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Expert-verified Found in: Page 425 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # For Duffing's equation given in Problem 13, the behaviour of the solutions changes as r changes sign. When ${\mathbit{r}}{\mathbf{>}}{\mathbf{0}}$ , the restoring force ${\mathbit{k}}{\mathbit{y}}{\mathbf{+}}{\mathbit{r}}{{\mathbit{y}}}^{{\mathbf{3}}}$ becomes stronger than for the linear spring${\mathbf{\left(}}{\mathbit{r}}{\mathbf{=}}{\mathbf{0}}{\mathbf{\right)}}$. Such a spring is called hard. When ${\mathbit{r}}{\mathbf{<}}{\mathbf{0}}$, 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 ${\mathbit{r}}{\mathbf{=}}{\mathbf{-}}{\mathbf{1}}$. Notice that for the initial conditions ,${\mathbit{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{0}}{\mathbf{,}}{{\mathbit{y}}}^{{\mathbf{\text{'}}}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{1}}$ the soft and hard springs appear to respond in the same way for small.(b) Keeping ${\mathbit{k}}{\mathbf{=}}{\mathbit{A}}{\mathbf{=}}{\mathbf{1}}$ and ,${\mathbit{\omega }}{\mathbf{=}}{\mathbf{10}}$ change the initial conditions to ${\mathbit{y}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{1}}$and ${{\mathbit{y}}}^{{\mathbf{\text{'}}}}{\mathbf{\left(}}{\mathbf{0}}{\mathbf{\right)}}{\mathbf{=}}{\mathbf{0}}$. Now redo Problem 13 with ${\mathbit{r}}{\mathbf{=}}{\mathbf{±}}{\mathbf{1}}$.(c) Based on the results of part (b), is there a difference between the behavior of soft and hard springs for small? Describe.

(a) The required polynomial is,${p}_{3}\left(t\right)=t+\frac{{t}^{2}}{2}-\frac{{t}^{3}}{6}$

(b) For $r=1,{p}_{3}\left(t\right)=1-\frac{{t}^{2}}{2}-\frac{33{t}^{4}}{8}$ and for $r=-1,{p}_{3}\left(t\right)=1+\frac{{t}^{2}}{2}-\frac{101{t}^{3}}{24}$

(c) For small values of t , both the soft and hard springs show the same behaviour.

See the step by step solution

## Step 1:To Find the Taylor polynomial of degree

The formula for the Taylor polynomial of degree n centered at ${x}_{0}$, approximating a function $f\left(x\right)$possessing n derivatives at ,${x}_{0}$ is given by

${p}_{n}\left(x\right)=f\left({x}_{0}\right)+{f}^{\text{'}}\left({x}_{0}\right)×\left(x-{x}_{0}\right)+{f}^{\text{'}\text{'}}\left({x}_{0}\right)×\frac{{\left(x-{x}_{0}\right)}^{2}}{2!}+\cdots +{f}^{n}\left({x}_{0}\right)×\frac{{\left(x-{x}_{0}\right)}^{n}}{n!}$

The differential equation is given as

${y}^{\text{'}\text{'}}+ky+r{y}^{3}=A\mathrm{cos}\omega t$

By substituting given values $k=A=1,r=-1$and $\omega =10$, differential equation becomes

${y}^{\text{'}\text{'}}+y-{y}^{3}=\mathrm{cos}10t$

It is given that for the function $y\left(x\right)$,

$y\left(0\right)=0\text{and}{y}^{\text{'}}\left(0\right)=1$

The Taylor's polynomial cantered around ${x}_{0}=0$ is given by

${p}_{n}\left(x\right)=y\left(0\right)+{y}^{\text{'}}\left(0\right)×\left(x-0\right)+{y}^{\text{'}\text{'}}\left(0\right)×\frac{{\left(x-0\right)}^{2}}{2!}+\cdots +{y}^{n}\left(0\right)×\frac{{\left(x-0\right)}^{n}}{n!}$

We need the value of$y\left(0\right),{y}^{\text{'}}\left(0\right),{y}^{\text{'}\text{'}}\left(0\right)$ and ${y}^{\text{'}\text{'}\text{'}}\left(0\right)$etc for finding the behaviour of spring. The first two are provided by the initial conditions.

The value of ${y}^{\text{'}\text{'}}\left(0\right)$can be deduced from the differential equation itself and the values of the lower derivatives.

$\begin{array}{l}{y}^{\text{'}\text{'}}\left(0\right)=\mathrm{cos}10t-y\left(0\right)+{y}^{3}\left(0\right)\\ {y}^{\text{'}\text{'}}\left(0\right)=1\end{array}$

Now since${y}^{\text{'}\text{'}}=\mathrm{cos}10t-y+{y}^{3}$ holds for some interval around ${x}_{0}=0$, we can differentiate both sides to derive,

$\begin{array}{c}{y}^{\text{'}\text{'}\text{'}}=-10\mathrm{sin}10t-{y}^{\text{'}}+3{y}^{2}{y}^{\text{'}}\\ {y}^{\text{'}\text{'}\text{'}}\left(0\right)=-10\mathrm{sin}10t-{y}^{\text{'}}\left(0\right)\left(1-3{y}^{2}\left(0\right)\right)\\ =-1\end{array}$

The Taylor polynomial for the soft spring when $r<0$ in the form of equation is given by.${p}_{3}\left(t\right)=t+\frac{{t}^{2}}{2}-\frac{{t}^{3}}{6}\cdots \cdots$

## Step 2:To Find the oscillatory equation

${x}_{0}=0$First, we will find the oscillatory equation in case where $r>0$,the restoring force is stronger and the spring is hard. The formula for the Taylor polynomial of degree centered at ${x}_{0}$, approximating a function $f\left(x\right)$ possessing n derivatives at ,${x}_{0}$ is given by,

${p}_{n}\left(x\right)=f\left({x}_{0}\right)+{f}^{\text{'}}\left({x}_{0}\right)×\left(x-{x}_{0}\right)+{f}^{\text{'}\text{'}}\left({x}_{0}\right)×\frac{{\left(x-{x}_{0}\right)}^{2}}{2!}+\cdots +{f}^{n}\left({x}_{0}\right)×\frac{{\left(x-{x}_{0}\right)}^{n}}{n!}$

The differential equation is given as

${y}^{\text{'}\text{'}}+ky+r{y}^{3}=A\mathrm{cos}\omega t$

By substituting given values $k=A=1,r=1$and $\omega =10$, differential equation becomes

${y}^{\text{'}\text{'}}+y+{y}^{3}=\mathrm{cos}10t$

It is given that for the function $y\left(x\right)$,

$y\left(0\right)=1\text{and}{y}^{\text{'}}\left(0\right)=0$

The Taylor's polynomial cantered around ${x}_{0}=0$is given by

${p}_{n}\left(x\right)=y\left(0\right)+{y}^{\text{'}}\left(0\right)×\left(x-0\right)+{y}^{\text{'}\text{'}}\left(0\right)×\frac{{\left(x-0\right)}^{2}}{2!}+\cdots +{y}^{n}\left(0\right)×\frac{{\left(x-0\right)}^{n}}{n!}$

We need the value of $y\left(0\right),{y}^{\text{'}}\left(0\right),{y}^{\text{'}\text{'}}\left(0\right)$and ${y}^{\text{'}\text{'}\text{'}}\left(0\right)$etc for finding the behaviour of spring. The first two are provided by the initial conditions.

The value of ${y}^{\text{'}\text{'}}\left(0\right)$can be deduced from the differential equation itself and the values of the lower derivatives.

$\begin{array}{l}{y}^{\text{'}\text{'}}\left(0\right)=\mathrm{cos}10t-y\left(0\right)-{y}^{3}\left(0\right)\\ {y}^{\text{'}\text{'}}\left(0\right)=-1\end{array}$

Now since ${y}^{\text{'}\text{'}}=\mathrm{cos}10t-y-{y}^{3}$holds for some interval around ${x}_{0}=0$, we can differentiate both sides to derive,

$\begin{array}{c}{y}^{\text{'}\text{'}\text{'}}=-10\mathrm{sin}10t-{y}^{\text{'}}-3{y}^{2}{y}^{\text{'}}\\ {y}^{\text{'}\text{'}\text{'}}\left(0\right)=-10\mathrm{sin}10t-{y}^{\text{'}}\left(0\right)\left(1+3{y}^{2}\left(0\right)\right)\\ =0\end{array}$

Similarly,

$\begin{array}{c}{y}^{4}=-100\mathrm{cos}10t-{y}^{\text{'}\text{'}}-6y{\left({y}^{\text{'}}\right)}^{2}\\ {y}^{4}\left(0\right)=-100\mathrm{cos}10t-{y}^{\text{'}\text{'}}\left(0\right)-6y\left(0\right){\left({y}^{\text{'}}\left(0\right)\right)}^{2}\\ {y}^{4}\left(0\right)=-99\end{array}$

The Taylor polynomial for the hard spring when $r>0$in the form of equation is given by,

${p}_{3}\left(t\right)=1-\frac{{t}^{2}}{2}-\frac{33{t}^{4}}{8}\cdots \cdots$

Now, we will find the oscillatory equation in case where ,the restoring force is weaker and the spring is soft. The formula for the Taylor polynomial of degree n

centered at${x}_{0}$ , approximating a function $f\left(x\right)$possessing n derivatives at , ${x}_{0}$is given by

${p}_{n}\left(x\right)=f\left({x}_{0}\right)+{f}^{\text{'}}\left({x}_{0}\right)×\left(x-{x}_{0}\right)+{f}^{\text{'}\text{'}}\left({x}_{0}\right)×\frac{{\left(x-{x}_{0}\right)}^{2}}{2!}+\cdots +{f}^{n}\left({x}_{0}\right)×\frac{{\left(x-{x}_{0}\right)}^{n}}{n!}$

The differential equation is given as

${y}^{\text{'}\text{'}}+ky+r{y}^{3}=A\mathrm{cos}\omega t$

By substituting given values$k=A=1,r=-1$and $\omega =10$, differential equation becomes

${y}^{\text{'}\text{'}}+y-{y}^{3}=\mathrm{cos}10t$

It is given that for the function , $y\left(x\right)$.$y\left(0\right)=1\text{and}{y}^{\text{'}}\left(0\right)=0$

The Taylor's polynomial centered around ${x}_{0}=0$is given by

${p}_{n}\left(x\right)=y\left(0\right)+{y}^{\text{'}}\left(0\right)×\left(x-0\right)+{y}^{\text{'}\text{'}}\left(0\right)×\frac{{\left(x-0\right)}^{2}}{2!}+\cdots +{y}^{n}\left(0\right)×\frac{{\left(x-0\right)}^{n}}{n!}$

We need the value of $y\left(0\right),{y}^{\text{'}}\left(0\right),{y}^{\text{'}\text{'}}\left(0\right)$and ${y}^{\text{'}\text{'}\text{'}}\left(0\right)$ etc. for finding the behaviour of spring. The first two are provided by the initial conditions.

The value of ${y}^{\text{'}\text{'}}\left(0\right)$can be deduced from the differential equation itself and the values of the lower derivatives.

$\begin{array}{l}{y}^{\text{'}\text{'}}\left(0\right)=\mathrm{cos}10t-y\left(0\right)+{y}^{3}\left(0\right)\\ {y}^{\text{'}\text{'}}\left(0\right)=1\end{array}$

Now since${y}^{\text{'}\text{'}}=\mathrm{cos}10t-y+{y}^{3}$ holds for some interval around ${x}_{0}=0$, we can differentiate both sides to derive,

$\begin{array}{c}{y}^{\text{'}\text{'}\text{'}}=-10\mathrm{sin}10t-{y}^{\text{'}}+3{y}^{2}\left({y}^{\text{'}}\right)\\ {y}^{\text{'}\text{'}\text{'}}\left(0\right)=-10\mathrm{sin}10t-{y}^{\text{'}}\left(0\right)\left(1-3{y}^{2}\left(0\right)\right)\\ =0\end{array}$

Similarly,

$\begin{array}{c}{y}^{4}=-100\mathrm{cos}10t-{y}^{\text{'}\text{'}}+6y{\left({y}^{\text{'}}\right)}^{2}\\ {y}^{4}\left(0\right)=-100\mathrm{cos}10t-{y}^{\text{'}\text{'}}\left(0\right)+6y\left(0\right)\left({y}^{\text{'}}{\left(0\right)}^{2}\right)\\ =-101\end{array}$

The Taylor polynomial for the hard spring when$r>0$ in the form of equation is given by

${p}_{3}\left(t\right)=1+\frac{{t}^{2}}{2}-\frac{101{t}^{4}}{24}\cdots \cdots$

## Step 3:The initial condition

From problem 13 and $14\left(a\right)$, the solution describing the behaviour of both soft and hard spring, for the initial conditions $y\left(0\right)=0$and ${y}^{\text{'}}\left(0\right)=1$, is approximated by the polynomial

${p}_{3}\left(t\right)=t+\frac{{t}^{2}}{2}-\frac{{t}^{3}}{6}+\cdots$

For small , the higher order terms will be negligible and the polynomial can be written as

${p}_{3}\left(t\right)\approx t$

Similarly, from problem $14\left(\text{b}\right)$, the solution describing the behaviour of soft and hard spring, for the initial conditions $y\left(0\right)=1$ and ${y}^{\text{'}}\left(0\right)=0$ , are approximated by the polynomial,

$\begin{array}{ll}{p}_{3}\left(t\right)=1-\frac{{t}^{2}}{2}-\frac{33{t}^{4}}{8}+\cdots & \left[\text{ForHardSpring}\right]\\ {p}_{3}\left(t\right)=1+\frac{{t}^{2}}{2}-\frac{101{t}^{4}}{24}+\cdots & \left[\text{ForSoftSpring}\right]\end{array}$

For small t , the higher order terms will be negligible and both the polynomial can be written as

${p}_{3}\left(t\right)\approx 1$

We can see from the above analysis that, in both the cases the soft and hard spring show same behaviour for small , the difference however is that for the initial conditions $y\left(0\right)=0$and ${y}^{\text{'}}\left(0\right)=1$, the springs behave in direct proportion to the small , whereas for the initial conditions$y\left(0\right)=$ 1 and , ${y}^{\text{'}}\left(0\right)=0$the springs behave as constant with value 1 , to the small t

## Step 4:Final proof

(a) the required Taylor’s polynomial is,${p}_{3}\left(t\right)=t+\frac{{t}^{2}}{2}-\frac{{t}^{3}}{6}$

(b) For $r=1,{p}_{3}\left(t\right)=1-\frac{{t}^{2}}{2}-\frac{33{t}^{4}}{8}$and for$r=-1,{p}_{3}\left(t\right)=1+\frac{{t}^{2}}{2}-\frac{101{t}^{3}}{24}$

(c) For small values of t , both the soft and hard springs show the same behaviour. ### Want to see more solutions like these? 