Question

Need help with Ordinary Differential Equations assignment. I will give a thumbs up to anyone that provides solution quickly.

Find the general solution of the given differential equation: 2y' + 3y' + y = 21

Answer 1

of We have to find the general ealution 2y+3y' ty= at3 Let us take dy = by ie. a = D. d4 then our D.E. become : - = 0 (20² +30+1 ) y = 2+² So, uniliary equation is 20² + 3D + 1 =0 i.e. 202 + 20+ 0 + 120 1) + I(D +1) (20 + 1) (0+1)=0 a D=-1/4-1 - 2t y = C, e t Hence, th x e the mi Now Particular integral can be obtained by (201730-71) ** (Rom) (01 = 2(044) 10+1 t - (0+₂ 5" (18+15+ = (0+1 2 3" [1-0 +0% 0% +2" + ... ] + (0++)' [43+'+ 6= 6 +04.04.] +(3+0) [4-34+0+ -] = 2 (1+20) " ( + - 3 + + 6+-6]

= 2 [1-20 +4D²-80² + 160"+.. ] , t²_3+²+67-6) = a1 + 3+ +67-6 -2 1 3+²-67 +6) +4167-6) -816) ) - 2 +² 3+² +67-6-64² +12+ - 12 + 24 7 - 24-48) - 21 -9+'+42+ -90) = 2+3 18+'+847 - 180 So, general enlution y(t) = CS. + P.I. . = -112 Z e t t é tat - 18t + 847-180

If you have any doubt feel free to ask in the comment section.