Chapter 3.7, Problem 1E

Determine the recursive formulas for the Taylor method of order $2$ for the initial value problem

$y^{\prime }=\cos \left(x+y\right)$, $y\left(0\right)=\pi$.

To determine

To find:

The recursive formulas for the Taylor method of order $2$.for the differential equation $y^{\prime }=\cos \left(x+y\right),y(0)=\pi$.

Answer to Problem 1E

Solution:

The recursive formulas for the Taylor method of order $2$ is $y_{n+1}=y_n+h\cos \left(x_n+y_n\right)+\frac{h^2}{2}-\sin \left(x+y\right)+-\sin \left(x_n+y_n\right)\cos \left(x_n+y_n\right)$.

Explanation of Solution

Formula used:

The recursive formulas for Taylor method of order second is,

$\begin{array}{l}{x}_{n+1}=x_n+h\\ y_{n+1}=y_n+hf\left(x_n,y_n\right)+\frac{h^2}{2}{f}_2\left(x_n,y_n\right)\end{array}$

Calculation:

Consider the differential equation.

$y^{\prime }=\cos \left(x+y\right)$ ……$(1)$

The differential equation is the function of $x$ and $y$ so,

$y^{\prime }=f\left(x,y\right),y\left(x_0\right)=y_0$ ……$(2)$

Compare equation $(1)$ and equation $(2)$.

$\begin{array}{l}f\left(x,y\right)=\cos \left(x+y\right)\\ x_0=0\\ y_o=\pi \end{array}$

Differentiate the function $f\left(x,y\right)=\cos \left(x+y\right)$ with respect to $x$,

$f_x\left(x,y\right)=-\sin \left(x+y\right)$

Differentiate the function $f\left(x,y\right)=\cos \left(x+y\right)$ with respect to $x$,

$f_y\left(x,y\right)=-\sin \left(x+y\right)$

To calculate,

$f_2\left(x,y\right)=f_x+f_y\left(f\left(x,y\right)\right)$ ……$(3)$

Substitute $-\sin \left(x+y\right)$ for $f_x$, $-\sin \left(x+y\right)$ for $f_y$ and $\cos \left(x+y\right)$ for $f\left(x,y\right)$ in equation $(3)$.

$f_2\left(x,y\right)=-\sin \left(x+y\right)+-\sin \left(x+y\right)\cos \left(x+y\right)$

Consider the recursive formulas for Taylor method of order second.

$y_{n+1}=y_n+hf\left(x_n,y_n\right)+\frac{h^2}{2}{f}_2\left(x_n,y_n\right)$ ……$(4)$.

Substitute $\cos \left(x_n+y_n\right)$ for $f\left(x_n,y_n\right)$ and $-\sin \left(x+y\right)+-\sin \left(x+y\right)\cos \left(x+y\right)$ for $f_2\left(x_n,y_n\right)$ in equation $(4)$.

$y_{n+1}=y_n+h\cos \left(x_n+y_n\right)+\frac{h^2}{2}-\sin \left(x+y\right)+-\sin \left(x_n+y_n\right)\cos \left(x_n+y_n\right)$

Therefore, the recursive formulas for the Taylor method of order $2$ is $y_{n+1}=y_n+h\cos \left(x_n+y_n\right)+\frac{h^2}{2}-\sin \left(x+y\right)+-\sin \left(x_n+y_n\right)\cos \left(x_n+y_n\right)$.

Conclusion:

Thus, the recursive formulas for the Taylor method of order $2$ is $y_{n+1}=y_n+h\cos \left(x_n+y_n\right)+\frac{h^2}{2}-\sin \left(x+y\right)+-\sin \left(x_n+y_n\right)\cos \left(x_n+y_n\right)$.