Question

If Euler’s method with step size\[{\bf{h = }}\frac{{\bf{1}}}{{\bf{n}}}\], where nis a positiveinteger, is used to approximate the solution to the initial value problem\[\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ + y = 0,y(0) = 1}}\] what formula (expressed as a function of n) do you obtain for the approximation of y (1 ) ? What is the exact value of y (1 )?

Short answer

\[{y_0} = {\left( {1 - \frac{1}{n}} \right)^n}\], \[{\bf{y}}\left( {\bf{1}} \right) = {{\bf{e}}^{{\bf{ - 1}}}} \approx 0.3679\]

Step-by-step solution

Finding the value of f (x)

Here,

\[\begin{array}{c}\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = - y}}\\\frac{{\bf{1}}}{{\bf{y}}}\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = - 1}}\\\int {\frac{{\bf{1}}}{{\bf{y}}}\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = }}\int {{\bf{ - dx}}} } \\{\bf{ln}}\;{\bf{u = ln}}\;{\bf{y = - x + c}}\end{array}\]

So,\[{\bf{y = }}{{\bf{e}}^{{\bf{( - x + c)}}}}\]

By using the initial conditions \[{\bf{y(0) = 1,}}\;{\bf{c = 0}}\]then

\[{\bf{y(x) = }}{{\bf{e}}^{{\bf{ - x}}}}\]

Using Euler’s method.

f(x,y)=-y

\[\begin{array}{c}{{\bf{y}}_{{\bf{n + 1}}}}{\bf{ = }}{{\bf{y}}_{\bf{n}}}{\bf{ + h}}\;{\bf{f(x,y)}}\\{\bf{ = }}{{\bf{y}}_{\bf{n}}}{\bf{ - }}\frac{{\bf{1}}}{{\bf{n}}}{{\bf{y}}_{\bf{n}}}\\{{\bf{y}}_{\bf{1}}}{\bf{ = }}\frac{{{\bf{n - 1}}}}{{\bf{n}}}\\{{\bf{y}}_{\bf{2}}}{\bf{ = }}{\left( {\frac{{{\bf{n - 1}}}}{{\bf{n}}}} \right)^{\bf{2}}}\\{\bf{.}}\\{\bf{.}}\\{\bf{.}}\\{{\bf{y}}_{\bf{k}}}{\bf{ = }}{\left( {\frac{{{\bf{n - 1}}}}{{\bf{n}}}} \right)^{\bf{k}}}{\bf{.}}{{\bf{y}}_{\bf{o}}}\end{array}\]

\[\begin{array}{l}{\bf{y(1) = }}{\left( {\frac{{{\bf{n - 1}}}}{{\bf{n}}}} \right)^{\bf{n}}}\\{y_0} = {\left( {1 - \frac{1}{n}} \right)^n}\end{array}\]

Finding the values for another n

Comparing the \[{\bf{y(1) = }}{{\bf{e}}^{{\bf{ - 1}}}}{\bf{ = 0}}{\bf{.3678}}\]as n increases

\[\begin{array}{*{20}{l}}{For{\rm{ }}n = 10,{\rm{ }}then{\rm{ }}0.3486,{\rm{ }}error = 0.0192}\\{For{\rm{ }}n = 100,{\rm{ }}then{\rm{ }}0.36603,{\rm{ }}error = 0.0018}\\{For{\rm{ }}n = 500,{\rm{ }}then{\rm{ }}0.3675,{\rm{ }}error{\rm{ }} = 0.0004}\\{For{\rm{ }}n = 1000,{\rm{ }}then{\rm{ }}0.3676,{\rm{ }}error = 0.002}\\{For{\rm{ }}n = 5000,{\rm{ }}then{\rm{ }}0.36784,{\rm{ }}error = 0.000037}\end{array}\]

Therefore the solution is\[{y_0} = {\left( {1 - \frac{1}{n}} \right)^n}\], \[{\bf{y}}\left( {\bf{1}} \right) = {{\bf{e}}^{{\bf{ - 1}}}} \approx 0.3679\].