Question

Find the general solution of the following differential equation: (1) ?′′ + 5?′ + 6? = 2????*?^? (2) ?′′ + 2?′ + ? = ? + ?e^(-t).

(please solve Question No.7 only)

7. (30 points) Find the general solution of the following differential equation: (1) y" + 5y' + 6y = 2etsint (2) y" + 2y + y=t+te-t 8. (10 points) Use the method of variation of parameters to find a particular solution of y" + y = 1/sin (t), 0 <t</2.

Answer 1

dy + by = 2 et sint lyr sy't by - Ret sant writing of as we have Dy + 5Dy & by: 2 et sint. - 10+50+6) y 2 et sint which is fast. If the form folye x where x is a function oft Now, to find the complementary function, we take right hand side as good replace to by min flo). ie fm) = o. (auxiliary equation). = m² +5m+6=0 m² 3 m + 2m +6=0 m (m+3)+2(m+3)=0. - (m+3) (m + 2) = 0. m = -2, 3. Now, ifm and n are the roots y the aviliary equation then complementary function, CF is given by, So, we have ff = y = c et est [since my hand Now, we have particelle integral, 1-3 - ie. P. I = (2e + sint) 025D +6 12 cht)sace Prof Dyev (0+1²5(0+1)+6 V. a = 1) flDta) where is a functionalt yage yet 1 (Here, we have

PIJ FODE et - e P. I = et (2 sint) D41 +2D + 5D +5+6 - (2 sint). D47D+12 Now, P.I. = sin at is given by, VP-1 - sinat f(D). In this type, we replace the term De by -a? So, P. I = et (2 sint). Chere a= 1) - 1 ²70 +12 (2 sint) 70+11 + (70-11) 12 sint) (70+11) (7D-11) 7D-11 (2 sint) Yg D²121 Again replacing De by-a² ie 1² P. I- et 7-11 12 sint) = et 170-11) (25int) 491-12) - 121 -170 - et (14D lente) – 22 sint) 170 - et ( 14 cost - 22 sint). 770 = - 7 et cost + 11 et sint 85 85 CFTP.I. Now, general solution le y = 0, ce + C₂ et 7 é cost tll et sint 85 -

(2) g + 2y 'ty iedy + 2 dy dt dt ty=trte t+tet. => 01 4+te+ complašrentary function, Cfl (6+atem =t+tet e-t Writing of as D, Dy + 2D y + y = is flm) = 0. where Dis replaced by mix f (D) = 0. m'42m +1 = 0 lauxiliary epx). = m+ 1)2 = 0 m =-1,-1. Now, it m be repeateel root y auniliowym So, Here, cf = (C, + (at) et (m=-1). And Particular integral, P. It D'42D+1 D²+2D+1. 1 (t). D42D+ Now, the auxiliary then -+ té Now, P. I oft - 042D + t (+1² (I+D) 2 + 11-2D+3D², ...) t (since (1+x)n. +1 -0. -7-2 [since olt) -1 *D*(+) = 0]-Ó

PI. Joc tet tet 0+ 1)2 t "since pa ps of fim eatv (OH)2 eat 2004) where is a function of t et Ilt), D D -t means +14) [6 integration St dt = 1² =e $142) +3 [ St°dt = $ 2 3 -t 3 6 So, P.I = t-2 + etts [ from eq (1) 2(2)]. 6 Now, the general solution is given by, = CF + P. I. where it and I are complementary function de particular integral. y = k tablet + 1-2 + 6413 19 = 9e acteut-2 the -t 3