1 First Order Ordinary Differential Equation

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1 1 Ordinary Differential Equation and Partial Differential Equations S. D. MANJAREKAR Department of Mathematics, Loknete Vyankatrao Hiray Mahavidyalaya Panchavati, Nashik (M.S.), India. 1 First Order Ordinary Differential Equation Basic Definitions (1) Ordinary Differential Equation: An ordinary differential equation expresses a relation between an independent variable, a dependent variable and one or more differential coefficients of the dependent with respect to the independent variable. (2) Partial Differential Equation: A partial differential equation involves one dependent and two or more independent variables, together with partial differential coefficients of the dependent with respect to the independent variables. (3) Order and Degree of Differential Equation: The order of a differential equation is the order of the highest differential coefficient which is involved. When an equation is polynomial in all the differential coefficients involved, the power to which the highest differential coefficient is raised is known as the degree of the equation. (4)Linear Differential Equation: When, in an ordinary or partial differential equation, the dependent variable and its derivatives occur lo the first degree only, and not as higher powers or products, the equation is said to be linear. The coefficients of a linear equation are therefore either constants or functions of the independent variable or variables. Questions on Differential Equation (1) The solution of the Differential Equation 2xy + 6x + (x 2 4)y = 0 is (a) y = 3 (c)y = 3 + A (b)y = 6 x 2 4 A (d)y = 3 x 2 4 A x 2 4 A x 2 +4 Answer: (a) Hint: By Rearranging the terms, we get (x 2 4)y = (2xy + 6x) and then by Variable Seperable method one can solve the differential equation.

2 2 (2) The Solution of the Initial Value Problem sin(x)dx + ydy = 0, where y(0) = 1 is (a)y = 2 sin(x) 1 (b) y = 2 cos(x) 1 (c)y = 2 sin(x) + 1 (d) y = 2 cos(x) + 1 Hint: Again by Variable Separable method then integrate and find out the integrating constant by the given Initial condition. (3) Let y 1 and y 2 be the solution of dy dx = y + 17 with Initial Condition y 1(0) = 0 and y 2 (0) = 1 then (a)y 1 and y 2 will never Intersect (b) y 1 and y 2 will Intersect at x = 17 (c)y 1 and y 2 will Intersect at x = 1 (d) None of the Above Answer: (a) Hint: Use the intial condition to find constants and draw the the graph of the two different solutions. (4) The Diff. Eq. y = 60(y 2 )) 1 5 ; x > 0 y(0) = 0 has (a) Unique Solution (b) Two Solution (c) No Solution (d) Infinite Number of Solutions Hint: Variable Separable Method (5) The solution of the exact Differential equation (3x 2 y 2 +x 2 )dx+(2x3y+y 2 )dy = 0 is (a) x 3 y 2 + x3 3 + y3 3 = c (b) x3 y 2 x3 3 + y3 3 = c (c)x 3 y 3 + x3 3 + y3 3 = c (d) x3 y 2 + x 3 + y3 3 = c Answer: (a) Hint: The Equation is given to be exact. Use the formula to find the solution. (6) Which function is a solution to yy sin ty = t 2 sin t cos t (a) y = t cos t (b) y = t sin t (c) y = cos t (d) y = sin t (e) y = t 2 sin t cos t (f) None of the Above Answer:?

3 3 Hint: Substitute directly in the equation (7) Consider the initial value problem yy + t sin(y) = 0; y(0) = y 0 For which values of y 0 is a solution guaranteed? (I) y 0 = 0 (II) y 0 = π 2 (III) y 0 = π (a)i only (b)ii only (c) III only (d) I and II (e) II and III (f) All Answer:? Hint: By variable Separable method solve it and find out the constant or interval in which the point lies. (8)For which value of α the solution of the equation y + tanxy = secx with y(0) = α is bounded above by 1 Hint: Find out the solution by variable separable method and by using IC find out the upper bound. (9)Which of the following functions can be used as an integrating factor to turn the following non-exact equation into an exact equation? (a) x 2 (b)yx 2 (c) x (d) None of the Above Hint: just multiply and check the condition of Exactness. (10)The integrating factor for the differential equation y x 2 + y x (a) x (b) x 2 (c) x 1 (d) None of Above Hint: Use the formula to find Integrating Factor = 0 to be exact is (11)The interval where the solution of the differential equation xy +2y = 4x 2. exist is (a) [0, 1] (b) (, ) (c) ( 2, 4) (d) x 0

4 4 (12) Find the general solution of y 6y + 9y = t 3 e 3t Answer: y(t) = e3t 2t + C 1e 3t + C 2 te 3t Hint: Find Auxiliary equation to find roots by using which write y h then by using Variation of Parameter method find v 1 andv 2 so that the solutio is y(t) = y h + y p here roots are repeated (13) Which two out of the following four conditions must be satisfied by a differential equation in order for the Principle of superposition to apply to its solution? (I) Constant Coefficients (II)Homogeneous (III)linear (IV)Ordinary (a) I and II (b) II and III (c)i only (d)iv only Hint: Use the principle of Superposition condition (14) Find out whether the set of function {e t, e 2t, e 3t } is linearly independent or not? Hint: Find out the Wronskian Note: here determinant of 3 3 matrix (15) Which of the following initial value problems is guaranteed to have a unique solution near the initial value? (I) dy + ty = t + 1 dy, y(1) = 0 (II) + ty = t + 1, y(0) = 1 dt t dt t (III) dy = y 1 3, y(0) = 0 dt (a) Only II (b) only I (c)only III (d) None of the Above Hint: Use Fundamental Existence and Uniqueness theorem to find the point (16) Find the solution to the initial value problem t 2 dy dt Then the closest value for y(2) is + ty = 1 with y(1) = 1 (a) 1.9 (b) 0 (c) 0.85 (d) 0.90 Answer: (c)

5 5 Hint: Solve the first order differential equation and by using IC find out the constant then put the value t = 2 in the solution (17)Which of the following differential equations is not separable? (a) y = xy (b) e x+y y = 2x (c) y + y = e x e y (d) y = ye x] (18) The value(s) of k which make(s) y(x) = ke kx a solution of 2y 4y = 0 is(are) (a) 2 (b) 3, 4 (c) 1 (d) 0,2 (18) One of the following is not a differential equation (a) xy = 3y cos(x)(b)txx = 2t (c) x 2 y = cos x (d) All of the above are differential equations (19) Determine order and linearity of the differential equation (y ) 2 y = 2t (a) Linear (b)first order and Linear (c) First order but Non Linear (d) None of the Above (20) Find the value of α for which the solution of the differential equation y 2y 3y = 0 with the IC y(0) = α andy (0) = 1 remains finite as x Hint: Find out the solution and use the condition to find value of α (21) The solution of the differential equation y + xy = xy 3. is (a) y = x + 3 (b) y = x 3 1 (c) y = ± (d) None of the Above 1+Ce x2 Answer: (c) Hint: Bernoulli s Equation (22)The solution of the differential equation y = 4y 3 2 for zero constant is (a) x = 1 3y 3 (b) x = 1 2y 3

6 6 (c) y = 1 3x 3 (d) None of the Above Answer: (a) Hint: By using reduction of order i. e. v = y and then solve the first order differential equation (23) Find the general solution to 2t 2 y + ty 3y = 0 given that one solution is y 1 (t) = t 1 Hint: By using given solution find out the other solution by using the formula y 2 = vy 1 where it can be found directly by putting the value y 2 = vy 1 in equation then reduce it to first order and solve it. (24) Find a fundamental set of solutions for y + 4y + 3y = 0 with x 0 = 0 Hint: Find the solution directly by using roots of Char. eq. then use the value of x 0 = 0 to find two linearly independent equations (25) The particular solution of the equation y 4y 12y = 3e 5t is (a)e 5t (b)3e 5t (c) 3 7 e5t (d)- 3 7 e5t Answer: Particular solution is Ae 5t, look at right hand side. (26) Find a particular solution for the following differential equation. y 4y 12y = 2t 3 t + 3 Hint: See the right hand side which is polynomial in t (27) The solution of the second order nonlinear equation y = 2t(y ) 2 with initial conditions y(0) = 2, y (0) = 1 is (a)y = (ln t 1 ln t + 1 ) + 2 (b) y = 1 (ln t 1 ln t + 1 ) (c)y = 1 (ln t 1 + ln t + 1 ) + 2 (d) None of the Above 2 Hint: In the equation the term y is missing so put v = y and solve it by using given IC (28) The general solution of the differential equation y 2y + y = et 1+t 2 is Hint: Find the roots of Char. Eq. Here we cannot compare directly so use variation of parameter method to find Particular Solution.

7 7 (29) Explain the method of Variation of Parameter. (30) Explain the method Undetermined Co - efficient. (31)Let V be the set of all bounded solution of the differential equation y 4y +3y = 0 then V contains (a) Infinitely many Solutions (b)more than one complex valued solutions (c)only Trivial Solution (d) None of the above Hint: See the condition of Bounded ness. (32) Show that the function f(x, y) = y does not satisfies Lipschitz condition on x < 1 and 0 y 1 (33)Find the real-valued general solution of y + 5y = 0. (34) If dy = (x + y + dx 1)2 (x + y 1) 2 and y( 1) = 0 then y( 1) is 4 4 (a) 1 (b) 0 (c) 1 (d) None 2 Answer : (c) Hint: put w = x + y then solve the new eq. by using variable separable form, the given IC then find the constant. put (35) Which of the following is an integrating factor of diff. eq. ty + (t + 2)y = t 3 (a) t (b) te t (c) t 2 e t (d) None Answer: (c) Hint: Use the property of Integrating factor (36) Solve the IVP y (3) 5y 22y + 56y = 0 with y(0) = 1, y (0) = 2, y (0) = 4 Hint: By finding roots directly - 4, 2, 7 (37) Write the following second order differential equations as a system of first order, linear differential equations. 2y 5y + y = 0 with y(3) = 6, y (3) = 1

8 8 Hint: put X 1 = y(t) then take the derivative on both sides name it by X 2 and solve it. (38) Solve the following IVP. ( ) 1 2 Y = AX where A = and X(0) = 3 2 ( 0 4 Hint: find the roots of the equation det.(a λi) = 0 and by using each eigen value find eigen vector. (39) solve the initial value problem. x (t) + 2x2y = e 2t, x(0) = 0 y (t)2x + 3y = 0, y(0) = 0 Hint: Convert them into System of Linear equations and solve it by above method. (40) Explain the method solving system of first order linear equations. (41)Consider the IVP y = xy 1 3, y(0) = 0 Then (a)f(x, y) = xy 1 3 does not satisfy Lipschitz Condition w.r.t. y in any nbd. of y = 0 (b)there is no solution to IVP (c) there exist unique solution to IVP (d) Exist more than one Solution to IVP Hint: Use the Definition (42)Consider IVP y = y 2, y(0) = 1 then there exist a unique solution of IVP on (a) R (b) (-, 1) (c) (-2,2)(d)(-1, ) Hint: Use the property of Fundamental theorem of existence and uniqueness. ( ) 0 (43) Consider the System Y (t) = AY (t), t > 0 with Y (0) = 1 where A is 2 2 constant matrix with real entries satisfying T r.(a) = 0 and det(a) > 0 Then y 1 and y 2 are (a) monotonically decreasing (b) monotonically increasing (c) Oscillating function of t (d)constant function of t (44) Solve the system of Linear equation ( ) 1 3 Y (t) = AY (t) + B where where A = and B = 3 1 ) ( 7 5 )

9 9 (45) Solve the Diff. Equation y λ2y = sin 2x (46) Find out the largest interval on which solution of diff. equation y + 2(x + 1)y 2 = 0, y(0) = 1 8 exist Hint: variable separable form solve and find value of a constant : interval :: connected (47) The derivative map D : R[x] R[x] is (a)one to one (b) onto (c) E : R[x] R[x] such that D(E(f)) = f, f (d) D : R[x] R[x] such that E(D(f)) = f, f Answer : (C) Hint: give an example or use the property of inverse (48)The differential equation y + y = 0 has (a)no solution (b)infinitely many solutions (c)two solutions (d)only one solution Hint: Find the roots and write the solution (49)Solve the initial value problem y + y x 2 = 3x, y(3) = 4 (50) Find the general solution of (x 2 + 1)y + 6xy = x

10 10 2 Partial Differential Equation Basic Concepts Let z = f(x, y) be a function of two independent variables x and y. Then, partial derivative of z w.r.t. x keeping y constant is z z and partial derivative of z w.r.t. y keeping x constant is. x y Notation: p = z x and q = z y Definition 2.1 An equation involving partial derivatives is called a partial differential equation (PDE). Definition 2.2 A partial differential equation is said to be linear, if the dependent variable and its partial derivatives occur only in the first degree and not multiplied. Definition 2.3 A partial differential equation is said to be non linear, if it is not linear. (1) The PDE obtain by eliminating arbitrary constants a and b from the equation z = ax + by + c is (a)z = px + qy + p + q (b)z = px + qy + p q (c)z = px + qy + p q (d) z = px + qy Hint: eliminate by differentiating w.r.t x and w..r.t. y separately and substitute (2)The PDE obtain by eliminating function from the equation z = f(x 2 y 2 ) is (a) px + qy = 0 (b)py + qx = 0 (c)px qy = 0 (d)p + q = 0 Hint: eliminate by differentiating w.r.t x and w..r.t. y separately and substitute (3) The solution of partial Diff. Eq. y 2 p xyq = x(z 2y) is (a) x 2 + y 2 = C 1 and y 2 yz = C 2 (b)x 2 y 2 = C 1 and y 2 + yz = C 2 (c) x 2 + y = C 1 and yz = C 2 (d) x y = C 1 and y z = C 2 Answer: (a) Hint: write auxiliary equation and then compare 1 and 2 and last two.

11 11 (4) Solve (y + z)p (x + z)q = x y Hint: Same procedure as above (5)The relation z = (a + x)(b + y) represents the partial differential equation (a) z = p (b) z = pq q (c) z = p q (d) None Hint: eliminate by differentiating w.r.t x and w..r.t. y separately and substitute (6) The solution of p + q = z is (a) f(xy, y ln z) = 0 (b)f(x + y, y + ln z) = 0 (c) f(x y, y ln z) = 0 (d) None Answer:? Hint: write auxiliary equation and find out the combination + y2 b 2 (7)The relation 2z = x2 a 2 (a) z = p + q (b) z = p q represents the partial diff. Eq. (c)2z = px + qy (d)z = px qy Answer: (c) Hint: eliminate by differentiating w.r.t x and w..r.t. y separately and substitute (8) The solution of z x x + z y y = z a is (a) z = e y a f(x y) (b)z = e y a f(x) (c)z = e y f(y) (d)z = e a f(x + y) (9)The partial diff. Eq. y 2 zp + zx 2 q = xy 2 (a)f(x 3 + y 3, x 2 + y 2 ) = 0 (b)f(x 3 y 3, x 2 y 2 ) = 0 (c)f(x 2 + y, x y) = 0 (d) None Hint: write auxiliary equation and find out the combination (10) The general solution of q = 3(p) 2 is (a) z = ax + 3ay + c (b) z = ax + 3ya 2 + c (c) z = ax 2 + 3ay + c (d) None

12 12 Hint: The equation is of the form f(p, q) = 0 so the solution is z = ax + by + c then solve it. (11) Elimination of a and b from z = ae by sin bx gives the partial diff. Eq. (a)z x x z y y = 0 (b) z x + z y = 0 (c) z x + z = 0 (d) None Answer: (a) Hint: Diff. w.r.t. x and y respectively and find out the combination. (12)The equation pe y = qe x gives the general solution (a) z = ae x be x (b)z = e x e x (c)z = a(e x + e x ) + b (d) None Answer: (c) Hint: f(x, p) = g(y, q) form (13) For PDE (p 2 + q 2 )y = qz (a)no Singular Solution Set (b) z = yx 2 is singular solution (c) z = 0 is singular solution (d) z 2 = (axb) 2 + a 2 y 2 is complete solution Answer: (c) and (d) Hint: Find out the complete solution by using auxiliary equation (14) Given cy = c 2 x + 1 then (a) Elimination of c gives y = ±2 x (b)elimination of c gives y 2 = 4x (c)elimination of c gives y = 4x (d)elimination of c gives y = 3x + 2 Answer: (a) and (b) Hint: Diff. w.r.t. x and y respectively and find out the combination. (15) Given sin x, cos x, sin 2x then Wronskian is (a) 3 sin 2x (b) zero (c) The variables are lin. ind. (d) The variable are lin. dep. Answer: (a) and (c) Hint: Use the definition of Wronskian.

13 13 (16) The complete integral for the PDE z = px + qy sin pq is (a)z = ax + qy sin aq (b)z = ax + by sin ab (c) z = ax by + sin ab (d) z = bx + ay + sin ab Answer:? Hint: Find the type of the PDE (17) Solution of p 2 + px + qy + xy = 0 is (a) (2y + x 2 + c 1 )(x + ln y) = 0 (b)(2y + x 2 c 1 )(x + ln y C 2 ) = 0 (c) (2y + x 2 + c 1 ) = 0 (d) None Hint: Use Charpit s Method (18) The complete integral of q(p cos x) = cos y is (a) z = ax + cos x + cos y + b (b)z = ax + sin x + sin y a a + b (c)z = ax + sin x + cos y a + b (d) z = ax + cos x + sin y a + b Hint: The equation is of the form f(p, q) = 0 (19) Solve dx cos(x+y) = dy = dz sin(x+y) z Hint: Find out the Combination (20) The partial differential equation formed by eliminating arbitrary functions from the relation z = f(x + at) + g(x at) is (a) z tt a 2 z xx = 0 (b) a 2 z t + z x = 0 (c)az tt + z xx = 0 (d)none Answer: (a) Hint: Diff. w.r.t. x and y respectively and solve it simultaneously.

14 3 Cauchy Problem for First order Partial Differential Equation 14 Given the First order Partial Differential Equation of the form f(x, y, z, p, q) = 0 which is conti. in all variables in a certain region D with the existence of a function φ(x, y)) such that φ(x, y), φ x, φ y are conti. in xy plane with some initial condition then there exist a solution also called as curve at a point of which the direction ratios of normal (p, q, 1) is such that f(x, y, z, p, q) = 0. Example based on cauchy problem: Solve the Cauchy problem for zp+q = 1 with initial conditions x 0 = λ, y 0 = λ,z 0 = λ 2, 0 λ 1. Solution: Given f(x, y, z, p, q) = zp + q 1 = 0 f p = z and f q = 1 dx 0 = 1 = dy 0 dλ dλ Now we get following ordinary differential equations, Since, dz = z x dx + z y dy dz = z dt x dx + z dt y dy dt dz dt = pz + q dz dt = 1 A here dx dt = f p = z dy dt = f q = 1 B integrating equation A and B, we get y = t + C 1 and z = t + C 2 now from the given IC C 1 = λ and C 2 = λ 2 y = t + λ I z = t + λ 2 II Now dx dt = f p = z = t + λ 2

15 15 integrating on both sides, we get x = t2 2 + λ 2 + C 3 and from third initial condition C 3 = λ x = t2 + λ + λ 2 2 Now put the values of λ from I in the above equation we get t = x y y 2 1 Now, λ = y t = y x y y = y x y 2 1 substitute it in equation II we get the solution. Exercise: Solve q + zp = 0 with x 0 = λ

16 4 Classification of Second order Partial Differential Equation Consider the general second order partial differential equation in the form Rz xx + Sz xy + T z yy + Dx + Ey + F = 0 I Then it is (a) Elliptic if S 2 4RT < 0 (b) Parabolic if S 2 4RT = 0 (c) Hyperbolic if S 2 4RT > 0 at given point (x, y) in the region D Characteristic Equation and Characteristic Curves Second order PDE Consider the general second order partial differential equation in the form Rr + Ss + T t + f(x, y, z, p, q) = 0 Then, quadratic equation in λis given by Rλ 2 + Sλ + T = 0 (Case I) if S 2 4RT > 0 then equation is hyperbolic and corresponding to which we get, Two characteristic equations dy + λ dx 1 = 0 and dy + λ dx 2 = 0 solving which we get two families of curve. (Case II) If S 2 4RT = 0then equation is parabolic and corresponding to which we get, Two characteristic equations dy dx + λ = 0 solving which we get only one families of curve. (Case III) If S 2 4RT < 0 so that it has complex roots. hence there are no real characteristics and we get two families of complex characteristics.

17 6 Methods of Separation of variables to solve Second order PDE Given the Second order PDE if one can separate the variables then the solution can be written in the form of U(x, t) = X(x)T (t) and use the equations to find separate solutions. 17

18 18 (21) The solution of the equation 3u x + 2u y = 0 with u(x, 0) = 4e x is (a) 4e ( x+ 3y 2 ) (b)e ( x+ 3y 2 ) (c) 4e ( x+ 2y 3 ) (d) None Answer: (a) Hint: Variable Separable form (22) The solution of the equation u t t = c 2 u x x = 0 is (a)(c 1 cos αx + C 2 sin αx)(c 3 cos αt + C 4 sin αt) (b)(c 1 cos αx + C 2 sin αt)(c 3 cos αt + C 4 sin αx) (c)(c 1 cos αx C 2 sin αx)(c 3 cos αt + C 4 sin αt) (d)(c 1 cos αx + C 2 sin αx)(c 3 cos αt C 4 sin αt) (23) Given yu x x + (x + y)u x y + xu y y = 0 Then (a) Then the canonical equation for the given pde is 2(x y) 2 u ξη + 2(x y)u η = 0 (b) No canonical form of the given equation exist (c)the general solution of the equation is u = 1 y x f(y 2 x 2 )d(y 2 x 2 ) + g(y x) (d) No general solution exist Answer: (a),(c) Hint: Find S 2 4RT and solve it (24) Which of the following is/ are true? (a) Two dimensional Laplace equation is elliptic (b) One dimensional wave equation is hyperbolic (c) One dimensional heat equation is parabolic (d) None Answer: (a) Hint: write the equation and find S 2 4RT (25) Find out the cauchy problem px + qy = z on D = {(x, y, z)/x 2 + y 2 0, z > 0} with IC x 2 + y 2 = 1, z = 1 (26) The pde u yy yu xx = 0 has (a) Two families curve of real roots for y < 0 (b) No real curve root for y > 0 (c) Vertical lines as family of curves for y = 0 (d) branches of quadratic curves for y 0 Hint: Find out S 2 4RT and solve it

19 19 (27) The solution of Z xxx = 0 has the solution (a) z = (1 + x + x 2 )f(y) (b)z = (1 + y + y 2 )f(x) (c)z = f 1 (x) + yf 2 (x) + y 2 f 3 (x) (d)z = f 1 (y) + xf 2 (y) + x 2 f 3 (y) Answer: (d) Hint: Just integrate (28) The partial differential equation representing variable heat flow in three dimension, is (a) u t = c 2 (u x + u y + u z ) (b)u t = c 2 (u xx + u yy + u zz ) (c)u t t = c 2 (u xx u yy u zz ) (d) None Hint: Definition (29) The differential Equation f xx + 2f xy + 4f yy = 0 is classified as, (a)elliptic (b)hyperbolic (c) parabolic (d) None Answer: (a) Hint: find S 2 4RT (30) λ = cnπ in a solution of one dimensional wave equation is a (a) Eigen functions (b) Eigen Values (c) Set of Spectrum (d) Fourier Functions Hint: Fourier series solution of one dimensional wave equation (31) The equation z xx + z yy = 12(x + y) has the solution (a) z = f 1 (y + ix) + f 2 (y ix) (b) z = f 1 (y + ix) + f 2 (y ix) + (x + y) 2 (c) z = (x + y) 3 (d) None of the Above Hint: find S 2 4RT gives complex roots

20 20 (32)Pick the region in which the partial differential equation yu xx + 2xyu xy + xu yy = u x + u y is hyperbolic (a) xy 1 (b) xy 0 (c) xy > 1 (d) xy > 0 Answer:? Hint: Find S 2 4RT (33)The complete integral of PDE x(1 + y)p = (1) + x)q is (a) z = a(ln(xy) + x + y) + b (b)z = a(ln(xy) + x) + b (c) z = a(1 + x) + (1 + y) (d) z = ax + by + (a + b)xy Answer :? Hint: rearrange, we get the equation of the form f(x, p) = g(y, q) (34)The equation x 2 (y 1)Z xx x(y 2 1)Zxy + y(y 2 1)z yy + z x = 0 is hyperbolic in the entire xy-plane except along (a)x - axis (b)y - axis (c) a line parallel to y - axis (d) a line parallel to x - axis Answer:? Hint: Find S 2 4RT (35) The solution of px + qy = 0 is of the form (a) f( y ) (b) f(x + y) x (c) f(x y) (d)f(xy) Answer: (a) Hint: Equation of the form f(x, p) = g(y, q) (36) Which of the following/s is/are true about Strum - Liouville Problem (SLP ) (a)all Eigen values of SLP are real and non -negative (b) Eigen functions corresponding to different Eigen Values are orthogonal w.r.t to weight function (c)slp has always an Eigen function (d) For each Eigen Value of SLP there exist only one linearly independent Eigen function. Answer: (a), (b) and (d) Hint: theorem based on SLP

21 21 (37) Green s function of the BVP y = 0, y(0) = Y (1) = 0 is given by (a)g(x, t) = x(1 t); 0 x < t (b)g(x, t) = t(1 x); t < x 1 (c) G(x, t) = x 2 (1 t); 0 < t (d)g(x, t) = x(1 t 2 ); t < x 1 Answer: (a) and (b) Hint: Solve the BVP (38) Given y = f(x at) + F (x + at) then (a) y x = f (x at) + F (x + at) (b) y t = af (x at) + af (x + at) (c) y xx = 1 a 2 y tt (d) Elimination of arbitrary functions gives pde of second order Answer: All are correct (39) Elimination of a function f from z = f( y x ) gives a partial differential equation (a) xp + q = 0 (b) p + q = 0 (c) p + yq = 0 (d) px + qy = 0 Answer: (d) Hint: It is an Euler Equation with n = 0 (40) If y 1 and y 2 are the solutions of some initial value problem and the Wronskian W (y 1, y 2 ) = 0 then y 1 and y 2 are (a) Linearly Independent (b) Linearly dependent (c) Proportional (d) None of the above

22 22 7 Greens Function Greens function G(x, t) for the boundary value problem is defined in the square [a, b] [a, b] and possesses the following fundamental properties: (I) G(x, t) is continuous in [a, b] [a, b] (II) G(x,t) is continuous in each of the triangles a x t b and a t x b x ; Moreover G(t +,t) - G(t,t) = 1 x x p 0 (t) (III)G(x, t) satisfies the boundary condition on the square. 8 Lipschitz Condition of order 1 We define the function f(x) to satisfy a Lipschitz condition of order 1on the interval [a, b] if there exists a constant K (dependent on both f and the interval) such that f(x 1 ) f(x 2 ) < K x 1 x 2 if K 0 called as a Lipschitz constant. If k = 1 it is called as short map and if, 0 < k < 1 it is nothing but Contraction Map. Some Properties and Examples of Lipschitz Condition: (I) The function f(x) = x is a continuous functions that are not everywhere differentiable. (II) If a function f(x) is differentiable with bounded derivative then it satisfies Lipschitz Condition. 9 Lipschitz Condition of order α We define the function f(x) to satisfy a Lipschitz condition of order 1on the interval [a, b] if there exists a constant K (dependent on both f and the interval) such that f(x 1 ) f(x 2 ) < K x 1 x 2 α if K 0 called as a Lipschitz constant. If k = 1 it is called as short map and if, 0 < k < 1 it is nothing but Contraction Map. We denote it as : f Lip(α) on the given interval Some Properties and Examples of Lipschitz Condition: (I) If f Lip(α) then it is uniformly continuous. (II)If α > 1 then f is constant.

23 23 (41) Suppose f is function that satisfies f(x 1 ) f(x 2 ) K x 1 x 2 α, α > 0 Then which of the following is true (a) If α = 1 then f is differentiable (b) If α > 0 then f is uniformly continuous (c)if α > 1 then f is constant (d) f must be polynomial. Answer:? Hint: Use the definition (42) For the IVPy = f(x, y) with y(0) = 0 which of the following statements are true? (a) f(x, y) = xy satisfies Lipschitz condition and IVP has unique solution. (b)f(x, y) = xy does not satisfies Lipschitz condition and IVP has no unique solution. (c) f(x, y) = y satisfies Lipschitz condition and IVP has unique solution. (d) f(x, y) = y does not satisfies Lipschitz condition and IVP has no unique solution. Answer: (c) Hint: Use the Properties (43)For the IVPy = f(x, y) with y(0) = 0 which of the following statements are true? (a) If f(x, y) satisfies Lipschitz condition, then IVP has unique solution. (b) IVP may not be unique although f(x, y) is continuous. (c)ivp does not satisfies Lipschitz condition still IVP has unique solution. (d) IVP does not satisfies LIpschitz condition and so, IVP has no solution. Answer: (a), (b) and (d) Hint: Use the definition and Properties. example x 1 3 (44) If f(x, y) = y 2 3 on rectangle x 1 and y 1 then (a) This function satisfies Lipschitz condition. (b) This function does not satisfies Lipschitz condition on a rectangle. (c) f y Constant for y = 0 (d) f y Constant for y = 0 (45) Every function f which satisfies Lipschitz Condition is (a) Is of Bdd. Varation (b) Incrasing function (c) Decreasing function (d) None All The Best