Question

1. Consider the differential equation: 49) – 48 – 24+246) – 15x4+36” – 36" = 1-3a2+e+e^+2sin(2x)+cos - *cos(a). (a) Suppose that we know the characteristic polynomial of its corresponding homogeneous differential equation is P(x) = x²(12 - 3)(1? + 4) (1 - 1). Find the general solution yn of its corresponding homogeneous differential equation. (b) Give the form (don't solve it) of p, the particular solution of the nonhomogeneous differential equation 2. Find the general solution of the equation. (a) v" - 9 = 2 2 + (b) y + 4y + 4y =*+2+ 2 (C) y," + x + y = 3 sin 32 (d) y" + 9 = cos(3.c) (e) y - 3y" + 3y - y = 12e" 3. Solve the following initial value problems. (a) y" + 3' - 2y = 14 + 22 – 2x², y(0) = 0, 7/0) = 0 (b) y" + - 2y = -6 sin(2.c) - 18 cos(2x), y(0) = 2, '(0) = 2

Answer 1

solution: 1 Guven differential equation is - (9) (8) (7) 24/8) -1549+36 4" - 364" = (6) (4) Lacul 2601 - 1-3xt e pritet ta sin (aw + cosu - e-acas (ww). (a) the corresponding homogeneous Die has the following characteristic polynomial - PA) = x2(92-3)*(*%+4)(1-1). So, the characteristic roots are - P(x) =0 → 82(2-3)*(x2+4) (1-1) = 0 = x=0, (²-3) = 0 , 8+4=0, x-1=0 = 1=0,0, +13, tv3 , triit. io Jhe. general solution of the corresponding hom- -ogeneous Dis. is - yh = [G,+Czuje"* +(63+(4m) e usu+(+ (xe von + (C7 ces ex + (sinox)+ Cell =) 4h = Citlaht (l3t lun) eV 3M + (CG ton) e v3u + + = 2x + (sin 2x + ( * ci's are constants.

(6) The form of the particulari solution of the non homo geneous Dié is - yo como o + 40% so +269 2007 7 1-3rdt etike tellt es in (on)+con - e Mecas (wu)}. @> (a) y" -94 = 22-24 +23x The corresponding homogeneow Die is – mq zo + m² =9 = m = $3. so, yn = 4,64 +623. And, ypa (de-Px + ed) sa on ente tóxq px = a + x mp3 [1:89=0) c ena + =-de-pn +

i. The general solution is - y = Yn typ = 0, e3" + re-3x - 2/3 - 2x + 2 e34 2 (y" +44'+ 4y = 4+2+ e-Ru The corresponding homogeneous Dię.is — mit um +4=0 =) (m+2) =0 =) m = -2,-2. So, y = (0,+(2x) e-?. - +2.- "Yp = conta? (x+2+e=) - atleta +2.comen totalmente = *(1+3+* . Ju to forget 60.21 210+2) = (-4.1 +0....)+&+*.v e- = ☆ tegi tegemine moo Therefore, general sol" is- y = Yh+Y6 Chat an eziyyet neta y =

3) a) y"+y'- 2y = 14+ 21- Ru”; y(0) = 0, 4'10) = 0. The corresponding hom.D-€ IŠ - m+m-2=0 => [m+ 2)(m-1)= 0 => m=1,-2. So, yoh = Cielit ce-pre And yo. (14424–3x2) = 14 coll +2 I (1-x2) u = 14.out one-oo-otors" (n=22) =-7-01+ 010+(0)+...f(xer?) = -7 -7 61–12+ (1-21 +0–3] + (6 + + *) (1-x)+ 0 } =-7-760-29+4–20) +(2+0+ oltof =-7- (n=4"-3-2-) --++? +1 = 12-6 1. So, general solution is -

y = "ht Up => y = 0,9"+ Cap-2 +2²-6. Quien initial conditions are — y (0)=0 & 4'(o)=0 => G176-6=0 =) -262 +0=0 =) citla =6 =) C3d62- - 0 . Using ® in on : 262 +₂=6 32 = 6 =) Ca =2 © C = 6 -2 =4. Thus, the required selection is - 4 = 0, et lae-20 +8²-6 leny = 484 +2e=px +r?-6. - x-