Exercises 5.219.From the first equation, we conclude thaty=x−4x. Substitution into the second equationyields(x−4x) =−2x+ (x−4x)⇒x−5x+ 6x= 0.The characteristic equation,r2−5r+ 6 = 0, has rootsr= 2,3, and so a general solution isx(t) =c1e2t+c2e3t⇒y(t) =(c1e2t+c2e3t)−4(c1e2t+c2e3t)=−2c1e2t−c2e3t.We find constantsc1andc2from the initial condition.1 =x(0) =c1e2(0)+c2e3(0)=c1+c2,0 =y(0) =−2c1e2(0)−c2e3(0)=−2c1−c2⇒c1=−1,c2= 2.Therefore, the answer to this problem isx(t) =−e2t+ 2e3t,y(t) = 2e2t−2e3t.21.To apply the elimination method, we write the system using operator notation:D2[x]−y= 0,−x+D2[y] = 0.(5.5)Eliminatingyby applyingD2to the first equation and adding to the second equation gives
This is the end of the preview.
access the rest of the document.