TAM3B DIFFERENTIAL EQUATIONS Unit : I to V

TAM3B DIFFERENTIAL EQUATIONS Unit : I to V

Unit I -Syllabus Homogeneous Functions and examples Homogeneous Differential Equations Exact Equations First Order Linear Differential Equations Reduction of order TAM3B-Differential Equations 2

Homogeneous Function A functionf(x,y)iscalledhomogeneous n of degree nif f(tx,ty) = t f(x,y) Example1: 2 f(x,y) = x + xy isahomogeneousfunctionof degree2 Example2 : f(x,y) = sin(x / y) isahomogeneousfunctionof degree0 TAM3B-Differential Equations 3

Homogeneous Differential Equations ThedifferentialequationM( x,y) dx +N( x,y) dy = 0 issaidtobehomogeneous if MandNarehomogeneousfunctions of samedegree Theequationcanbe writteninthe form dy = f ( x,y ) dx -M x,y where f ( x,y ) =,isahomogeneousof degree0 N x,y ( ) ( ) TAM3B-Differential Equations 4

Procedure for Solving Homogeneous Equation dy Step1:Rewrite thegivenequationinthe form = dx dy dz Step2:Put y = zx and = z + x dx dx Step3:Seperate the variableszandx Step 4:Integrateandget zandx y Step5:Replace zby x ( ) ( ) -M x,y N x,y TAM3B-Differential Equations 5

Illustration Problemsforpractice Problem1:Solve(x + y)dx -(x - y)dy = 0 2 2 Problem2:Solve(x - 2y )dx + xydy = 0 Problem3:Solvexy = 2x +3y Problem4:Solvex 2 y = y 2 + 2xy TAM3B-Differential Equations 6

Exact Equation M The expressionmdx + Ndyissaidtobeexact if = y Example: 2 x + dy ydx 0 y + = 2 M N M= yn, = x+ = 1, = 1 y y x M N = Theequationisexact y x N x TAM3B-Differential Equations 7

Solving an Exact Equation Step 1: Check whether the given equation is exact or not. If exact proceed to step 2 Step2: I1 = M ( x, y) dx, treatingyasconstant Step3: I2 = N( x, y) dy, Neglectingthe'x'terms Step4:Solutionis I1+ I2 = c TAM3B-Differential Equations 8

Illustration Problemsforpractice Problem1:Test theequationfor exactnessand solve it if it is exact e dx + ( xe + 2 y) dy = 0 y Problem2:Test theequationfor exactnessand solve it if it is exact ( y + y cos xy) dx + ( x + xcos xy) dy = 0 Problem 3:Test theequationfor exactnessand 2 solve it if it is exact (2y 4x + 5) dx (4 2y + 4 xy) dy = 0 y TAM3B-Differential Equations 9

Solving First Order Linear Equation The general first order linear equation is dy Pxy ( ) Qx ( ) dx + = Step1:Integrating Factor (. IF) = e Pdx Step2:Find Q(. I F) dx Step 3: Solution is yif (. ) = QIF (. ) dx+ c TAM3B-Differential Equations 10

Illustration Problems for practice dy Problem1: Solve x 3y = x dx 4 Problem2: Solve (2 y x ) dx = xdy 3 Problem3: Solve 1 y + y = 1 + e 2x Problem4: Solve (1 + x ) dy + 2xydx = cot xdx 2 TAM3B-Differential Equations 11

Reduction of Order The general second order differential equation has the form Fxyy (,,, y ) = 0 There are two special types of second order equations that can be solved by first order methods. Type 1: Dependent variable missing Type 2: Independent variable missing TAM3B-Differential Equations 12

Dependent Variable Missing If y is not present then the general second order Differential equation can be written as f( xy,, y ) = 0 dp Step1:Put y = p and y = dx Step2:Substitute in the given equation and seperate the variables Step3:Integrate and get the value of p in terms of x dy Step4:Replace pby and seperate the variables yand x dx Step 5:Integrate to get the solution. TAM3B-Differential Equations 13

Independent Variable Missing If x is not present then the general second order Differential equation can be written as gyy (,, y ) = 0 dp Step1:Put y = p and y = p dy Step2:Substitute in the given equation and seperate the variables Step3:Integrate and get the value of p in terms of y dy Step4:Replace pby and seperate the variables yand x dx Step 5:Integrate to get the solution. TAM3B-Differential Equations 14

Problems for practice 2 Problem1: Solve yy + ( y ) = 0 Problem2: Solve xy = y + ( y ) 3 2 = Problem3: Solve y k y 0 Problem4: Solve xy + y = 4x TAM3B-Differential Equations 15

Problems for practice Problem5: Solve xy = y + 2xe y x Problem6: Check for exactness and solve if it is exact 2 cos xcos ydx + 2sin xsin y cos ydy = 0 Problem7: Solve xy = 2 xy + ( y ) 2 Problem8: Solve xy + y = x y 4 3 TAM3B-Differential Equations 16

Unit II- Syllabus Second order linear differential equations Types of Solutions Complementary Function and its types Particular Integral and its types Variation of Parameters TAM3B-Differential Equations 17

Second order linear Differential Equations The general linear differential equation of order n is of the form n dy n-1 d y n 1 n-1 n +a +...+a y = f(x), dx dx where are a 1,a 2,...,a n are real constant. This equation can also be written in operator form as n n-1 ( 1 n) ( ) D +a D +...+a y = f x (1) TAM3B-Differential Equations 18

Types of Solution The solution consists of two parts (i) Complementary funtion (ii) Particular integral. i.e., y = y + y where y is a complementary function, c p c y is particular integral p TAM3B-Differential Equations 19

Complementary Function To find complementary function we have to form the auxillary equation which is obtained by putting D = m and f(x)=0. Thus the auxillary equation of (1) is given by m am am a n n n 1 n 2 + 1 + 2 +... + = 0 (2) Equation (2) is an ordinary algebraic equation in m of degree 'n'. By solving this equation we get n roots for m, say m, m,..., m 1 2 n TAM3B-Differential Equations 20

Types of Complementary Functions Type(i): If all the roots m,m,...,m are real and different then the 1 2 n mx 1 m2x mx 3 complementary function (C.F) y = Ae +Be +Ce +... Type(ii): If any two roots are equal say m = m = m then the complementary fu ( ) c 1 2 mx nction (C.F) y = Ax +B e Type(iii): If the roots are imaginary say m = α ± iβ the the complementary function (C.F) y = e c c αx ( Acosβx +Bsinβx) TAM3B-Differential Equations 21

Particular Integral When the R.H.S of the given differential equation is zero, we need not find particular integral. In otherwords the complementary function gives the complete solution. When R.H.S of a given differential equation is a function of ( ) ( ) ax ax x say e,sinax or cosax, Algebraic function, e f x,xf x, we have to find particular integral. TAM3B-Differential Equations 22

Type 1 ax If f(x) = e, then the particular integral is given by ( ) 1 P.I = f D ( ) 1 f a ( ) ( a) ax ( ) ax = e, provided f a 0 1 f ( ) ax If f a = 0, then P.I = x. e, provided f a 0 2 1 ax If f ( a ) = 0, then P.I = x. e, provided f ( a) 0 f a e ( ) TAM3B-Differential Equations 23

Type 2 1 If f ( x ) = sinax or cosax, then P.Iis given by P.I = f D 2 2 ( ) ( ) In f D replace D by - a,provided f D 0. ( ) ( ) ( ) 1 f ( D) ( D) ( ) ( ) ( ) 2 2 If f D = 0, then P.I = x. sinax (or) cosax,(d -a ) provided f D 0. 1 f 2 If f D = 0, then P.I = x. si nax (or) cosax sinax (or) cosax 2 2 Again replace D by - a in f D provided f D 0 and so on. TAM3B-Differential Equations 24

Type 3 If f(x) = a x +a x n n-1 0 1 n ( ) ( ) ( n n-1 ) 0 1 n -1 n n-1 ( ) ( 0 1 n) -1 +...+a where f(x) is a pure algebraic function then 1 P.I = a x +a x +...+a f D = f D a x +a x +...+a Expand f D by using Binomial theorem in ascending powers of D then operete on a x +a x +...+a n n-1 0 1 n TAM3B-Differential Equations 25

Type 4 ( ) ax If f x = e X where X is some function of x, then 1 P.I = f D ( ) ax.e X 1 f D+a ax = e..x ( ) If X is either sinax or cosax then proceed as Type 2 If X is some algebraic expression then proceed as Type 3 TAM3B-Differential Equations 26

Method of Variation of Parameters 2 Consider D y +a1dy +a2y = X, where X is a function of x Let the complementary function C.F = c f +c f 1 2 1 2 1 2 11 2 2 where c,c are constants and f,f are functions of x. Then P.I = Pf +Qf fx fx 2 1 where P = - dx and Q = dx ff -f f ff -f f 12 1 2 12 1 2 Hence complete solution y = c f +c f +P.I 11 2 2 TAM3B-Differential Equations 27

Illustration Problems for practice Problem1:Solve y + 3y 10y = 6e 4x Problem2:Solve y 2y = 12x 10 Problem3:Solve y 3y + 2y = 14sin 2x 18cos 2x Problem4: Solve y + 4y = tan 2x TAM3B-Differential Equations 28

Problems for practice Problem :Solve y + a y = bx 2 5 cos Problem 6: Solve 2 2x ( 13 12) 5 D D+ y = e + e x Problem 7: Solve 2 2 ( 5 6) 2 D D+ y = x x+ Problem 8: Solve 2 x 2 (4 4 5) cos D + D+ y = e x+ e x Problem 9: Solve y 2y + y = 2x TAM3B-Differential Equations 29

Unit III- Syllabus Linear Systems Theorems Relating the solutions of the system Theorems Relating the Wronskian Solution of Homogeneous Linear System With Constant Coefficients TAM3B-Differential Equations 30

Linear System The linear system dx = a1() tx+ bty 1() + f1() t dt dy = a2() tx+ b2() ty+ f2() t dt is homogeneous if f () t and f () t are identically zero 1 2 TAM3B-Differential Equations 31

Theorem A [ ] [ ] If t is any point of the interval ab,, and if x y are o 0 0 any numbers whatever, then the linear system has one x= xt () and only one solution y = yt () valid throughout ab,, such that xt ( ) = x and yt ( ) = y 0 0 0 0 TAM3B-Differential Equations 32

Theorem B dx = a1() tx+ bty 1() dt If the homogeneous system has two solutions dy = a2() tx+ b2() ty dt x= x1() t x= x2() t and on [ ab, ], then y = y1() t y = y2() t x= cx 1 1() t + cx 2 2() t y = cy 1 1() t + cy 2 2() t is also a solution on for any constants c and c [ ab, ] 1 2 TAM3B-Differential Equations 33

Theorem C x= x () t x= x () t 1 2 If the two solutions and of the y = y1() t y = y2() t dx = a1() tx+ bty 1() dt homogeneous system have a Wronskian W(t) dy = a2() tx+ b2() ty dt x= cx 1 1() t + cx 2 2() t that does not vanish on [ ab, ] then y = cy 1 1() t + cy 2 2() t is the general solution of the system TAM3B-Differential Equations 34

Theorem D If Wt () is the Wronskian of the two solutions x= x1() t x= x2() t and of the homogeneous system y = y1() t y = y2() t dx = a1() tx+ bty 1() dt then W(t) is either identically zero dy = a2() tx+ b2() ty dt or nowhere zero on [ ab, ] TAM3B-Differential Equations 35

Theorem E x= x1() t x= x2() t If the two solutions and of the y = y1() t y = y2() t dx = a1() tx+ bty 1() dt homogeneous system are linearly independent dy = a2() tx+ b2() ty dt x= cx 1 1() t + cx 2 2() t on [ ab, ] then is the general solution of the system y = cy 1 1() t + cy 2 2() t on this interval. TAM3B-Differential Equations 36

Theorem F x= x () t x= x () t y = y1() t y = y2() t dx = a1() tx+ bty 1() dt homogeneous system are linearly independent dy = a2() tx+ b2() ty dt x= xp () t on [ ab, ] and if is any particular solution of the non homogeneous y = yp () t 1 2 If the two solutions and of the x= cx 1 1() t + cx 2 2() t + xp () t system on this interval then is the y = cy 1 1() t + cy 2 2() t + yp () t general solution of the nonhomogeneous system on [ ab, ] TAM3B-Differential Equations 37

Solution of Homogeneous Linear System With Constant Coefficients dx = ax 1 + by 1 dt Consider the system dy = ax 2 + by 2 dt mt Step1: Assume x = Ae and y = Be Step 2: Substituting in (1) we get Ame = a Ae + b Be mt mt mt 1 1 mt mt mt = 2 + 2 Bme a Ae b Be mt (1) TAM3B-Differential Equations 38

Solution of Homogeneous Linear System With Constant Coefficients mt Step 3:Dividing by e we get the linear algebraic system ( a1 m) A+ bb 1 = 0 aa+ ( b mb ) = 0 Step 4 :Find 2 2 a1 m b1 a b m 2 2 2 m a1 b2m ab 1 2 ab 2 1 = 0, to get the quadratic equation ( + ) + ( ) = 0 Step 5: Solve the quadratic equation and find the value of m 1 and m 2 TAM3B-Differential Equations 39

Types of Solutions Distinct real roots: The general solution is mt 1 m2t x= c1ae 1 + c2ae 2 mt 1 m2t y = cbe 1 1 + c2be 2 Distinct complex roots: The general solution is at x = e [ c1( A1cos bt A2sin bt) + c2( A1sin bt + A2cos bt)] at y = e [ c1( B1cos bt B2sin bt) + c2( B1sin bt + B2 cos bt)] Equal real roots: The general solution is mt x = c1ae + c2( A1+ A2t ) e mt y = c1be + c2( B1 + B2t) e mt mt TAM3B-Differential Equations 40

Illustration Problems for Practice Problem1:Find the Wronskian of dx = x+ y dt Problem 2: Solve dy = 4x 2y dt dx = 3x 4y dt Problem 3: Solve dy = x y dt 3t 2t x= e x= e and y = e y = 2e 3t 2t TAM3B-Differential Equations 41

Problems for Practice dx = 3x+ 4y dt Problem 4: Solve dy = 2x+ 3y dt dx = 2x dt Problem 5: Solve dy = 3y dt dx = 5x+ 4y dt Problem 6: Solve dy = x+ y dt TAM3B-Differential Equations 42

Unit-IV-Syllabus Partial Differential Equations Formation of PDE by eliminating arbitrary constants and functions Types of Solutions Lagrange Equation TAM3B-Differential Equations 43

Partial Differential Equations A partial differential equation is an equation involving a function of two or more variables and some of its partial derivatives. Therefore a partial differential equation contains one dependent variable and more than one independent variable. EXAMPLES: 2 u 1. = 2 x 3 u y u u 2. + = 0 x y ( u - dependent variable; x,y -independent variable) ( u - dependent variable; x,y -independent variable) TAM3B-Differential Equations 44

Notations 2 z z z p=, q=, r =, 2 x y x t z = and s = 2 y z xy 2 2 TAM3B-Differential Equations 45

Formation of PDE by eliminating arbitrary constants Let f ( x,y,z,a,b ) = 0 (1) be an equaion which contains two arbitrary constants 'a' and 'b'. To eliminate two constants we need atleast three equations. Partially differentiating eqn (1) w.r.t. x and y we get two more equations. From these equations we can eliminate the two constants 'a' and 'b'. Similarly for eliminating three constants we need four equations and so on. TAM3B-Differential Equations 46

Formation of PDE by eliminating arbitrary functions Formation of Partial Differential Equations by Elimination of Arbitrary Functions : The elimination of one arbitrary function from a given relation gives a partial differential equation of first order while elimination of two arbitrary function gives a second or higher order partial differential equations TAM3B-Differential Equations 47

Illustration Problems for practice Problem1:Form the partial differential equations by eliminating the constants from z = ax + by + ab Problem 2:Form the partial differential equations by eliminating the constants from z = x + a y + b 2 2 ( )( ) Problem 3:Form the partial differential equations by eliminating the arbitrary functions from z = x + y + f ( xy) Problem 4:Form the partial differential equations by eliminating xy the arbitrary functions from z = f( ) z TAM3B-Differential Equations 48

Types of Solutions (a) A solution in which the number of arbitrary constants is equal to the number of independent variables is called Complete Integral (or) Complete Solution (b) In Complete integral if we give particular values to the arbitrary constants we get Particular Integral. ( xyzab) ( ) (c) Singular Integral: Let f x,y,z,p,q = 0 be a partial differential equation whose complete integral is φ,,,, = 0, Differentiating partially w.r.t. a and b and then φ φ equate to zero we get = 0 and = 0 a b The eliminant of a and b is called Singular Integral. TAM3B-Differential Equations 49

Lagrange Equation The equation of the form Pp +Qq =R is known as Lagrange's equation, where P,Q and R are functions of x, y and z. To solve this equation it is enough to solve the subsidiary equations dx dy dz = = P Q R If the solution of the subsidiary equation is of the form ( ) ( ) u x,y = c and v x,y = c then the solution of the 1 2 ( ) Lagrange's equation is φ u,v = 0 TAM3B-Differential Equations 50

Methods for Solving Lagrange Equation To solve the subsidiary equation we have two methods (i) Method of grouping (ii) Method of multipliers. (i)in the method of grouping consider any two ratios dx dy dz of the auxillary equation = =, integrate to P Q R get the value of c and c 1 2 The solution is φ( c, c ) = 0. 1 2 TAM3B-Differential Equations 51

Method of Multipliers If we were not able to group the ratios we will use method of multipliers to find c 1 1 1 and c 1 2 Consider the multipliers to be ( l, m, n ) and ( l, m, n ) then the auxillary equation will be 1 1 1 2 2 2 dx dy dz l1dx + m1dy + n1dz l2dx + m2dy + n2dz = = = = P Q R lp+ mq+ nr lp+ mq+ nr 1 2 1 2 2 2 2 Choose the multipliers in such a way that the denominator becomes zero in the ratios so that numerator can be equated to 0 to find c and c. Otherwise choose the required number of multipliers to find c and c by method of grouping. TAM3B-Differential Equations 52

Illustration Problems for Practice Problem1: Solve px + qy = z Problem 2: Solve y zp + zx q = xy 2 2 2 Problem 3: Solve xp + yq = 0 Problem 4: Solve xp+ yq= z 2 2 2 TAM3B-Differential Equations 53

Problems for Practice Problem 5: Solve ( ) = 2 2 zx y xp yq Problem 6: Solve ( y+ z) p+ ( z+ xq ) = x+ y Problem 7: Solve ( y+ z) p ( x+ zq ) = x y Problem 8: Solve 2 2 2 2 2 2 ( ) ( ) ( ) xz y p+ yx z q= zy x Problem 9: Solve ( mz ny) p + ( nx lz) q = ly mx Problem10: Solve ( y z) p+ ( z xq ) = x y TAM3B-Differential Equations 54

Unit V-Syllabus Charpit s method Special types of Functions Type I Type II Type III Type IV TAM3B-Differential Equations 55

Charpit s Method This is the general method of solving a partial differential equation of first order f( xyz,,, pq, ) = 0 The auxillary equation is dp dq dz dx dy = = = = f + pf f + qf pf + qf f f x z y z p q p q Where f f f f f fx =, fy =, fz =, fp =, fq = x y z p q TAM3B-Differential Equations 56

Procedure Step1:Find f, f, f, f f x y z p, q Step 2: Write the auxillary equation dp dq dz dx dy = = = = f + pf f + qf pf + qf f f x z y z p q p q Step 3:Integrate the simplest differential equation Step 4: Solve the integral obtained in step 3 and use the given equation to find pand q Step 5: Substitute in dz = pdx + qdy, Step 6:Integrate to get the solution TAM3B-Differential Equations 57

Illustration Problems for Practice Problem1: 2 2 ( ) p + q x = pz Problem 2: 2 6yz 6pxy 3qy + pq = 0 Problem 3: 2 2 ( ) p + q y = qz Problem 4: z 2 = pqxy TAM3B-Differential Equations 58

Special Types of First Order Equations Type I:Equaions Involving pand qonly f( pq, ) = 0 Step1:Put p= ain the given equation Step 2:Find qin terms of ' a' Step 3: Substitute p and q in z = pdx + qdy Step 4:Integrate to get the complete integral TAM3B-Differential Equations 59

Special Types of First Order Equations Type II:Equations Not Involving the Independent Variables f(, z pq, ) = 0 Step1:Put x + ay = u. Let z = f ( x + ay) be the solution dz dz Step 2: Substiute p= and q= a in the given equation du du Step 3: Seperate the variables zand u Step 4:Integrate and get z Step 5:Put u = x + ay to get the solution TAM3B-Differential Equations 60

Special Types of First Order Equations Type III: Seperable Equaions f( xp, ) = f( yq, ) 1 2 Step1: Consider f( xp, ) = f( yq, ) = k 1 2 f ( x, p) = k and f ( y, q) = k 1 2 Step 2:Find pin terms of xand k, and qin terms of yand k Step 3: Substitute p and q in dz = pdx + qdy Step 4:Integrate to get the complete integral TAM3B-Differential Equations 61

Special Types of First Order Equations Type IV: Clairaut's Form z = px + qy + f ( p, q) Step1:Put p= aand q= bto get the complete integral z = ax + by + f ( a, b) z z Step 2:To get the singular integral find and equate a b to zero eliminate aand b TAM3B-Differential Equations 62

Illustration Problems for Practice Problem1: z = px + qy 2 pq Problem 2: z = px + qy + pq Problem 3: z = px + qy + log pq Problem 4: z = px + qy + p q 2 2 Problem 5: p + q = 4 2 2 TAM3B-Differential Equations 63

Problems for Practice Problem 6: Solve p + q = 1 Problem 7:Solve pq = 1 Problem 8:Find the complete integral of p(1 + q) = qz Problem 9:Find the complete integral of the PDE pz + q = 1 2 2 2 TAM3B-Differential Equations 64

Problems for Practice Problem10:Find the complete integral of the PDE p y(1 + x ) = qx 2 2 2 Problem11:Find the complete integral of the PDE 2 2 p q x y + = + Problem12:Find the complete integral of the PDE pq = xy TAM3B-Differential Equations 65

Problems for Practice Problem13:Find the complete integral of the PDE p + q = pq Problem14:Find the complete integral of the PDE zpq = p + q Problem15:Find the complete integral of the PDE 2 2 2 2 2 2 2 2 pq + xy = xq( x + y) TAM3B-Differential Equations 66