Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

Could someone tell me the steps of solving this 2nd order differential equation?
(x^2)y'' - x y' + y = 8(x^3)
Is it Euler's equations for the left side and method of undertermined coeffficients for the right side? Thanks!!!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

yeah or you can use vartiation of parameters

- anonymous

thanks but how do you solve for the left side of the equation which now looks: y'' - (1/x) y' + y/(xsquared) = 8x

- anonymous

you leave the left side as it is and solve for the characteristic equation using cauchy eulers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

and i would use variation of parametesr after because i actually don't think undetermined coefficients work

- anonymous

so apply euler right off the bat to the left side which is : y'' - 2y' + y = 0
and use variation of parameters to get the particular solution?

- anonymous

yes but when you apply variation of parameters remember to get a coefficient of 1 on y'' so that your g(t)=x

- anonymous

what do you mean by getting a coefficient of 1 on y''?

- anonymous

because the form to use variation of parameters is y''+q(t)y'+p(t)y=g(t)

- anonymous

yeahh my bad...thanks a lot

- anonymous

Could you verbally tell me the steps of solving this system of differential equations as well? Thanks!!!
x'= x-y+z
y'= x+y-z
z'= 2x-y

- anonymous

you take the laplace of each equation and then cancel to try to get X Y or Z by itself and then inverse laplace that to solve for that and plug back in to solve for other two.. it's an algebraic nightmare so i suggest you use wolfram to help simplify things :P

- anonymous

Okok thanks at least i know how to attack it now. Thanks a lot. Do you engineers use DE everyday?

- anonymous

DE is probably one of the most useful maths compared to the other ones. But technically I haven't even taken my DE course yet, I just learned it on my own right now. I have it next semester, so I can plan on skipping class :P. And it does show up quite a bit in engineering, though in real life, most DE are most likely solved by a computer. It is still crucial to understand how they work and the meaning behind them though

- nikvist

- anonymous

what method is that nikvist? i've never seen that method before =o

- nikvist

this is the product rule

- anonymous

why is y = xz?

- nikvist

this is substitution \[y(x)=x\cdot z(x)\]

- nikvist

replace x=0 in starting differential equation, you will get y=0
\[\Rightarrow\quad y(x)=x\cdot z(x)\]

- anonymous

thanks a lot! do you know how to use power series here instead?

- anonymous

yeah i know product rule is involved but isnt there a name for the method?

- anonymous

spacenight: what he did is called "reduction of order" but in order to do that we needed to know 1 of the 2 solutions. He just assumed y1(x) = x here.

- anonymous

so i don't think he started off the right way

- anonymous

http://www.physicsforums.com/showthread.php?t=151779

- anonymous

lol oh yeah i know reduction of order, i just didnt know why y=xz is an assumed solution.. hm that's weird, i'd prefer to just do it the other way, that's how it was done in our homework for us

Looking for something else?

Not the answer you are looking for? Search for more explanations.