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Differential Equations - Solved Assignments - Semester Spring 2005

Assignment 1 (Spring 2005) Maximum Marks 60 Due Date 07, April 2005 Assignment Weight age 2% Question 1 (a) Define initial value and boundary value problem elaborate with the help of examples (at least one). 05 Solution INITIAL VALUE PROBLEM Differential equation of first order or greater in which the dependent variable y or its derivative are specified at one points such as ( )( )0 0,dyf x ydxy x y== If equation of the second order ( ) ( ) ( ) ( )( ) ( )22 1 0 2'0 0 0 1, 'd y dya x a x a x y g xdx dxy x y y x y+ + == = Where0y and 1' y are arbitrary constants Is called the initial value problem and the ( )0 0y x y = ( )'0 1, ' y x y = is called the initial conditions BOUNDARY VALUE PROBLEM Differential equation of order two or greater in which the dependent variable y or its derivative are specified at different points such as ( ) ( ) ( ) ( )( ) ( )22 1 0 20 0 1 1,d y dya x a x a x y g xdx dxy x y y x y+ + == = Where0y and 1' y are arbitrary constants Is called the boundary value problem and the ( ) ( )0 0 1 1, y x y y x y = = is called the boundary conditions NOT THE PART OF THE SOLUTION (MOTIVATION OF THE CONDITIONS) Basically when we solve the differential equations then we find some constant in the solution so we need to elaborate these constants to solve this problem we formulate two kinds of conditions(given above) which basically helps us to elaborate the constants finding in the solution of the differential equation. Still the question is how many constants are present in the differential equation. Is there any hard and fast rule? yes! It depends that what is the order of the differential equation. If we have the first order DE then its solution contains must one constant similarly if second order DE then its solution contains must two constantsand so on and definitely to elaborate the constant we always need same number of condition. Question 1 (b) Solve the following differential equation. 2 20; (0) 1xdy ydxxdx ydy yx y+ + = =+ 10 Solution ( )( )( )2 22 2 2 22 22 22 2 2 23 2 2 32 2 2 22 2 3( ) 0 , 0 10 ( , )( , )=,2 1 2xdy ydxxdx ydy yx yy xx dx y dyx y x yf yM x y xx x yf xN x y yy x yy xM x N yx y x yx xy y x y y xM Nx y x yxy x y y xMy+ + = =+| | | | + + = | |+ +\ . \ .c = = c +c = +c +| | | |= = + | |+ +\ . \ .| | | | + + += = | |+ +\ . \ . + c =c ( )( )( ) ( )( )( )( )( )( ) ( )( )( ) ( )( )222 23 3 2 2 3 3 222 23 3 2 2 3 3 222 22 222 22 2 2 322 23 3 2 2 3 3 222 23 3 22 2 2 2 22 2 2 2 22 1 22 2 2 2 22 2xy yx yx y xy x y x y xy yMy x yM x y xy x y x y xy yy x yM y xnowy x yxy x y x x y y xNx x yx y xy x y x y xy xNx x yN x y xy xx+ ++ + c =c +c + +=c +c =c ++ + + +c =c ++ + + + +c =c +c + +=c ( )( )2 3 3 222 22 222 22 2 2 y x y xy xx yN y x Mx yx y+ +c c= =c c+ 2 2( , )soour givenequationis exactnowas weknowf yM x y xx x yc = = c + Integrating both sides with respect to x 2 2 2 22112 221'2 22( , )1( , ) ( )1( , ) tan ( )21sin tan( , ) tan ( ) (1)21( )1f x y Mdxyf x y x dx xdx y dx yx y x yx xf x y y yy ydx xcex a a ax xf x y yyf xyx y yyfuuuu=| |= = + |+ +\ .| | | |= + | |\ . \ .| |= |+ \ .| |= + |\ .| | c = + |c \ .+c 2'2 2 2'2 22 2'2 2 2 222121( )( )( , ) ( )'( )( )212( , ) tan2 2tan2y xyy y x yf xyy y xas we knowthatf xN x y yy x ywe will getx xy yx y y xy yyynow putting this valueinequationnosox y yf x yxx ycxuuuuu| | = + |c + \ .c = +c +c = = +c ++ = ++ +==| |= + |\ .|= \212212 21 2tan2 2 2 2yapplying initial condition we getcx y yxtt |+|.= +| |+ = + |\ . Question 2 (a) Define the order and degree of DE also write down the order and degree of following DE. ( )( ) ( )2 " ' 222 "' ' 21 ( 1) 02 19 4 (4 25) 0x y xy x yx y xy x+ + =+ + = 04 Solution Order of Differential Equation: Order of the differential equation is the highest order derivative in a differential equation e.g.3y''+(y') =0 has order two. Degree of the Differential Equation: Degree of the differential equation is the power of the highest order derivative in the differential e.g. y''+y=2 has degree one. Although, (y'') 3+y'+y=0 has degree three. Order (1) 2 (2) 3 Degree (1) 1 (2) 2 Question 2 (b) Find I.F from 3rd and 4th method of exactness i.e. (xM+yN) & (xM-yN) respectively if possible. If not then explain the reason. ( ) ( )3 3 5 44 6 0 x y xy dx x x y dy + + = 05 Solution Integrating factor for 3rd case can not be found due to that 0 xM yN + = is not homogenous So first case is not applicable as M=3 34 x y xy , N=5 46 x x y + ( ) ( )3 3 5 44 2 3 5 4 24 +y 6 4 + 6xM yN x x y xy x x yxM yN x y x y x y x y+ = ++ = + To check it is homogeneous Say ( )( ) ( )( )( ) ( )( ) ( )4 2 3 5 4 25 4 5 2 3 6 5 6 4 25 4 2 3 5 4 254 + 6 ,, , ,, 4 + 6, 4 +t 6, ,nxM yN x y x y x y x y f x yf tx ty t f x y for somereal number nbutf tx ty t x y t x y t x y t x yf tx ty t x y x y x y tx yf tx ty t f x y+ = + === += += For 4th case it is not possible because condition ( ) ( ) 0 yf xy dx xg xy dy + = is not being fulfilled here. As ( ) ( )( ) ( )3 3 5 43 2 4 34 6 04 6 0x y xy dx x x y dyy x xy dx x x x y dy + + = + + = But 3 24 x xy and 4 36 x x y + are not the function of xy NOTE (RMEMBER) ( ) f xy Means each term should have the product of xy e.g. ( )2 2f xy xy x y = + And ( )2 2 4 4f xy xy x y x y = + + Both are functions of xy Question 2 (c) Define ordinary and partial differential equation as well as check whether the following are ordinary differential equation or partial differential equation. ( )221 4 2d x dy dzdu du du+ + = Where x, y and z are dependent variable and u is independent variable ( )222 3 4d x dxdu du+ = Where x is dependent variable and u is independent variable ( )2 22 23 3x xu vc c+ =c c Where x is dependent variable, u and v are independent variable ( )2 22 24 4u vx yc c+ =c c Where u, v are dependent variable, x and y are independent variable 06 Solution ORDINARY DIFFERENTIAL EQUATION (ODE) If an equation contains only ordinary derivatives of one or more dependent variables depending on only one independent variable is known as ODE. PARTIAL DIFFERENTIAL EQUATION (PDE) If an equation contains partial derivatives of one or more dependent variables depending on two or more independent variables is known as PDE (1) ODE (2) ODE (3) PDE (4) PDE Question 3 (a) Define explicit and implicit solution. Check whether ( )( ) 1 1 1 y x c = i.e. solution of the DE01 1dx dyx y+ = can be defined explicitly or not? If it can be define then write down the explicit form? 05 Solution Explicit solution: A solution of the differential equation of the form y= f (x) is called the explicit solution of the differential equation or roughly speaking if we can separate the dependent an independent variable in the solution then we say solution is explicit. For example dy / dt = y2-1/x has the solution y = 3+x2/3-x2. It is explicit solution because dependent and independent variable are separate here of it is form of y = f(x). Implicit solution: A solution of the differential equation of the form G (x ,y)=0 is called the implicit solution of the differential equation or roughly speaking if you cannot separate the dependent an independent variable then that solution is said to be implicit. For example, dy/dt = 1+1/y2 has the solution y tan-1 (y) = t+c. it is implicit solution becau

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