does anyone understand Differential equations

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does anyone understand Differential equations

Mathematics
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yes
einstein did i think
newton faked it lol

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Other answers:

lol
i just begun D.E. and i am totally lost. find a function, y=f(x) satisfying the given differential equation and the prescribed initial condition. dy/dx= 2x+1; y(0)=3 where do i start
with integration id assume
multiply both sides by dx .. then integrate both sides
\[y' = 2x+1\] \[\int y' = \int 2x+1\] \[y=x^2+x +C\]
plug in x=0 and y=3 to solve for C
this is a seperable DE
yeah, semantics :)
:D:D
you do these just like seperable this is 2 sections before seperable
Initial Value problem?
intergrals as general and particular solutions
forget what i said, its done the way amistre showd u
yay!!
hehehe
ok appreciate it!
\[Y(x)=\int y'\ dx\] \[Y(x)=\int (2x+1)dx\] stuff like that
it just shows you the equation and intial condition and you have to choose what to do. So when you work it out it becomes what you are doing with the integral
\[\frac{dy}{dx}=2x+1\]\[dy=(2x+1)dx\]now integrate both sides\[\int\limits{dy}=\int\limits{(2x+1)dx}\]\[y=x^2+x+c\]
yea, just like that, thanks,
:)
i need help with intergals, it has been 2 yrs since i have done them. x times sq. root of x^2+9
let u=x^2+9 du=2x so integration becomes\[\frac{1}{2} \int\limits{ \sqrt u}du\]
integrate sqrt u then substitute the x back
du=2xdx **
I am really sorry, i have to leave now. Post a new question and someone else could help you
thanks
is \[^{?} \int\limits_{?}^{?} xe ^{-x}\] u substitution or by parts integral
or simply recognize that your missing a -1 in front
i missed read it
by parts with that one
alright
e^-x + x -e^-x | -1 e^-x | = x e^-x - e^-x +0 -----
+C if youaint got bounds
i missed a negative on there .... bad fingers, bad!!
\[-xe^{-x}-e^{-x}\ or\ -e^{-x}(x+1)\]

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