Question

In the study of the vacuum tube, the following equation is encountered:

${\mathit{y}}^{\mathbf{"}}\mathbf{+}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{)}\mathbf{(}{\mathbf{y}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{)}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{.}$

Find the Taylor polynomial of degree 4 approximating the solution with the initial values$\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$,${\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.

Short answer

The required polynomial is, $p_3\left(x\right)=1-\frac{x^2}{6}$.

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

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

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

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

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 four 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.

$\begin{array}{c}{y}^{\text{'}\text{'}}=-y-\left(0.1\right)(y^2-1)y^{\text{'}}\\ y^{\text{'}\text{'}}\left(0\right)=-y\left(0\right)-\left(0.1\right)\left(y^2\left(0\right)-1\right)y^{\text{'}}\left(0\right)\\ =-1\end{array}$

Now since $y^{\text{'}\text{'}}=-y-\left(0.1\right)(y^2-1)y^{\text{'}}$holds for some interval around $x_0=0$, we can differentiate both sides to derive,

$\begin{array}{c}{y}^{\text{'}\text{'}\text{'}}=-y^{\text{'}}-\left(0.1\right)[(y^2-1)\left(y^{\text{'}\text{'}}\right)+2y{\left(y^{\text{'}}\right)}^2]\\ y^{\text{'}\text{'}\text{'}}\left(0\right)=-y^{\text{'}}\left(0\right)-\left(0.1\right)\left[({\left(y\right(0\left)\right)}^2-1)\left(y^{\text{'}\text{'}}\left(0\right)\right)+2y\left(0\right){\left(y^{\text{'}}\left(0\right)\right)}^2\right]\\ =0\end{array}$

Similarly,

$\begin{array}{c}{y}^4=-y^{\text{'}\text{'}}-\left(0.1\right)[(y^2-1)\left(y^{\text{'}\text{'}\text{'}}\right)+2y\left(y^{\text{'}}\right)\left(y^{\text{'}\text{'}}\right)+2{\left(y^{\text{'}}\right)}^3+2y\left(2y^{\text{'}}\right)\left(y^{\text{'}\text{'}}\right)]\\ y^4\left(0\right)=-y^{\text{'}\text{'}}\left(0\right)-\left(0.1\right)[(y{\left(0\right)}^2-1)\left(y^{\text{'}\text{'}\text{'}}\left(0\right)\right)+2y\left(0\right)\left(y^{\text{'}}\left(0\right)\right)\left(y^{\text{'}\text{'}}\left(0\right)\right)+2{\left(y^{\text{'}}\left(0\right)\right)}^3+2y\left(0\right)\left(2y^{\text{'}}\left(0\right)\right)\left(y^{\text{'}\text{'}}\left(0\right)\right)]\\ =1\end{array}$

Step 2:Final proof

Hence by substituting the Taylor polynomial of degree four for the solution is given by.$p_n\left(x\right)=1-\frac{x^2}{2!}+\frac{x^4}{4!}$