Consider the initial$-$value problem $y^{\text{'}}=y,y(0)=9.$

$(a)$ Use Euler's method with each $of$ the following step sizes $to$ estimate the value $ofy(0\mathrm{.}4).$

$(i)$

$h=0\mathrm{.}4$

$y(0\mathrm{.}4)=$

$(ii)$

$h=0\mathrm{.}2$

$y(0\mathrm{.}4)=$

$(iii)$

$h=0\mathrm{.}1$

$y(0\mathrm{.}4)=$

$(b)We$ know that the exact solution $of$ the initial$-$value problem $in$ part $(a)isy=9e^x.$ Draw, $as$ accurately $as$ you can, the graph $of$

$y=9e^x,0x0\mathrm{.}4,$ together with the Euler approximations using the step sizes $in$ part $(a).$ Use your sketches $to$ decide whether

your estimates $in$ part $(a)$ are underestimates $or$ overestimates.

The estimates are

$(c)$ The error $in$ Euler's method $is$ the difference between the exact value and the approximate value. Find the errors made $in$ part $(a)in$

using Euler's method $to$ estimate the true value $ofy(0\mathrm{.}4),$ namely, $9e^{0\mathrm{.}4}.(Round$ your answers $to$ four decimal places.$)$

$h=0\mathrm{.}4$ error $=|($ exact value $)-($ approximate value $)|,=$

$h=0\mathrm{.}2$ error $=|($ exact value $)-($ approximate value $)|,=$

$h=0\mathrm{.}1$ error $=|($ exact value $)-($ approximate value $)|,=$

What happens $to$ the error each time the step size $is$ halved?

Each time the step size $is$ halved, the error estimate appears $tobe$

$(approximately).$