Q17E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The Taylor method of order 2 can be used to approximate the solution to the initial value problem\({\bf{y' = y,y(0) = 1}}\) , at x = 1. Show that the approximation \({{\bf{y}}_{\bf{n}}}\) obtained by using the Taylor method of order 2 with the step size \(\frac{{\bf{1}}}{{\bf{n}}}\) is given by the formula\({{\bf{y}}_{\bf{n}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}} \right)^{\bf{n}}}\). The solution to the initial value problem is\({\bf{y = }}{{\bf{e}}^{\bf{x}}}\), so \({{\bf{y}}_{\bf{n}}}\)is an approximation to the constant e.

Step-by-Step Solution

Answer

proved

Step 1: Find the value of \({{\bf{f}}_{\bf{2}}}{\bf{(x,y)}}\)

 

Here \({\bf{f(x,y) = y}}\),  \({{\bf{x}}_{\bf{o}}}{\bf{ = 0,}}{{\bf{y}}_{\bf{o}}}{\bf{ = 1}}\)

 

Apply the chain rule.

 

\({{\bf{f}}_{\bf{2}}}{\bf{(x,y)}} = \frac{{\partial {\bf{f}}}}{{\partial {\bf{x}}}}{\bf{(x,y)}} + \frac{{\partial {\bf{f}}}}{{\partial {\bf{y}}}}{\bf{(x,y)f(x,y)}}\)

Now 

 

\(\begin{array}{l}\frac{{\partial {\bf{f}}}}{{\partial {\bf{x}}}}{\bf{(x,y) = }}0\\\frac{{\partial {\bf{f}}}}{{\partial {\bf{y}}}}{\bf{(x,y) = }}1\end{array}\)

 

So, the equation is \({{\bf{f}}_{\bf{2}}}{\bf{(x,y) = y}}\)

Step 2: Apply the recursive formulas for order 2

The recursive formula is

 

\(\begin{array}{l}{{\bf{x}}_{{\bf{n + 1}}}}{\bf{ = }}{{\bf{x}}_{\bf{n}}}{\bf{ + h}}\\{{\bf{y}}_{{\bf{n + 1}}}}{\bf{ = }}{{\bf{y}}_{\bf{n}}}{\bf{ + hf(}}{{\bf{x}}_{\bf{n}}}{\bf{ + }}{{\bf{y}}_{\bf{n}}}{\bf{) + }}\frac{{{{\bf{h}}^{^{\bf{2}}}}}}{{{\bf{2!}}}}{{\bf{f}}_{\bf{2}}}{\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{ + }}{{\bf{y}}_{\bf{n}}}{\bf{) + }}.....\frac{{{{\bf{h}}^{\bf{p}}}}}{{{\bf{p!}}}}{{\bf{f}}_{\bf{p}}}{\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{ + }}{{\bf{y}}_{\bf{n}}}{\bf{)}}\end{array}\)

 

\(\begin{array}{l}{{\bf{x}}_{\bf{1}}}{\bf{ = }}\frac{{\bf{1}}}{{\bf{n}}}\\{{\bf{y}}_{\bf{1}}}{\bf{ = }}\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}} \right)\end{array}\)

 

Apply the same procedure for other values up to n.

 

\(\begin{array}{c}{{\bf{y}}_{\bf{2}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}} \right)^{\bf{2}}}\\{{\bf{y}}_{\bf{3}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}} \right)^{\bf{3}}}\\{\bf{.}}\\{\bf{.}}\\{{\bf{y}}_{\bf{n}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}} \right)^{\bf{n}}}\\{{\bf{y}}_{\bf{n}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}} \right)^{\bf{n}}}{\bf{n}} \in {\bf{N}}\end{array}\)

Hence it is proved that \({{\bf{y}}_{\bf{n}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}}