Question

is necessarily the number -b/2a)]. 1. Find the general solution to the following homogeneous differential equations. (a) y" - 2y + y = 0 (b) 9y" + 6y + y = 0 (c) 4y" + 12y +9y = 0 (d) y' - 6y +9y = 0 2. Solve the the following initial value problems. (a) 9y" - 12y + 4y = 0 with y(0) = 2 and y(0) = -1 (b) y' + 4y + 4y = 0 with y(-1) = 2 and y/(-1) = 1 4y" +12y + Oy 0 with y(0) 1 and y(0) = -4 4y" + 4y +y=0 with y(0) = 1 and y(0) = 2 Case 3 (complex roots): ar2 + br +c=0 has two complex roots n = X +if and r2 = X-in where 1 and 1 are real numbers. The general solution is y(t) = ciecos(pt) + coe sin(at) where ci and ca are (real) constants determined by an initial condition. 3. Find the general solution to the following homogeneous differential equations. (a) y' - 2y + 2y = 0 (b) y' + 2y + 2y = 0 (c) y" + 6y' + 13y = 0 (d) y" + 2y + 1.25y = 0 4. Solve the the following initial value problems. (a) y + 4y + 5y =0 with y(0) 1 and y'(0) = 0 (b) y + y +1.25y = 0 with y(0) 3 and y'(0) = 1 y' + y = 0 with y(7/3) = 2 and y(1/2) = -4 (d) y" + 2y + 2y = 0 with y(7/2) = 2 and y'(/4) = -2

homo 2nd order linear equations

Answer 1

on. 0 (a) y" - 2y'+y = 0 This is homogeneous differential equation Let ya tent trial solution. be m y'- A ent Ament Then mit get Amtemt A eht y" - Am²ent substitutin ing these values in 6 we - 2 Amemt + A g" mt (in - 2m+ 1) TO m-2m + 1 = 0 mc !! .. General solution is y =(4,+56) e" where i and a constants. are arbitrary gyll & 64't y = 0 Auniliary equation is gm' + 6 m+ 1 =0 (3 m+1) = 0 -> m=5,- General solution is y = (C, +62 t) e 3 are arbitrary constants. و را

© is a 2. (a) gy" - 124'+ 4y = 0 yo) = ; y' ola-1 homogeneolas differential equation. Auriliary equation gx - 12) + 4 =0 (32-2) ? itutitzan a > > 3? 3 :. General schition is 2/3 ta y = (2, + C, t) e a bit constants. Y(0) = 2 >> Y CO Y' (t) = رارا are CI -1 2 wali C e +(c +, t. ) et 3 yo) = -|- a tcx = -1 3 4-7 => 2 Hmo 3 P is the given required solution for initial value problem.

y" + 44'+ 4y co is 4 [ ) = 2, y' E1 = 1 Auriliary equation for m2 + 4 mt 4 (m+2) = m = -2, are E C e-zt : General Solution is y = (?, +47) e-27 where C. arbitrary constants Ty' (t) +((,+Ctje zimny y! (-1) =! > ae²+(c, c.le (-2)=1 ge + zce' = 1 "(-1)=2 (c, -C2 ) e² (;) 46e² 3 L - 2 ü - -] = 4e2 From (in G 7 G 4e² required solution 4e² 2 is

y' (t) = (- sint te rust) ezt +(e, cost te sint) e 2+(-2) =) 4 m) = 1 Ctc (-2) =) -2= = q = 3 - at ::- y = (cost est + 3 sinthe required Solution of diff. equation giren