Question

Exercise 3: Solve the following differential equation (with initial conditions) for the three cases below..

Solve the following differential equation (with initial conditions) for the three cases below (by hand!). You may use whatever method you find simplest. You may check your work in MATLAB or Python. 2ä + 3c – 2x = f(t), x(0) = 1, ¿(0) = 3. (a) For f(t) = 0. Note that this is just the homogeneous ODE, 2ä + 3i – 2x = 0. (b) For f(t) = 5e-t/2. (c) For f(t) = -4ta. Hint: try xp = A + Bt + Ct? and solve for A, B, and C. In all cases, be sure to make sure that your initial conditions are still satisfied!

Answer 1

a) let aurillary equation corresponding to homogencouy equahan 23 +32 4 -21 20 is, 2m² +3m-2=0 si.c. f(t)=0}. 2m² + 4m-m-2=0 am (m+2)-1(m+2)=0. . (2m - 1) (m+2) = 0 m= 1 -2. As roots are real & distinct, solution is, xc = C, ett/24 G ezt coefficients, b] let flt) = 5et/2. By method of undetermined let up= A e-t/2 - xp = -1 A et/2 > Xp = TA8212 substituting abore values in Pico equation. 2. 1 Apt2 - 3 A 6t12 - 2A etl2 = 5et/2 » -3A et/h 25 et/2 .. = -3A = 5 A = -5 - particular solution is Apa 35 et/2. General solution is, X(t) = Xe txp = Get/2t6 3

. 200) : so. D -> G+62 - 5/3 = - - City = 8/3 - 2 I by 0 , x (t) = 1.6 et 12 - 26 e 27 ts et. As ico)-3, 2 .4-262+ = 3: → C-4C2 = 13/3 - . solving ② & 4 we get 4433 a particular solution with initial conditions is, O act)- 3 ett 1 e 2 - 5 et 12

c) let flt) = -4t2 let Xp= A + B t + ct² - ip = Bt at ca > X = 2C. substituting above values in giren equation, we get, - 40 + 3B + GC t - 2 A - B t - act² = -4t². = (40+3B-2A) + (66-2B) t+(-20) t = -4t2. » -2C = -4 = 2 66-2B=0 B = 60 4 40+ 3B-2A=0 A3 13 >> Up = 13+66 +2+ General solution is - x = x ctxp = Ciecht(20 2t + 13t6tt2t².a » ä(t)= Icetl2_262 e 2t +6+4t. (0)=1 (0) = 3 C+(₂ = -132 1 4-2 (₂ + 6 =3 ² leic 4 6 =-6.. solving abore equations, * = -615 - - 5415 - General solution is, gires, alt) = Get + Gent +13 +66 +24² a(t) = (-6/5) 12 (5415) ezt +13 +66 +2+²