# Using the Euler's method (with h=10^(−n)) to find y(1) y′=sin(x)/x

Using the Euler's method (with $h={10}^{-n}$) to find $y\left(1\right)$
${y}^{\prime }=\frac{\mathrm{sin}\left(x\right)}{x}$
Since
${y}^{\prime }\left(x\right)=\frac{\mathrm{sin}\left(x\right)}{x}$
then
$f\left(x,y\right)=\frac{\mathrm{sin}\left(x\right)}{x}$
I know
${y}_{n+1}={y}_{n}+h\cdot f\left({x}_{n},{y}_{n}\right)$
and given $y\left(0\right)=0$, so
${x}_{0}={y}_{0}=0$
Therefore,
${x}_{1}={x}_{0}+h=0+{10}^{-0}=1⇒{y}_{1}=y\left({x}_{1}\right)=y\left(1\right)$
Then,
${y}_{1}=0+{10}^{-0}\cdot f\left({x}_{0},{y}_{0}\right)$
${y}_{1}=f\left(0,0\right)$
But
$f\left(0,0\right)=\frac{\mathrm{sin}\left(0\right)}{0}=\frac{0}{0}$
How am I supposed to do it?
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Alden Holder
You have to use boundary condition $y\left(0\right)=1$ (i.e. consistent boundary condition) to keep the right hand side continuous. In other words, you have to solve a bit different problem

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.)