Chapter 8.1, Problem 14E

Soft versus Hard Springs. For Duffing’s equation given in Problem 13, the behavior of the solutions changes as $r$ changes sign. When $r>0$, the restoring force $ky+r{y}^3$ becomes stronger than for the linear spring $\left(r=0\right)$. Such a spring is called hard. When $r<0$, the restoring force becomes weaker than the linear spring and the spring is called soft. Pendulums act like soft springs.

a. Redo Problem 13 with $r=-1$. Notice that for the initial conditions $y\left(0\right)=0$, $y^{\prime }\left(0\right)=1$, the soft and hard springs appear to respond in the same way for $t$ small.

b. Keeping $k=A=1$ and $\omega =10$, change the initial conditions to $y\left(0\right)=1$ and $y^{\prime }\left(0\right)=0$. Now redo Problem 13 with $r=\pm 1$.

c. Based on the results of part (b), is there a difference between the behavior of soft and hard springs for $t$ small? Describe.

13. Duffing’s Equation. In the study of a nonlinear spring with periodic forcing, the following equation arises:

$y^{″}+ky+r{y}^3=A\cos \omega t$.

Let $k=r=A=1$ and $\omega =10$. Find the first three nonzero terms in the Taylor polynomial approximations to the solution with initial values $y\left(0\right)=0$, $y^{\prime }\left(0\right)=1$.