Department of Mathematics Jain Global campus, Jakkasandra Post, Kanakapura Taluk, Ramanagara District -562112 Mathematics-I Dr Kokila Ramesh HoD, Department of Mathematics Jain (Deemed-to-be-University) 1 Differential Equations • Exact and Reducible to Exact differential equations • Linear and Bernoulli’s differential equations • Higher order linear differential equations with constant Coefficients • Method of variation of parameters • Cauchy and Legendre’s differential equations Department of Mathematics, Jain (Deemed-to-be-University) 2 Introduction β’ A differential equation in which the dependent variable and its derivatives occur only in the first degree and are not multiplied together is called a linear differential equation. β’ The linear differential equation with constant co-efficient of ππ‘β order is of the form ππ π¦ π π−1 π¦ π π−2 π¦ π0 π + π1 π−1 + π2 π−2 +. . . . . . . . . . . . . +ππ π¦ = π π₯ … … … (1) ππ₯ ππ₯ ππ₯ Where, π0, π1, π2 , π3 , . . . . . . . ππ are constants and Q(π₯) is functions of π₯ only. β’ If π(π₯) = 0, then the equation (1) is known as homogeneous differential equation, otherwise, it is a non-homogeneous differential equation. β’ Operator D: Let us denote ∴ π ππ₯ ππ¦ ππ₯ = π·, = π·π¦, π2 ππ₯ 2 ππ¦ 2 ππ₯ 2 = = ππ 2 π· ,…………., ππ₯ π 3 ππ¦ π·2 π¦,ππ₯ 3 = π·π = π·3 π¦ … … . . Equation (1) becomes (π0 π·π + π1 π·π−1 +. . . . . +ππ )π¦ = π(π₯), i.e.π(π·)π¦ = π(π₯) where,π(π·) = π0 π·π + π1 π·π−1 +. . . . . +ππ , is a polynomial in D β’ The symbol D is called the differential operator and it stands for the operation of differentiation. Department of Mathematics, Jain (Deemed-to-be-University) 3 Solution of the HDE β’ Solution of the differential equation is defined as the function of the independent variables, which satisfies the given differential equation. β’ The General Solution of the higher order linear differential equation with constant coefficients of the form π(π·)π¦ = π(π₯) is given by CF+PI, where CF=complementary function and PI-Particular Integral β’ The Complementary Function of the differential equation is found using the homogeneous part of the equation π π· π¦ = 0 β’ The Particular Integral of the differential equation is found using the nonhomogeneous part of the equation π(π₯) Department of Mathematics, Jain (Deemed-to-be-University) 4 Continued…. Rules for finding the Complementary Function: Let π π = 0 be the auxiliary equation using which we find CF Case(i): Let π1 , π2 , π3 … … ππ be the roots of the auxiliary equation which are real and distinct then CF is given by πΆπΉ = π1 π π1π₯ + π2 π π2π₯ +. . . . . . . . . . . . . . . . +ππ π πππ₯ Case(ii): Let π1 = π2 = π3 = β― = ππ = π be the roots of the auxiliary equation which are real and equal then CF is given by πΆπΉ = (π1 + π2 π₯+. . . . . . . . . . . . . . . . +ππ π₯ π−1 )π ππ₯ Case(iii): Let one pair of the roots are complex π1 , π2 = πΌ ± ππ½ and rest of the roots are real and distinct, then CF is given by πΆπΉ = π πΌπ₯ π1 πππ π½π₯ + π2 π πππ½π₯ +. . . . . . . . . . . . . . . . +ππ π πππ₯ Case(iii): Let two pair of the roots are complex and equal ie π1 , π2 = πΌ ± ππ½, π3 , π4 = πΌ ± ππ½ and rest of the roots are real and distinct, then CF is given by πΆπΉ = π πΌπ₯ (π1 +π2 π₯)πππ π½π₯ + (π3 +π4 π₯)π πππ½π₯ +. . . . . . . . . . . . . . . . +ππ π πππ₯ Department of Mathematics, Jain (Deemed-to-be-University) 5 Continued…. Rules for finding the Particular Integral: ππΌ = π 1 π· π π₯ Case(i): π π₯ = π ππ₯ 1 1 ππ₯ 1 ππ₯ ππΌ = π π₯ = π = π if π(π) ≠ 0 π π· π π· π(π) If π(π) = 0, then If π′(π) =0, we get 1 π ππ₯ π(π·) 1 = π₯ π′(π) π ππ₯ if π′(π) ≠ 0 1 ππ₯ π π(π·) 1 = π₯ 2 π′′(π) π ππ₯ , provided π′′(π) ≠ 0 and so on. Case(ii): π π₯ = sin ππ₯ + π ππ cos(ππ₯ + π) 1 1 ππΌ = π ππ ππ₯ + π = π ππ ππ₯ + π ππ π −π2 ≠ 0 2 2 π(π· ) π(−π ) 1 If π −π2 = 0, π‘βππ ππΌ = π₯ π′(−π2) π ππ( ππ₯ + π) if π ′ −π2 ≠ 0 1 If π ′ −π 2 = 0, π‘βππ ππΌ = π₯ 2 π′′(−π2) π ππ( ππ₯ + π) if π ′′ −π2 ≠ 0 and so on Department of Mathematics, Jain (Deemed-to-be-University) 6 Continued…. Case(iii): π π₯ = π₯ π 1 π₯ π = [π(π·)]−1 π₯ π. π(π·) Expand [π(π·)]−1 in ascending powers of D as far as the term in π·π and operate on π₯ π term by term. Since (π + 1)π‘β and the higher derivatives of π₯ π are zero, we need not consider terms by π·π . ππΌ = Important Formulae: 1. (1 − π·)−1 = 1 + π· + π·2 +. . . . . . . . . . . . 2. (1 + π·)−1 = 1 − π· + π·2 −. . . . . . . . . . . . 3. (1 − π·)−2 = 1 + 2π· + 3π·2 + 4π·3 . . . . . . . . . . . . 4. (1 + π·)−2 = 1 − 2π· + 3π·2 − 4π·3 . . . . . . . . . . . . Case(iv): π π₯ = π ππ₯ π, π€βπππ π ππ sin ππ₯ + π ππ cos ππ₯ + π πππ₯ π 1 ππΌ = π ππ₯ π π(π·) In such cases, first take π ππ₯ term outside the operator, by shifting D to (π· + π). 1 ⇒ ππΌ = π ππ₯ π(π·+π) π Depending upon the nature of π we will solve further. 7 Continued… Working procedure to solve Higher order Linear Differential equations with constant coefficients β’ Write the given differential equation in operator D form β’ Identify the auxiliary equation π π = 0 and find the roots of the equation β’ Based on the nature of roots of the auxiliary equation, write CF β’ If the given DE is homogeneous in nature, then Complete Solution(CS)=CF β’ If the given DE is non-homogeneous in nature, then Complete Solution(CS)=CF+PI, where PI is found based on the RHS of the given equation 8 Problems Problem 1: Solve π·2 + 4 π¦ = π 4π₯ Solution: The auxiliary equation(AE) is π2 + 4 = 0 The roots of the AE are π = ±2π Hence the complementary function is π¦π ππ πΆπΉ = π1 πππ 2 π₯ + π2 π ππ 2 π₯ The particular integral is π¦π ππ ππΌ = π 1 π· 1 π(π₯) = (π·2+4) π 4π₯ Replacing D by 4, we get ππΌ = 1 1 4π₯ 4π₯ π = π (42 + 4) 20 ∴The Complete Solution(CS) π¦ = πΆπΉ + ππΌ π¦ = π1 πππ 2 π₯ + π2 π ππ 2 π₯ + 1 4π₯ π 20 9 Continued… Problem 2: Solve π·2 − 2π· + 2 π¦ = πππ π₯ − 1 Solution: The A.E is π2 − 2π + 2 = 0 π = 1 ± π are the roots of the AE Therefore πΆπΉ = π π₯ π1 πππ π₯ + π2 π πππ₯ 1 1 ππΌ = π π₯ = 2 πππ π₯ − 1 π π· π· − 2π· + 2 Replace π·2 by −12 , we get 1 1 π¦π = πππ π₯ − 1 = πππ π₯ − 1 −1 − 2π· + 2 1 − 2π· 1 + 2π· 1 + 2π· = πππ π₯ − 1 = πππ π₯ − 1 1 − 2π· 1 + 2π· 1 − 4π·2 Now replace π·2 by −12 , we get 1 + 2π· 1 = πππ π₯ − 1 = 1 + 2π· πππ π₯ − 1 5 1 − 4 −1 1 ∴ ππΌ = πππ π₯ − 1 − 2 π ππ π₯ − 1 5 ∴ πΆπ π¦ = πΆπΉ + ππΌ 1 π₯ π¦ = π π1 πππ π₯ + π2 π πππ₯ + πππ π₯ − 1 − 2 π ππ π₯ − 1 5 Department of Mathematics, Jain (Deemed-to-be-University) 10 Continued… Problem 3: Solve π·2 + π· π¦ = π₯ 2 + 2π₯ + 25 Solution: π΄πΈ ππ π2 + π = 0 π = 0, −1 Therefore πΆπΉ = π1 π 0π₯ + π2 π −π₯ 1 1 ππΌ = π π₯ = 2 π₯ 2 + 2π₯ + 25 π π· π· +π· 1 = (1 + π·)−1 π₯ 2 + 2π₯ + 25 π· 1 = (1 − π· + π·2 β β― … . . ) π₯ 2 + 2π₯ + 25 π· 1 2 = π₯ + 2π₯ + 25 − 2π₯ + 2 + 2 π· 3 π₯ = ΰΆ± π₯ 2 + 25 ππ₯ = + 25π₯ 3 π₯3 ππΌ = + 21π₯ 3 Hence CS y=CF+PI π₯3 0π₯ −π₯ π¦ = π1 π + π2 π + + 25π₯ 3 Department of Mathematics, Jain (Deemed-to-be-University) 11 Continued… Problem 4: Solve (π·2 − 2π· + 4)π¦ = π π₯ πππ π₯ . Solution: π΄πΈ ππ π2 − 2π + 4 = 0 π =1±π 3 Hence πΆπΉ = π π₯ (π1 πππ 3π₯ + π2 π ππ 3π₯) 1 ππΌ = 2 π π₯ πππ π₯ . (π· − 2π· + 4) Shift D to D+1 we get 1 π₯ ππΌ = π πππ π₯ (π· + 1)2 − 2(π· + 1) + 4 1 π₯ =π 2 πππ π₯ π· +3 Now replace π·2 by −12 1 1 π₯ πππ π₯ ππΌ = π π₯ πππ π₯ = π −12 + 3 2 Hence CS y=CF+PI 1 π¦ = π π₯ π1 πππ 3π₯ + π2 π ππ 3π₯ + π π₯ πππ π₯ 2 Department of Mathematics, Jain (Deemed-to-be-University) 12 2 Continued… Problem 5: Solve (π· + 5π· − 6)π¦ = πππ 4 π₯ πππ π₯ Solution: AE is π2 + 5π − 6 = 0 Roots are π = 1, −6 CF = π1 π π₯ + π2 π −6π₯ 1 1 1 ππΌ = 2 πππ 4 π₯ πππ π₯ = 2 πππ 5 π₯ + πππ 3 π₯ π· + 5π· − 6 (π· + 5π· − 6) 2 1 1 1 1 = πππ 5 π₯ + 2 πππ 3 π₯ = π1 + π2 2 2 (π· + 5π· − 6) (π· + 5π· − 6) 2 1 Consider π1 = 2 πππ 5 π₯ π· +5π·−6 Replacing π·2 by−52 , we get 1 1 5π· + 31 5π· + 31 π1 = πππ 5 π₯ = πππ 5 π₯ = πππ 5 π₯ = πππ 5 π₯ −25 + 5π· − 6 5π· − 31 5π· − 31 5π· + 31 25π·2 − 961 Now replacing π·2 by−52 , we get 5π· + 31 1 1 π1 = πππ 5 π₯ = − −25 π ππ 5 π₯ + 31 πππ 5 π₯ = 25 π ππ 5 π₯ − 31 πππ 5 π₯ −625 − 961 1586 1586 1 Now consider π2 = 2 πππ 3 π₯ π· +5π·−6 2 2 Replacing π· by−3 , we get π2 = 1 (−9+5π·−6) πππ 3 π₯ = 1 5π·−15 πππ 3 π₯ = π·+3 5 π·−3 π·+3 πππ 3 π₯ = 1 π·+3 5 π·2 −9 πππ 3 π₯ Replacing π·2 by −32 ,we get 1 π·+3 −1 3 3 π2 = πππ 3 π₯ = π· + 3 πππ 3 π₯ = π ππ 3 π₯ − πππ 3 π₯ 5 −9 − 9 90 90 90 1 1 ∴ π2 = π ππ 3 π₯ − πππ 3 π₯ 30 30 Department of Mathematics, Jain (Deemed-to-be-University) 13 Thank you! Dr Kokila Ramesh Department of Mathematics Department of Mathematics, Jain (Deemed-to-be-University) 14
