Exercises 4.3c2from the initial conditions.y(0) =−2=c1+c2,y0(0) = 0 =−3c1+c2⇒c1=−1/2,c2=−3/2.Thus, the solution to the initial value problem isy(t)=−e−3t2−3et2,and, ast→∞, it approaches−∞.(e)As in the previous problem, we have negative stiFnessk=−6. But this time the initialdisplacement,y(0) = 1, as well as the initial velocity,y0(0) = 1, is positive. So, themass will start moving in the positive direction, and will continue doing this (due to thenegative stiFness) with increasing velocity. Thus our prediction is thaty(t)whent.Indeed, the roots of the characteristic equation in this problem arer=−2 and 3, andso a general solution has the formy(tc1e−2t+c2e3t. To satisfy the initial conditions,we solve the systemy(0) = 1 =
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