Question

Find the general solutions to the following non-homogeneous Cauchy-Euler equation using variation of parameters. 22" + tz' + 362 = - tan (6 Int) z(t)= (Use parentheses to clearly denote the argument of each function.)

Answer 1

2 + Given #23"+*3! +363 tan (Gents 420277D7387 a Etant later where but t=en os x = ent to co t2D2 oco, whese o = de from eqcij Tolot) +0+36) 3 = -dan (Gxj (02_0+0+3612 Janon 02+36) 3 -tenore Using Medhod of datter Beiseimeter Aurilla zy 'equatters is 0+36 = 0 o replaced by my m²+36 0 m = +6 complementang solustasinya cosbx+csinsk let u= cos 6x Sinox Wenskian of these function is W.. W sinau Seos 60og2x t6stnou -6Singh 6 Color - 6 C cobre tsin²Gw -616 m² = -36 u ul equation il conpaising with " + P 31+Q3 =R. So R=-deinox Particulis integral of eqciis is yo Alxucx) + BCx) V (x2) Cin where Alle Team dre Scanned with Calisandera VR - B(x) = uk dol

Hence, Ax) = - = ots Sinor (atan 6x) di 6 Singh sindic cosor Sinbre dhe Cos on Il-costsx du Cos or st I due = f Stone - Copex I de == S (seafy - Cossx] de [Eln/Seafo than 6xl - Singre G B(x) = Coooxel-tenor) dae 6 Heussafcimos Jeden Scanner at der sinbredy (-Cossy) its cannes Cosby 36 CamScanner

ا) from eq bentlantseext toget-Stins] Cos Ox (cos 6x) Sin 6x a + st BE 36 3 = Į en Iseex ttannt - Cosou so solution is 3 ct 3p = G Cosbx + GSimbu + cossx ln sedi+tanbul partx 3= 4 cos ollost) + G sinblinit tar fojte entt en Sect Glat) + Jean (Genty CS Scanned which is required solution est teamscanner
Firstly we convert ordinary differential equation into Cauchy Euler's equation. Then used method of variation of parameter.