calculus question!
I have an exam tomorrow, please help :)

- anonymous

calculus question!
I have an exam tomorrow, please help :)

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- anonymous

use Euler's method.
given dy/dx=x+2y
x=1, y=2, find y if x=2, using n=5

- Jhannybean

@FutureMathProfessor

- anonymous

thank you, @Jhannybean

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## More answers

- anonymous

thank you everyone for coming. please help me out! :)

- Jhannybean

Mr. Moo will help you :) -rhyming-

- dumbcow

shoot i never learned Eulers method... i could solve this diff equ using integrating factor method but it might give diff answer since Eulers is an approximation ...
http://en.wikipedia.org/wiki/Eulers_method#Formulation_of_the_method

- dumbcow

Mr Moo ? haha i dont know if i like sound of that

- anonymous

awh :/ bummer.. well thanks for trying !
and please tell other people about this problem

- dumbcow

@satellite73 , @amistre64 , @Zarkon
any of you good at Eulers method

- bahrom7893

I sort of remember this.. If i can find my notes from 4 years ago that is.

- anonymous

I think it helps if you make a chart.. I remember my teacher drawing a chart on the board. I THINK. haha

- bahrom7893

So we have:
\[\frac{dy}{dx} = f(x;y)\]
\[y(x_0) = y(1) = 2\]

- bahrom7893

Let's approximate the solution to \[\frac{dy}{dx}\] near \[x=x_0\]
\[\frac{dy}{dx} at x=x_0 is = f(x_0; y_0)\]

- bahrom7893

And the tangent line is y - y0 = m(x-x0) (equation of a line).

- bahrom7893

From which we know that:
\[y = y_0 + f(x_0;y_0)(x_1-x_0) = 2 + (1+2*2)(x_1 - 1)\]

- bahrom7893

By the way, did you mean using n = 0.5 ?

- bahrom7893

oh wait, by n = 5 u mean in five steps

- bahrom7893

|dw:1371005922955:dw|

- anonymous

I am.. not sure. I wrote exactly what was on the review paper our teacher gave us.. haha, gosh, calculus is so difficult! :/

- bahrom7893

Sorry, n=5 means you have to split your interval (from x=1 to x=2, into 5 pieces).
So from my drawing up there \[x_1 = 1.2\]and
\[y_1 = y_0 + f(x_0;y_0)(x_1-x_0) = 2 + (1+2*2)(1.2 - 1) = 2 + 5*0.2 = 3\]

- bahrom7893

So now for the next approximation we have:
\[y_2 = y_1 + f(x_1;y_1)(x_2-x_1) = 3 + 7.2(1.4 - 1.2) = 4.44\]

- bahrom7893

I just made an excel spreadsheet if you want to track the values along with me. By the way:
\[x_{n+1} - x_{n} = 0.2\]
all the time in our case, because our step size is 0.2

##### 1 Attachment

- bahrom7893

Updated spreadsheet

##### 1 Attachment

- bahrom7893

Okay moving on to
\[y_3 = y_2 + f(x_2;y_2)(x_3-x_2) = 4.44 + 10.28(1.6 - 1.4) = 6.496\]

- bahrom7893

\[y_4 = y_3 + f(x_3;y_3)(x_4-x_3) = 6.496 + 14.592(1.6 - 1.4) = 9.4144\]

- bahrom7893

Btw the previous one should have been \[1.8-1.6\], but our answer was the same, since again all the differences are 0.2
\[y_5 = y_4 + f(x_4;y_4)(x_5-x_4) = 9.4144 + 20.6288(2.0 - 1.8) = 13.54013 \]

- bahrom7893

So your approximation is:
\[x_5=2; y_5 \approx13.54\]

- bahrom7893

dumbcow, check it over using integration. Thanks :)

- dumbcow

interesting ... i was curious how close it would be to exact answer of 19.07

- bahrom7893

hmm i don't think i made any errors.. but it wouldn't hurt to check over my work. If you get something else @mlddmlnog , just double check again. Your answer may be right

- bahrom7893

Hate this method because there's so much room for error.

- bahrom7893

http://tutorial.math.lamar.edu/Classes/DE/EulersMethod.aspx
Great tutorial on this.

- Zarkon

2
3
4.44
6.496
9.4144
13.54016
19.356224
27.5387136
39.03419904
55.16787866
77.79503012
109.5130422
153.958259
216.2215626
303.4301877
425.5622628
596.5871679
836.0620351

- bahrom7893

so wait, we had to continue for another step?

- Zarkon

19.356224

- anonymous

wow, I do not like this method either.
gosh, it is so much work, and it's confusing. but thank you everyone again, for helping me out!

- Zarkon

be happy you are not using the Rungeâ€“Kutta method

- bahrom7893

@mlddmlnog This guy's tutorial is also amazing http://www.youtube.com/watch?v=RGtCw5E7gBc

- bahrom7893

lol just googled that. Made me feel a bit better hahah

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