integrate..
?
\[\int\limits_{}^{}\frac{ dx }{ x+x ^{3} }\]
i think 42
wingspans
get off my question
\[\frac{ 1 }{ x+x ^{3} }\] \[=\frac 1x\cdot\frac{ 1}{ 1+x ^{2} }\] now use partial fractions
I factored too. Don't need partial fractions for the test tomorrow so I'm going to move on to another study question. thx
??????????????????????
@cshalvey
1/(x+x^2) = 1/(x*(x+1))
\[\frac 1x\cdot\frac{ 1}{ 1+x ^{2} }=\frac Ax+\frac{ Bx}{ 1+x ^{2} }\] \[1={ (1+x ^{2}) A+Bx^2}\] \[1={ A+(A+B)x^2}\] is x=0 we see 1=A \[1={ 1+(1+B)x^2}\] \[0=(1+B)x^2\] \[B=-1\] \[\frac 1x\cdot\frac{ 1}{ 1+x ^{2} }=\frac 1x-\frac{ x}{ 1+x ^{2} }\]
is x=0 * 'should be' when x=0
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.
Still nice to see it again for the future
anymore DE questions?
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
do you mean how to determine if a system is linear?
yes
A differential equation is linear if it can be written in the form \[y'+p(x)y=q(x)\]
ok I just have to learn to recognize that form or manipulate to get that form if possible.
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, ...., ...,
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
Thx for the tips
e^x is a function of x
hmm, The book says this isn't linear can you help explain why not
\[\frac{ dx }{ dt }+xt=e ^{x}\] changing x into y and t into x \[\frac{ dy }{ dx }+yx=e ^{y}\] to be linear it would have to be e^t
make sense?
oh, good catch.
I wasn't paying close enough attention to what the independent variable was
it can be confusing when they swap the lettes around
I really appreciate the help. I'm going to work through some undetermined coefficient problems. I might post another question if I get stuck
link me in when you do.
will do
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