Question

Euler's method estimate y(2)

Answer 1

h=0.5, y(1)=2, y'=t^2+1-y

Answer 2

I tried to use \[f_{n}=f_{n-1}+hf(x_{0},y_{0})\]

Answer 3

@ganeshie8

Answer 4

okay so what did you get for the value f(t_0,y_0)?

Answer 5

y'=f(t,y)
so i want to actually find f(1,2)=1^2+1-2=f(y0)
then we can find
y1=y0+hf(y0)
then we can find
y2=y1+hf(y1)
and so on...

Answer 6

f(1,2)=0

Answer 7

ok so y1=2+.5(0)=2
then
y2=y1+hf(y_1)

Answer 8

t1 would be 1.5 in that case and you already found y1 which is 2

Answer 9

\[t_n=t_0+nh\]

Answer 10

y2=2+0.5(1.25)

Answer 11

2.625

Answer 12

y3=2.625+0.5(f(2,2.625))?

Answer 13

yes

Answer 14

and we have
\[y_n \approx y(t_n) \text{ so } y_2 \approx y(t_2)=y(t_0+2*h)=y(1+2*.5)=y(1+1)=y(2)\]
so it looks like we only need to go just to y_2

Answer 15

we only needed to go to y2 because we were asked to estimate y(2)

Answer 16

i wonder if that is a good approximation

Answer 17

yes, it is

Answer 18

we could use runge trapezoidal

Answer 19

for better aprox

Answer 20

\[y'+y=t^2+1 \\ e^t y'+e^ty=e^t(t^2+1)\\ (e^ty)'=e^t(t^2+1) \\ e^t y=e^t(2t+1)-e^t(2+0)+C \\ y=2t+1-2+Ce^{-t} \\ y(1)=2 \text{ gives us } \\ 2=2+1-2+Ce^{-1} \\ 1=Ce^{-1} \\ C=e \\ y=2t-1+e^{-t+1} \\ y(2)=4-1+e^{-1}=3+e^{-1} \approx 3.368\]
if i did everything right

Answer 21

oops i made a booboo

Answer 22

\[y'+y=t^2+1 \\ e^t y'+e^ty=e^t(t^2+1)\\ (e^ty)'=e^t(t^2+1) \\ e^t y=e^t(t^2+1)-e^t(2t+0)+2e^t+C \\ y=t^2+1-2t+2+Ce^{-t} \\ y(1)=2 \text{ gives us } \\ 2=1^2+1-2+2+Ce^{-1} \\ 0=Ce^{-1} \\ C=0 \\ y=t^2-2t+3 \\ y(2)=2^2-2(2)+3=4-4+3=3\]

Answer 23

this was a linear differential equation

Answer 24

first order*

Answer 25

first order linear differential equaiton
you can solve these by finding an integrating factor

Answer 26

like the first line i multiplied by e^t so i could write the left hand side as (e^ty)'

because (e^ty)'=(e^t)'y+e^t(y)'
=e^t*y+e^t *y'

Answer 27

the other side perform integration by parts

but wait you should know this
i think euler's method comes after this