MHB 9.1.14 Use Euler's method to calculate

[karush]

$\tiny{9.1.14}$
$\textsf{Use Euler's method to calculate
the first 3 approximations to the given initial value problem for the specified increment size.}\\$
$\textsf{ Calculate the exact solution.}$
$y'=y^2(5+5x), y(1)=-1, dx=0.2$
$y_1=$
$y_1=0.4$
$y_2=0.6268$
$y_3=1.2694$
$y(exact)=-\frac{1}{x^2+5x-5}$

$\textsf{the given answers are book answers but i couldn't derive them}$

 

[MarkFL]

Given:

$$\d{y}{x}=f(x,y)$$

Euler's method is the recursion:

$$y_{n+1}\approx y_{n}+\Delta x\cdot f\left(x_n,y_n\right)$$

Now, in the given problem, we are given:

$$\d{y}{x}=y^2(5x+5)$$ where $$y(1)=-1$$

Let's first compute the exact solution...separate variables, use bounds as limits and switch dummy variables:

$$\int_{-1}^{y(x)}u^{-2}\,du=5\int_{1}^{x}v+1\,dv$$

Apply the FTOC:

$$-\left[\frac{1}{u}\right]_{-1}^{y(x)}=5\left[\frac{v^2}{2}+v\right]_{1}^{x}$$

$$-\left(\frac{1}{y(x)}+1\right)=5\left(\frac{x^2}{2}+x-\frac{3}{2}\right)$$

$$-\frac{1}{y(x)}=\frac{1}{2}\left(5x^2+10x-13\right)$$

$$y(x)=-\frac{2}{5x^2+10x-13}$$

Okay, now back to Euler's method...we begin with:

$$y_1=-1$$ and $$x_1=1$$ and we are told to use $$\Delta x=\frac{1}{5}$$

$$y_2\approx-1+\frac{1}{5}\left((-1)^2(5(1)+5)\right)=1$$

Compare to the exact value:

$$y\left(\frac{6}{5}\right)=-\frac{10}{31}$$

Next approximation:

$$y_3\approx1+\frac{1}{5}\left((1)^2\left(5\left(\frac{6}{5}\right)+5)\right)\right)=\frac{16}{5}$$

Compare to the exact value:

$$y\left(\frac{7}{5}\right)=-\frac{5}{27}$$

Can you continue, and explain why Euler's method isn't doing a very good job here?

 

[karush]

well just from observation the approx value increases while the exact value decreases

I've got to do more of these Euler's method so hope to understand it better

The examples to follow were ?