Question

In Problems 1 and 2 use Euler's method to obtain a four-decimal approximation of the indicated value. Carry out the recursion of (3) by hand, first using $\mathit{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}$and then using$\mathit{h}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{05}$.

1.${\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{+}\mathbf{1}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{;}\mathbf{}\mathit{y}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{2}\mathbf{)}$

Short answer

$Forh=0.1,y(1.2)=2.9800,Forh=0.05,y(1.2)=3.1151$

Step-by-step solution

To Find the equation of Euler’s method

$y_{n+1}=y_n+h*f\left(x_n,y_n\right)$

This problem says to use (3) from the book, the general form looks like so.

For$h=0.1$

For$h=0.1$: We have an initial starting point given, and we have the location we want to end at. We want$y\left(1.2\right)$, and we start at$\left(1,5\right)$.

$y_{n+1}\approx \left(y_n\right)+0.1\left(2\left(x_n\right)-3\left(y_n\right)+1\right)$

Now we will see what our equation looks like using$h=0.1$, and the given$f\left(x,y\right)$.

$y_1\approx y_o+0.1\left(2x_o-3y_o+1\right)$

We start at $n=0$; we will be using our initial starting point $\left(1,5\right)$. So $y_o=5$, and$x_0=1$.

Substitute the value

$\begin{array}{c}{y}_1\approx y\left(1\right)\\ =\left(5\right)+0.1\left[2\right(1)-3(5)+1]\\ =3.8\end{array}$

Hey! How about that using the formula we have our next y value.

Also adding h to our previous x value will give us our next x value (remember h is the distance of our steps so this makes perfect sense).

Doing so gives us$\left(1.1,3.8\right)$

$$\begin{gathered} y_2\approx y(1.1) \\ (1.1,3.8) \\ =(3.8)+0.1\left[2\right(1.1)-3(3.8)+1] \end{gathered}$$

Now instead of going on to plugin in$y(1.2)$remember that each

$=2.98$

time we plugin an x value we are getting our next y value.

$(1.2,2.98)$in other terms $y(1.2)\approx 2.98$

Since the next point we would plugin is $(1.2,2.98)$, it should become obvious that we have the y value that corresponds with $y(1.2)$. So we can stop!

If we did plugin $y(1.2)$in to our approximation formula we would end up with the y value that goes with $x=1.3$, so no need to do that.

For$h=0.05$

Since I already walked through$h=0.1$, I am going to pick up the pace a little here. Remember we want$y(1.2)$, and we start at$(1,5)$.

$$\begin{gathered} y_1\approx y\left(1\right)=\left(5\right)+0.05\left[2\right(1)-3(5)+1]=4.4 \\ n=0,h=0.05,x=1,y=5 \\ {y}_2\approx y(1.05) \\ =(4.4)+0.05\left[2\right(1.05)-3(4.4)+1] \\ =3.895 \end{gathered}$$

$n=1,h=0.05,x=1.05,y=4.4$Plugin:

Final proof

$$\begin{gathered} y_3\approx y(1.1) \\ =(3.895)+0.05\left[2\right(1.1)-3(3.895)+1] \\ =3.47075 \end{gathered}$$

$$\begin{gathered} Plugin:n=2,h=0.05,x=1.1,y=3.895 \\ {y}_4\gg y(1.15) \\ =(3.47075)+0.05\left[2\right(1.15)-3(3.47075)+1] \\ =3.1151375 \\ Plugin:n=3,h=0.05,x=1.15 \\ y=3.47075 \\ (1.2,3.1151375) \end{gathered}$$

Remember the last x-value, gives the next y-value, so at$x=1.2$the y-value would be what we got from x=1.15. Thus we can stop

That is to say: here.

y at x=1.2 is$3.1151375$

For$h=0.05,y(1.2)\approx 3.1151$

For$h=0.05,y(1.2)$ rounded to four decimal places (via problem

For$h=0.1,y(1.2)\approx 2.9800$

instructions) is$2.8179$.

For $h=0.1$we didn't round the result, but it is still an approximation.