Question

1. Second order ODE (25 points) a. Consider the following nonhomogeneous ODEs, find their homogeneous solution, and give the form (no need to determine coefficients) of nonhomogeneous solution. (12 points) i. 44'' + 3y = 4x sin ( *2) ii. J + 2 + 3 = eº cosh(22) b. Find the general solution of y" + 2Dy' + 2D'y = 5Dº cos(Dx) where D is a real constant with following steps i) Determine homogeneous solution, ii) Find nonhomogeneous solution with coefficients determined using the method of undetermined coefficients and iii) Arrive at the general solution. (13 points)

Answer 1

(1a) i)Given 4y"+3y=4xsin V3 x 2 consider the homogeneous equation 4y' +3y = 0 The auxillary equation is 4m +3 = 0 Consider the nonhomogeneous equation 4y" + 3y = 4x sin now we have to find the particular solution by using method of coefficients Therefore the form of the nonhomogeneous equation is (ii) Given y" +2y'+ y = ecosh (2x) =y+2y+y=elete*). * * * : cosh (x)- & the ") => y*+2y'+y=e* consider the homogeneous equation y' + 2y'+y = 0 The auxillary equation is m+ 2m +1=0 =m=-1,-1. so, y = qe* + ce *x +e> Consider the nonhomogeneous equation y' +2y' + y = now we have to find the particular solution by using method of coefficients consider y, = Ae* x +Belt Therefore the form of the nonhomogeneous equation is y, = Ae**x? +Bes

(b)Given y" +2Dy' +2Dºy= 5D cos(Dx) (i)consider the homogeneous equation y" +2Dy'+2Dºy = 0 The auxillary equation is m2 +2Dm+2D² = 0 =m=(-1+i)D,(-1-i)D. so, y, =ce* sin(Dx)+c%e* cos(Dx). (ii)now we have to find the particular solution by using method of underterminant coefficients Consider the non homogeneous equation y" + 2Dy' +2Dº y=5D’cos (Dx) consider y, = A sin (Dx)+ B cos (Dx) y' = AD cos(Dx) – BD sin (Dx) y," =-ADsin (Dx) – BDcos (Dx) by substituting the values in y +2Dy', +2D’y, = 5Dcos(Dx) =- ADP sin (Dx) - BDcos(Dx) +2AD? cos(Dx) – 2BD2 sin (Dx) +2D²A sin (Dx)+ 2DB cos (Dx ) = 5D cos(Dx) = ADP sin(Dx)+BDcos(Dx)+2AD'cos(Dx)-2BDsin (Dx)= 5D' cos(x) =(A-2B)D' sin (Dx)+(B+ 2A)D'cos(Dx) = 5D cos(Dx) by comparing the values we get A – 2B = 0, B+ 2A = 5 Solving we get A = 2, B = 1 now substituting the values we get y, = 2 sin (Dx)+cos(Dx). (ii)Therefore the general solution is y(x)=c,e-* sin (Dx)+cze* cos(Dx)+2 sin (Dx)+cos (Dx).