Question

6. Which pair of functions below cannot be a fundamental set of solutions? * (8 Puan) O 1, x, x? Inx O 5e + 5, 2eX + 1 0 2x, cos 3x + sin 3x, 1 3 cost + 6 sint , 5 cost + 10 sint O e-31, te-31

Answer 1

must be Fundamental set of solutions linearly independent. So their Wronskian is non zero. (1) 1,x², x² lnx 1 x3lna (x3lnr) Wronskian, W(x) = x² dx d² d (1) Jx da 2 ď d' (1) dx² dx2 (23) dx2(x 3 lux) 1 x² x²lux x² + 3x²lnx 0 3 x² 6x 5x +belix = 1. [3x245x+Gxlnx) — 62 (x++3x* {nx)] 15x² + 18x3 lue - 6x²_1803 lux 9x² to Since hence W (3) +0, is Linearly 1, x², x²lna independent.

(2) 50 +5, 2e²+1 Wronskian, W= se"+5 ze'+1 (2e "+1) dr d ď de (se*+5) Zeltl = lºse sex+5 5e² 2 et ex [ibe 16€*+10 -102 5] = 5 et W(x) 70 so 5e*+5, 2e+1 is Linearly independent

(3) em, Cos 3x + sin 3x, 1 Wronskian, W(x) = e 2x cos 3x + sin3n dice) 605 3x+sin 3x) da (1) dx dx 2x 2 d² (1053x+ sin3x) da² dx2 dx2 (1) 1 e 2x cosnt sinza 202x 3(653x - sin3x) 0 O 14e2x zu - (sin3x + cos2x) 1 [-18(sin 3x + 49 3x ) -12 (105 32 –sin3x)] e - (30 cos3x + 6 sin3x) e 2x - > W(*) 70 So ex cos3x+ sin3x, 1 core linearly independent (4) 3 cost tosint, 5 cast + 10 sint 2. 3 cost+bsint 5 cost + 10 sint Wronskian, Wit) = d dt (3ost +6 sint) d at (5 ost + 10 sint) 3 ust +6 sint -3 sint + bost Scost +10 sint - 5 sint + 10 cost cost + 2 sint - sint +2 lost 3x5 cost + 2 sint - sint +2 cost - - o

are a Since W(t)=0, hence 3 cost +bsint, 5 cost + 10 sint linearly dependent, hence they cannot be fund am mental set of solution. (5) e-37 te-3t e -3t te-3t Wronskian W(t) = dt * (e-31) Šte-3t) te-3t e-3t -3 e34 e3+(1-31) 1 e 37 stu + 1-37 -3 e-3t [1-36 +3t e-34 70 So e-3t te-3t 2 we linearly independent From (1), (2), (3), (4), (5), see that only the set {3 cost 3 cost + 6 sint 5 cst +10 sint's is Linearly dependent, so it cannot be a of solution. Hence, Answer: 3 cost+ 6 sint, 5 ust + 10 sint 2 fundament al set