Question

QUESTION 4 One solution of x' = The general solution is 1 2 e3t ОА. + C3 e3t Int x+b(e) is x(e) = [7] 0 [8]+ole [l] to (o) ( OB + Ci Int] et + e3t OD ci Int OC e3t Ci + C2 + C3 Int

Answer 1

Question Giver: a'- [11] Je + bc) One solutions is : 264) [] solution x'= (2 :]* + be We have: = Axtbit) STEP 1: solution of Associate x(t) = Axit) Homogenous system l.e., where, 2 1 As 1 2 Now computing Eigenvalues of A: stast foom forming from the diagonal Entries of a New Mataix by subtracting Entries of the given 1

matrix 2-) 1 2-7 Find the determinant of the obtained Matrix: 2-) 1 P -D 1 2-) 270 - 1=0 介介 2-) = +! → 2, = 3 & 12=1 T Eigenvalues. Now, computing coresponding Eigenvectoos (i) For 11,=3 (2-) [*] - 2-) solue the Now, Matrix Equations: {": 0) (0 - vt V2=0 } ☆ V = V₂. ² V-V₂ = 0

Therefore eigenveet Eigenveetor V, 1 ) For ^2=1 2-) 1 1 2) 1 1 1 2-7 solve the V, TV2=0 7 V +V2 =0 Now, Matrix Equation: [J] ] { VE-VE Therefore, cigenvector [-] solution of Home Homogenous [1]** ctor = Thus, the system is. 3t z(t) = C, + C2 General solution is : the Hence, x(t) = «[4]** *4[*]*+ [] 3t ANSWER... et ci -e X(t) = + + C2 3t e ent