Question

u'=u/4+u^2 do you multiply by the reciprocal to both sides so all your u's are on one side giving you integral of (4+u^2/u)=t+C?

Answer 1

looks cauchy

Answer 2

\[u'=\frac14u+u^2\]
\[u'-\frac14u=u^2\]
\[u^{-2}u'-\frac14uu^{-2}=1\]
\[u^{-2}u'-\frac14u^{-1}=1\]

Answer 3

then we can make the substitution to conform this to something are used to playing with:\[z=u^{-1}\]\[z^{-1}=u\]\[-z^{-2}z'=u'\]and plug in the subs
\[u^{-2}u'-\frac14u^{-1}=1\]\[u^{-2}(-z^{-2}z')-\frac14(z)=1\]just have to redefine that last u\[(z^{-1}=u)^{-2}\]\[z^{2}=u^{-2}\]now we got it all
\[z^{2}(-z^{-2}z')-\frac14(z)=1\]simplify as
\[-z'-\frac14z=1\]

Answer 4

i dont think you can pull the 1/4 like that because i forgot some () it is supposed to be
u'=u/(4+u^2)

Answer 5

that can either be undone with partial fractions, or using a trick form of adding zero

Answer 6

so would be u^2/2 + 4ln/u/ = t+C?

Answer 7

using the differential equations separation of variables

Answer 8

not the zero adding, but the multiplying by 1 :)
\[u'=\frac{u}{4+u^2}\]\[u'=\frac{2}{2}*\frac{u}{4+u^2}\]
\[\int u'=\frac{1}{2}\int \frac{2u}{4+u^2}\]
\[\int u'=\frac{1}{2}ln(4+u^2)+C\]

Answer 9

theres no need to even separate it that i can see

Answer 10

if you wanted to separate it for the practice, i spose it would go something like this:

\[\frac{du}{dt}=\frac{u}{4+u^2}\]
\[\int \frac{4+u^2}{u}du=\int dt\]
\[\int \frac{4}{u}+\int \frac{u^2}{u}du=t+C\]
\[4ln(u)+\frac12u^2=t+C\]

Answer 11

trying to get an explicit function out of that might be tricky tho :)

Answer 12

yeah we just have to find a solution with u(0)=1

Answer 13

\[\frac12=0+C\]

Answer 14

so it would be C=1/2 right? then plug that into the equation

Answer 15

correct

Answer 16

alright thanks

Answer 17

youre welcome