Please show all work!! 1. Nonhomogeneous linear system. Consider the nonhomogeneous systemx' = P(t)x+ g(t), g(t) = 0where the 3 x 3 matrix P(t) is continuous for -∞ < t < ∞. Assume that the columns of matrix X(t)are the three solutions for the corresponding homogeneous systemX(t) = [x₁(t), x₂(t),x3(t)] =(10et-1 e¹(1+t)-1e-t1(1(a) Compute the Wronskian of X(t). Then verify the inverse of X(t) is1-e-t0etett et70).−e¯¹(1+t) _e¯¹(2+t),(b) Use the method of variation of parameters to find the solution of the original nonhomogeneoussystem subject to the initial condition x₁(0) = 0, x2(0) = 0, and x3(0) = 0. (No need to findP(t).)

Given fundamental matrix X(t)=[1et0−1et(1+t)et−1ettet]Then the wronskian of X(t) isW(X)=|1et0−1et(1+t)et−1ettet|=[1et00et(2+t)et0et(1+t)et]=e2tNow [1et0−1et(1+t)et−1ettet][1−110e−t−e−te−t−e−t(1+t)e−t(2+t)]=[100010001]and [1−110e−t−e−te−t−e−t(1+t)e−t(2+t)][1et0−1et(1+t)et−1ettet]=[100010001]Hence[X(t)]−1=[1−110e−t−e−te−t−e−t(1+t)e−t(2+t)]Particular solution is Xp(t)=X(t)∫[X(t)]−1g(t)dt.[X(t)]−1g(t)=[1−110e−t−e−te−t−e−t(1+t)e−t(2+t)][00t]=[t−te−t(t2+2t)e−t]∫[X(t)]−1g(t)dt=[∫tdt−∫te−tdt∫(t2+2t)e−tdt]=[t22e−t(1+t)−e−t(t+2)2]Thus the particular solution isXp(t)=[1et0−1et(1+t)et−1ettet][t22e−t(1+t)−e−t(t+2)2]=[1+t+t22−3−2t−t22−4−3t−t22]General Solution isX(t)=[1et0−1et(1+t)et−1ettet][c1c2c3]+[1+t+t22−3−2t−t22−4−3t−t22]Using initial condition[000]=[110−111−101][c1c2c3]+[1−3−4]Solving c1=0 , c2=−1 , c3=4Hence solution isX(t)=[−et+1+t+t22(3−t)et−3−2t−t22(4−t)et−4−3t−t22]