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Mathematics 5 Online
OpenStudy (anonymous):

integrate..

OpenStudy (anonymous):

?

OpenStudy (anonymous):

\[\int\limits_{}^{}\frac{ dx }{ x+x ^{3} }\]

OpenStudy (anonymous):

i think 42

OpenStudy (anonymous):

wingspans

OpenStudy (anonymous):

get off my question

OpenStudy (unklerhaukus):

\[\frac{ 1 }{ x+x ^{3} }\] \[=\frac 1x\cdot\frac{ 1}{ 1+x ^{2} }\] now use partial fractions

OpenStudy (anonymous):

I factored too. Don't need partial fractions for the test tomorrow so I'm going to move on to another study question. thx

OpenStudy (raden):

??????????????????????

OpenStudy (unklerhaukus):

@cshalvey

OpenStudy (raden):

1/(x+x^2) = 1/(x*(x+1))

OpenStudy (unklerhaukus):

\[\frac 1x\cdot\frac{ 1}{ 1+x ^{2} }=\frac Ax+\frac{ Bx}{ 1+x ^{2} }\] \[1={ (1+x ^{2}) A+Bx^2}\] \[1={ A+(A+B)x^2}\] is x=0 we see 1=A \[1={ 1+(1+B)x^2}\] \[0=(1+B)x^2\] \[B=-1\] \[\frac 1x\cdot\frac{ 1}{ 1+x ^{2} }=\frac 1x-\frac{ x}{ 1+x ^{2} }\]

OpenStudy (unklerhaukus):

is x=0 * 'should be' when x=0

OpenStudy (anonymous):

Thanks for the refresher in partial fractions. I took a year off of school and it has been a year before that since calc 3. My diff eq test tomorrow will not have partial fractions however.

OpenStudy (anonymous):

Still nice to see it again for the future

OpenStudy (unklerhaukus):

anymore DE questions?

OpenStudy (anonymous):

Can you explain in laymens terms how to determine a linear equation. I've read the def'n and I think i kinda get it. but its still iffy

OpenStudy (unklerhaukus):

do you mean how to determine if a system is linear?

OpenStudy (anonymous):

yes

OpenStudy (unklerhaukus):

A differential equation is linear if it can be written in the form \[y'+p(x)y=q(x)\]

OpenStudy (anonymous):

ok I just have to learn to recognize that form or manipulate to get that form if possible.

OpenStudy (unklerhaukus):

when you look at the DE first check if it is; separable , if not check if it is linear , if not check if it is homogenous, ...., ...,

OpenStudy (anonymous):

this was a trick question\[\frac{ dx }{ dt }+xt=e ^{x}\]I originally thought linear but e^x is not a function of x, correct? That is why it is not linear

OpenStudy (anonymous):

Thx for the tips

OpenStudy (unklerhaukus):

e^x is a function of x

OpenStudy (anonymous):

hmm, The book says this isn't linear can you help explain why not

OpenStudy (unklerhaukus):

\[\frac{ dx }{ dt }+xt=e ^{x}\] changing x into y and t into x \[\frac{ dy }{ dx }+yx=e ^{y}\] to be linear it would have to be e^t

OpenStudy (unklerhaukus):

make sense?

OpenStudy (anonymous):

oh, good catch.

OpenStudy (anonymous):

I wasn't paying close enough attention to what the independent variable was

OpenStudy (unklerhaukus):

it can be confusing when they swap the lettes around

OpenStudy (anonymous):

I really appreciate the help. I'm going to work through some undetermined coefficient problems. I might post another question if I get stuck

OpenStudy (unklerhaukus):

link me in when you do.

OpenStudy (anonymous):

will do

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