Ask your own question, for FREE!
Mathematics 16 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (jhannybean):

@FutureMathProfessor

OpenStudy (anonymous):

thank you, @Jhannybean

OpenStudy (anonymous):

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

OpenStudy (jhannybean):

Mr. Moo will help you :) -rhyming-

OpenStudy (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

OpenStudy (dumbcow):

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

OpenStudy (anonymous):

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

OpenStudy (dumbcow):

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

OpenStudy (bahrom7893):

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

OpenStudy (anonymous):

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

OpenStudy (bahrom7893):

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

OpenStudy (bahrom7893):

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

OpenStudy (bahrom7893):

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

OpenStudy (bahrom7893):

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

OpenStudy (bahrom7893):

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

OpenStudy (bahrom7893):

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

OpenStudy (bahrom7893):

|dw:1371005922955:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!