anonymous
  • anonymous
does anyone understand Differential equations
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
lalaly
  • lalaly
yes
amistre64
  • amistre64
einstein did i think
amistre64
  • amistre64
newton faked it lol

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Mr.Math
  • Mr.Math
lol
anonymous
  • anonymous
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
amistre64
  • amistre64
with integration id assume
lalaly
  • lalaly
multiply both sides by dx .. then integrate both sides
amistre64
  • amistre64
\[y' = 2x+1\] \[\int y' = \int 2x+1\] \[y=x^2+x +C\]
amistre64
  • amistre64
plug in x=0 and y=3 to solve for C
lalaly
  • lalaly
this is a seperable DE
amistre64
  • amistre64
yeah, semantics :)
lalaly
  • lalaly
:D:D
anonymous
  • anonymous
you do these just like seperable this is 2 sections before seperable
lalaly
  • lalaly
Initial Value problem?
anonymous
  • anonymous
intergrals as general and particular solutions
lalaly
  • lalaly
forget what i said, its done the way amistre showd u
amistre64
  • amistre64
yay!!
lalaly
  • lalaly
hehehe
anonymous
  • anonymous
ok appreciate it!
amistre64
  • amistre64
\[Y(x)=\int y'\ dx\] \[Y(x)=\int (2x+1)dx\] stuff like that
anonymous
  • anonymous
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
lalaly
  • lalaly
\[\frac{dy}{dx}=2x+1\]\[dy=(2x+1)dx\]now integrate both sides\[\int\limits{dy}=\int\limits{(2x+1)dx}\]\[y=x^2+x+c\]
anonymous
  • anonymous
yea, just like that, thanks,
lalaly
  • lalaly
:)
anonymous
  • anonymous
i need help with intergals, it has been 2 yrs since i have done them. x times sq. root of x^2+9
lalaly
  • lalaly
let u=x^2+9 du=2x so integration becomes\[\frac{1}{2} \int\limits{ \sqrt u}du\]
lalaly
  • lalaly
integrate sqrt u then substitute the x back
lalaly
  • lalaly
du=2xdx **
lalaly
  • lalaly
I am really sorry, i have to leave now. Post a new question and someone else could help you
anonymous
  • anonymous
thanks
anonymous
  • anonymous
is \[^{?} \int\limits_{?}^{?} xe ^{-x}\] u substitution or by parts integral
amistre64
  • amistre64
or simply recognize that your missing a -1 in front
amistre64
  • amistre64
i missed read it
amistre64
  • amistre64
by parts with that one
anonymous
  • anonymous
alright
amistre64
  • amistre64
e^-x + x -e^-x | -1 e^-x | = x e^-x - e^-x +0 -----
amistre64
  • amistre64
+C if youaint got bounds
amistre64
  • amistre64
i missed a negative on there .... bad fingers, bad!!
amistre64
  • amistre64
\[-xe^{-x}-e^{-x}\ or\ -e^{-x}(x+1)\]

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