Given the equation dy/dt= t-(y^2) compute 2 approx. solutions corresponding to delta t = 1 and delta t = 0.5 on the interval [0,1]. I've included a photo of my work and was wondering if someone could check it for me to see if I did it correctly. I think the solution for delta = 1 and the graph is correct but for delta = 0.5 I'm not sure. Thank you
Hi
Your results are correct.
But, personnally, here is how I would write the answer (I do not be in the habit of making tables).
If you set $t_0=0$, $t_1=0.5$, $t_2=1$ and $y_0=y(0)$, $y_1=y(0.5)$, $y_2=y(1)$, (explicit) Euler's method consists in calculating $y_1$ and $y_2$ from the following equation :
$$
y_{i+1}=y_i+(t_{i+1}-t_i).(t_i-y_i^2)
$$
Since $y_0=y(0)=0$, we thus obtain
$$
y_1=y(0.5)=0.5
$$
and
$$
y_2=y(1)=0.625
$$
Yes, you are right, I did not say the opposite.
It's just a habit that I took to write in words the calculations to help understanding when I write scientific papers or when I teach. But if only the answer is expected, a table can be sufficient. It's only a personal opinion.