anonymous
  • anonymous
calculus question! I have an exam tomorrow, please help :)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
use Euler's method. given dy/dx=x+2y x=1, y=2, find y if x=2, using n=5
Jhannybean
  • Jhannybean
@FutureMathProfessor
anonymous
  • anonymous
thank you, @Jhannybean

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anonymous
  • anonymous
thank you everyone for coming. please help me out! :)
Jhannybean
  • Jhannybean
Mr. Moo will help you :) -rhyming-
dumbcow
  • 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
  • dumbcow
Mr Moo ? haha i dont know if i like sound of that
anonymous
  • anonymous
awh :/ bummer.. well thanks for trying ! and please tell other people about this problem
dumbcow
  • dumbcow
@satellite73 , @amistre64 , @Zarkon any of you good at Eulers method
bahrom7893
  • bahrom7893
I sort of remember this.. If i can find my notes from 4 years ago that is.
anonymous
  • anonymous
I think it helps if you make a chart.. I remember my teacher drawing a chart on the board. I THINK. haha
bahrom7893
  • bahrom7893
So we have: \[\frac{dy}{dx} = f(x;y)\] \[y(x_0) = y(1) = 2\]
bahrom7893
  • 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
  • bahrom7893
And the tangent line is y - y0 = m(x-x0) (equation of a line).
bahrom7893
  • 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
  • bahrom7893
By the way, did you mean using n = 0.5 ?
bahrom7893
  • bahrom7893
oh wait, by n = 5 u mean in five steps
bahrom7893
  • bahrom7893
|dw:1371005922955:dw|
anonymous
  • 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
  • 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
  • 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
  • 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
  • bahrom7893
Updated spreadsheet
1 Attachment
bahrom7893
  • 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
  • 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
  • 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
  • bahrom7893
So your approximation is: \[x_5=2; y_5 \approx13.54\]
bahrom7893
  • bahrom7893
dumbcow, check it over using integration. Thanks :)
dumbcow
  • dumbcow
interesting ... i was curious how close it would be to exact answer of 19.07
bahrom7893
  • 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
  • bahrom7893
Hate this method because there's so much room for error.
bahrom7893
  • bahrom7893
http://tutorial.math.lamar.edu/Classes/DE/EulersMethod.aspx Great tutorial on this.
Zarkon
  • 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
  • bahrom7893
so wait, we had to continue for another step?
Zarkon
  • Zarkon
19.356224
anonymous
  • 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
  • Zarkon
be happy you are not using the Runge–Kutta method
bahrom7893
  • bahrom7893
@mlddmlnog This guy's tutorial is also amazing http://www.youtube.com/watch?v=RGtCw5E7gBc
bahrom7893
  • bahrom7893
lol just googled that. Made me feel a bit better hahah

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