Question

z1(x) = 2x2 + tan x, z2(x) = x2 − 2x + tan x, z3(x) = x2 − 3x + tan x are solutions of a second order, linear nonhomogeneous equation L[y] = f (x). (a) Give a fundamental set of solutions of the corresponding reduced equation L[y] = 0. (b) Give the general solution of the nonhomogeneous equation L[y] = f (x).

Answer 1

1 to Solution : Given that 2,600) = 2X2 + tan, 3(x) = x2 - 2x +tan x are solutions of second order, Iz(x) = x² – 3x + tanse linear nonhomogeneous equation L[4] = f(e). a) To give a fundamental set of solutions the homogeneous equation [y] =0. 2,(2), 22(7), 23.6.) are linear combination ot ix and tan(a). L[y] =0 is second order linear homogeneous equations. Hence L[y] =0 has only two funda- mental solutions. Also, if we check coefficients of x, xe2 in all Z , Z2, Zz then they are distinct. But coefficient of tan(x) in each Zi, Zz, z is Same and that is 1. We know particular solution of L[4] =f() is appear it is in each solution. tan (x) is particular solution of L[4] =f(x), AISO are linearly independent. Hence, fundamental set of solutions to the homogeneous equation L[y] =0 is {x, x²}. b) By (a) {x, x²} is Fundamental set of solution to 1[4] =0. Ye = ax + C xx² as Hence 2 , x2 .. Also, by (a) we observe that =tan(x). & General solution of the non homogeneous equation L[4] = f(x) is y = Yc typ :: y = 12 + C2 x? + tan()