Nelson Jennings

2022-07-21

Using the Euler's method (with $h={10}^{-n}$) to find $y(1)$

${y}^{\prime}=\frac{\mathrm{sin}(x)}{x}$

Since

${y}^{\prime}(x)=\frac{\mathrm{sin}(x)}{x}$

then

$f(x,y)=\frac{\mathrm{sin}(x)}{x}$

I know

${y}_{n+1}={y}_{n}+h\cdot f({x}_{n},{y}_{n})$

and given $y(0)=0$, so

${x}_{0}={y}_{0}=0$

Therefore,

${x}_{1}={x}_{0}+h=0+{10}^{-0}=1\Rightarrow {y}_{1}=y({x}_{1})=y(1)$

Then,

${y}_{1}=0+{10}^{-0}\cdot f({x}_{0},{y}_{0})$

${y}_{1}=f(0,0)$

But

$f(0,0)=\frac{\mathrm{sin}(0)}{0}=\frac{0}{0}$

How am I supposed to do it?

${y}^{\prime}=\frac{\mathrm{sin}(x)}{x}$

Since

${y}^{\prime}(x)=\frac{\mathrm{sin}(x)}{x}$

then

$f(x,y)=\frac{\mathrm{sin}(x)}{x}$

I know

${y}_{n+1}={y}_{n}+h\cdot f({x}_{n},{y}_{n})$

and given $y(0)=0$, so

${x}_{0}={y}_{0}=0$

Therefore,

${x}_{1}={x}_{0}+h=0+{10}^{-0}=1\Rightarrow {y}_{1}=y({x}_{1})=y(1)$

Then,

${y}_{1}=0+{10}^{-0}\cdot f({x}_{0},{y}_{0})$

${y}_{1}=f(0,0)$

But

$f(0,0)=\frac{\mathrm{sin}(0)}{0}=\frac{0}{0}$

How am I supposed to do it?

Alden Holder

Beginner2022-07-22Added 15 answers

You have to use boundary condition $y(0)=1$ (i.e. consistent boundary condition) to keep the right hand side continuous. In other words, you have to solve a bit different problem

$\begin{array}{r}{y}^{\prime}(x)={\textstyle \{}\begin{array}{l}\text{}\text{}\text{}1\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{if}\text{}x=0\\ \frac{\mathrm{sin}x}{x}\text{}\text{}\text{}\text{}\text{}\text{}\text{otherwise}\end{array}\end{array}$

Otherwise it is not solvable. (The right hand side must be defined in the starting point 0, where $\frac{\mathrm{sin}x}{x}$ is not.)

$\begin{array}{r}{y}^{\prime}(x)={\textstyle \{}\begin{array}{l}\text{}\text{}\text{}1\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{if}\text{}x=0\\ \frac{\mathrm{sin}x}{x}\text{}\text{}\text{}\text{}\text{}\text{}\text{otherwise}\end{array}\end{array}$

Otherwise it is not solvable. (The right hand side must be defined in the starting point 0, where $\frac{\mathrm{sin}x}{x}$ is not.)

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