematics: RC6901B46 Books: Engineering mathematics ----B.S. Grewal Engineering mathematins-----H. K. Dass Differential Calculus-------Narayana shanty Submitted for – www.mycollegebag.in 12