Question

Problem #3: Consider the following statements. [6 marks] (1) The Laplace Transform of 11 ter cos(e) is well-defined for some values of s. (ii) The Laplace Transform is an integral transform that turns the problem of solving constant coefficient ODEs into an algebraic problem. This transform is particularly useful when it comes to studying problems arising in applications where the forcing function in the ODE is piece-wise continuous but not necessarily continuous, or when it comes to studying some Volterra equations and integro-differential equations. (iii) If P and Q are continuous on R, then we can apply the idea of the method of variation of parameters to y' + P(x)y = Q(x) and assume y = ujyi where yi solves y' + P(x)y = 0. That way, we would obtain the same result the integrating factor method would give. (iv) The following is an example of a piece-wise continuous function x #0 FI* -{ f(x) = 0 x = 0 (v) To compute solutions to a homogeneous second-order Cauchy-Euler equation, there are three different cases to be considered, depending on whether the characteristic equation associated with the ODE has roots that are real and distinct, or real and equal, or complex. If the equation is non-homogeneous, then we cannot use any method we covered to find a particular solution. (vi) Given a second-order linear ODE, the method of variation of parameters gives a particular solution in terms of an integral provided yi and y2 can be found. Determine which of the above statements are True (1) or False (2). So, for example, if you think that the answers, in the above order, are True False, False, True False,True then you would enter '1,2,2,1,2,1' into the answer box below (without the quotes).

Consider the following statements. (i) The Laplace Transform of 11tet2 cos(et2) is well-defined for some values of s. (ii) The Laplace Transform is an integral transform that turns the problem of solving constant coefficient ODEs into an algebraic problem. This transform is particularly useful when it comes to studying problems arising in applications where the forcing function in the ODE is piece-wise continuous but not necessarily continuous, or when it comes to studying some Volterra equations and integro-differential equations. (iii) If P and Q are continuous on R, then we can apply the idea of the method of variation of parameters to y′ + P(x)y  =  Q(x) and assume y  =  u1y1 where y1 solves y′ + P(x)y  =  0. That way, we would obtain the same result the integrating factor method would give. (iv) The following is an example of a piece-wise continuous function  f (x)  =  {   1 x   x  ≠  0 0 x  =  0 (v) To compute solutions to a homogeneous second-order Cauchy-Euler equation, there are three different cases to be considered, depending on whether the characteristic equation associated with the ODE has roots that are real and distinct, or real and equal, or complex. If the equation is non-homogeneous, then we cannot use any method we covered to find a particular solution. (vi) Given a second-order linear ODE, the method of variation of parameters gives a particular solution in terms of an integral provided y1 and y2 can be found. Determine which of the above statements are True (1) or False (2). So, for example, if you think that the answers, in the above order, are True,False,False,True,False,True then you would enter '1,2,2,1,2,1' into the answer box below (without the quotes).

Answer 1

integro - differential Equation . is well 3. i) The Laplace Transform of 11 te cos(eta) defined for some values of s. because 11 te cas(t) is continuous function in B and Laplace mansform of every continuous function exist. So True (1) False (2). ii) The Laplace Transform is an integral manskom (by definition). ממי # Fls)- SF(t)e-stat 4 (0) 23 It turns the problems of solvins constant # coefficient des into an algebraic problem Eg y'+2y=0 ^ [y't2y] 26 [0] =) ac y' + 2y] 20 = [[y] + 2 h[y] 20 => [sL [y] y(0))+26[y]z 0 = a[7] 2 3/4+2 = y(t) 2 3 e 2t t (applying inverse Laplace tx * The Laplace tx is applicable for piece-wise continuous huntion, but not necessarily continuous haulion. Wolterma Egnation is xe (t) = f[]+KC,8)z(s)ds Alinear Veltensa Equation is a convolution equation if x (t) = f(t) + Sk (t-s (-5) x (52ds The humilian K in the integral is the kennel. such equations can be analyzed and solved by Laplace mansform techniques to

general iii) Variabion of parameters is a method to solve inhomogeneous linear ODES. for ist order inhomogenous linear ODE, it is usually possible to find solutions via integrating factors y'+ P(X)y = Q(x) SP(x) dx Integrating fadon, 1.F2 - 1.84' + I. P(x) = Q(x). I (oy.I)2 010):1 2) ya x da Sp(x) dx y, solves Y + P(x)y zo Chomogenous eqr) CF29, and general solution is g. CF + PI heve ÞI +0 Since it is inhomogenous differenelin so y & uyi So false (2) So(x). eSP(x) dx SO iv) A piece-wise continuous funtion is continuous everywhere except some finite no of peints. Here f(x) = 1/o xto x=0 So f(x) is continuons to every where nEO ER but not at a=0. so f(x) is piece-wise continuous fuulien so True (1)

"(x)+. So this is 1 dx² have (V) Cauchy a Eulers equation of ondern has Qoy(x)=0 the form : anxh M(x) + an. X y a homogenous differential equation if we take dr z Y = Dy and day DCD-DY So afters turning it to linean homogenous ODE. , we to considers only the Characteristic equation : The cases for roots of characteristic Equation : (i) meal and distinct (1) real and equal (iii) complex for inhomogenous case, we have to find particular solution ( PI ) by different methods. so True (1) vi) Given a second-onder linean ODE tp.gly + qy z oo f(x) on the kind of roots of the characteristic equation of linear ODE We get different eg. if y2 Aenx+ Bean then Y 2 Ae? Berx. y, and Y. are Hie fundamental Solutions of the equation and they linear independent. Wronskian, W (site) 2 Y Y Yi Yi there particular solution, y p[m) 2 -Y,(x) S Yo (7f (x) dx + y(«) Sym? Fle) dx Troue (1) day Based / YoY₂ ra and Y₂2 are so "W(Y, M2) So The answers : ®, 1.®.0.0.1