Q13E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

${\mathit{y}}^{\mathbf{\text{'}\text{'}}}\mathbf{+}\mathit{k}\mathit{y}\mathbf{+}\mathit{r}{\mathit{y}}^{\mathbf{3}}\mathbf{=}\mathit{A}\mathit{c}\mathit{o}\mathit{s}\mathit{\omega }\mathit{t}$

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

Step-by-Step Solution

Answer

The Taylor polynomial for the first three non-zero terms in the solution is given by

$p_3\left(t\right)=t+\frac{t^2}{2}-\frac{t^3}{6}\cdots \cdots \cdot$

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

$p_n\left(x\right)=f\left(x_0\right)+f^{\text{'}}\left(x_0\right)\times (x-x_0)+f^{\text{'}\text{'}}\left(x_0\right)\times \frac{{(x-x_0)}^2}{2!}+\cdots +f^n\left(x_0\right)\times \frac{{(x-x_0)}^n}{n!}$

The differential equation is given as,

 $y^{"}+ky+r{y}^3=Acos\omega t$

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

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

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 centered around $x_0=0$  is given by,

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

Step 2:To Find the three non-zero terms

 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 value of the three non-zero terms. 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.

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

Since, $y^{\text{'}\text{'}}=\cos 10t-y-y^3$holds for some interval around $x_0=0$, we can differentiate both sides to derive,

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

The Taylor polynomial for the first three nonzero terms in the solution is given by

 $p_3\left(t\right)=t+\frac{t^2}{2}-\frac{t^3}{6}\cdots \cdots \cdot$

So, the required polynomial is,$p_3\left(t\right)=t+\frac{t^2}{2}-\frac{t^3}{6}\cdots \cdots$