Question

6. Solve the initial value problem y" + y = 0, y(0)=0, y'0=1 (a) -COS X (b) -sin x (c) -sin x + cos x (d) -sin x COS X (e) COS X (f) sin x (g) sin x-COS X (h) sin x + cos x 7. Find a particular solution yn of the differential equation (using the method of undetermined coefficients): y + y =p2 (a) 2e (b) 3e (c) 4e: (d) 6e (e) 2/2 (f) e2/3 (g) e2/4 (h) 28/6

Answer 1

Let, y=eme (EU) be the trial solution of 6. yo +y = 0 ->0 y (0)= 0, y'70)= / Solution- Let, y=emx (²0) be the trial solution of O. Then auxilary equation of is given by, m²+1=0 m²=-1 m = ti so, the complementery fonetion of ois, where e fez are arbitery constant y (2) 2 C, CB(A) + (2 sin(a); Le y'(a) = -C, Sinx tez eos x Now, Y(0)=0 gives C.1 +€2.0 =0 - 1 = 0 NOW, y (0) - 1 gives -2,0 +62:= 1 => 82 - 21 Thus from ② , 9(2)=sina The salution of the initial vahie Problem is Y(a) = sina so, The oorreet oftion astion is o. 6 y" + y's e2x solution: corresponding Romo gereens equatin of @is. y"+y'=0+ 9

eofae, en .: The auxilary equation of ② is given by, m²+m=0 mcm+)=0 => mao , m +1=0 a>m=-1 voo, the complementing fumetion of ois, constant . since, ex are not on the e.F. so, By undefermined coefficient que can take the b. Articulars integral as, Yp AeLX Now, y = 26228 => $t" = 44828 since, yp is also a solution of 0, then from 0. *" +9 = 22 => 4A 022 + 2A 021 zeza e-ҳ 2 6A een => => 6t al :.A= Then from ③ er sp=6 so, the O’is, e22 6 furticular salution of the differential equation Yp= so, The correct option is- (0)