Question

Differential Equations, please find the general solution

2 5. '(t) = - ( 11 1 21 1| c(t) 2 11/

Answer 1

x() = 2. 27 x (치 2 let n () = 씨 호 3) then - () (*) 72. 2. (t) 버 0 2.. 키지 (서게 (치 +의치 4275 (서 키 (D- 12 -2 22} = 0 (1) 지42길 473 +(D-2)길 칠 가3 =O (2)

& ns =20,tų + x3 7 (1) +12) (0-1) = 22 n 2 3 10²-30 +1) x₂ - (20-3) 2=0 7 (oriaz = 2c₂ett- + (0-1) 33 = 0 :: –(3) -M, -23 + (0-1) (0-2) -(0-1)*3.20 7 (52-30 +241) mq - (0-1 +2)^3 +0, > (52_30+ 1)^2 -(D+1)^2 = 0, (4 =1,4, +te ett -3e, et = 2[c 2e, et] (2)^2-(3) 2(D-2/42 = 2^3 + - (0-1) = 0 → (2.0 – 3) mg - (-1 +2) "3 = ( 2D-3)^2 - (0+1)^2 = 0 -(5) (4) - (5) ► 7 ( 6²50 +4 ) ₂ = 0 . (6) The ausiliary equation is, m² - 5m + 4 =0 ² m²_4m-m +4=0 n (m-4)-(m-4)=0 → m_4) (m-V = 0 7 m general solution of (6) is, dq (!)=e, et te est From 2 (5), (0+1)^3 = ( 2D-3)^2 = 200₂-32 (6+1). "3 = 2[cet .30 2017 G2 07]

(1) is, The ausiliary earn is, (m + 1) = 0 complementary solu Partictan solum **- tong +36 Cett (2 - 1 1 5 ) - Cze- teet where c. The homogenopus oan orresponding to mel = czert (D 41 [202244 -feet) = 2c, ent ne, et ..^3 (t) = ge-t + 20 ett 5 ma (0-2) Nq-"} x = (012) - 20₂3 aplilciet +4c, ett - 2010 -26,014 is creat 1-2-2 +2c Jet 5 2 at 5 ere 2ci est from (2), 20, et се 41 5 egett (1-2 Czert & ciet t ciet eige2, 3 ore constants dy of integration 8 >G 5 8 s ezett -ge-t cett elet sert teet n Cert to ett + 2 (t) = e, et (1) Šett +0,2941 -2, et tegert 5 ( (And)