Solve this differential equation: y\'\"+3y\"-4y\'=0: y\"(0)=2, y\'(0)=1, y(0)=3 Solution The characteristic equation is: r^3 + 3r^2 - 4r = 0 r(r^2 + 3r - 4) = 0 r(r+4)(r-1) = 0 The equation has 3 roots: r = 0 , 1 , -4 So the general soluttion is: y = Ae0x + Bex + Ce-4x = A + Bex + Ce-4x y(0) = A+B+C = 3 y\'(0) = B - 4C = 1 y\"(0) = B + 16C =2 From these equations we have: C = 1/20 , B = 6/5 , A = 7/4 So the final solution is: y = 7/4 + 6/5 ex + 1/20 e-4x.