Q18E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

If the Taylor method of order p is used in Problem 17, show that \({{\bf{y}}_{\bf{n}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{6}}{{\bf{n}}^{\bf{3}}}}}{\bf{ + }}.....{\bf{ + }}\frac{{\bf{1}}}{{{\bf{p!}}{{\bf{n}}^{\bf{p}}}}}} \right)^{\bf{n}}}\), n = 1, 2, …

Step-by-Step Solution

Answer

proved

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

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

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: Find the other values of \({{\bf{f}}_{\bf{n}}}{\bf{(x,y)}}{\bf{.}}\)

\(\begin{array}{c}{{\bf{f}}_{\bf{3}}}{\bf{(x,y) = y}}\\{{\bf{f}}_{\bf{4}}}{\bf{(x,y) = y}}\\{\bf{.}}\\{\bf{.}}\\{{\bf{f}}_{\bf{p}}}{\bf{(x,y) = y}}\end{array}\)

Step 2: Apply the recursive formulas for order n

 The recursive formula is

 

\(\begin{array}{l}{{\bf{x}}_{{\bf{n + 1}}}}{\bf{ = }}{{\bf{x}}_{\bf{n}}}{\bf{ + h}}\\{{\bf{y}}_{{\bf{n + 1}}}}{\bf{ = }}{{\bf{y}}_{\bf{n}}}{\bf{ + hf(}}{{\bf{x}}_{\bf{n}}}{\bf{ + }}{{\bf{y}}_{\bf{n}}}{\bf{) + }}\frac{{{{\bf{h}}^{^{\bf{2}}}}}}{{{\bf{2!}}}}{{\bf{f}}_{\bf{2}}}{\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{ + }}{{\bf{y}}_{\bf{n}}}{\bf{) + }}.....\frac{{{{\bf{h}}^{\bf{p}}}}}{{{\bf{p!}}}}{{\bf{f}}_{\bf{p}}}{\bf{(}}{{\bf{x}}_{\bf{n}}}{\bf{ + }}{{\bf{y}}_{\bf{n}}}{\bf{)}}\end{array}\)\(\begin{array}{c}{{\bf{y}}_{\bf{1}}}{\bf{ = }}\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{6}}{{\bf{n}}^{\bf{3}}}}}{\bf{ + }}.....{\bf{ + }}\frac{{\bf{1}}}{{{\bf{p!}}{{\bf{n}}^{\bf{p}}}}}} \right)\\{{\bf{y}}_{\bf{2}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{6}}{{\bf{n}}^{\bf{3}}}}}{\bf{ + }}.....{\bf{ + }}\frac{{\bf{1}}}{{{\bf{p!}}{{\bf{n}}^{\bf{p}}}}}} \right)^{\bf{2}}}\\{\bf{.}}\\{\bf{.}}\\{{\bf{y}}_{\bf{n}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{6}}{{\bf{n}}^{\bf{3}}}}}{\bf{ + }}.....{\bf{ + }}\frac{{\bf{1}}}{{{\bf{p!}}{{\bf{n}}^{\bf{p}}}}}} \right)^{\bf{n}}}\\{{\bf{y}}_{{\bf{n + 1}}}}{\bf{ = }}\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{6}}{{\bf{n}}^{\bf{3}}}}}{\bf{ + }}.....{\bf{ + }}\frac{{\bf{1}}}{{{\bf{p!}}{{\bf{n}}^{\bf{p}}}}}} \right){{\bf{y}}_{\bf{n}}}\;{\bf{n}} \in {\bf{N}}\end{array}\)

 

Hence it is proved that \({{\bf{y}}_{\bf{n}}}{\bf{ = }}{\left( {{\bf{1 + }}\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{2}}{{\bf{n}}^{\bf{2}}}}}{\bf{ + }}\frac{{\bf{1}}}{{{\bf{6}}{{\bf{n}}^{\bf{3}}}}}{\bf{ + }}.....{\bf{ + }}\frac{{\bf{1}}}{{{\bf{p!}}{{\bf{n}}^{\bf{p}}}}}} \right)^{\bf{n}}}\)